Suppose that \(\pi : X \to Y\) is a continuous map of differentiable manifolds (as topological spaces — not a priori differentiable). Show that \(\pi\) is differentiable if differentiable functions pull back to differentiable functions, i.e., if pullback by \( \pi \) gives a map \(\OO_Y \to \pi_*\OO_X\). (Hint: check this on small patches. Once you figure out what you are trying to show, you will realize that the result is immediate.)
Proof:
Recall that for every open subset \(V \subset Y\) and differentiable map \( f \in \OO_Y(V) \), our induced map on \( \pi_* \OO_X (V) = \OO_X ( \pi^{-1}(V) ) \) is \( f \circ \pi \). If we let \( (V, \psi) \) be a coordinate patch (where \(\psi : V \to \R^m\) differentiable coordinate functions) then by possibly restricting \( U = \pi^{-1}(V) \) we may assume without loss of generality that \( (U, \phi) \) is another coordinate patch on \(X\) (where \( \phi : U \to \R^n \) differentiable). Recall that a function \( g : X \to Y \) is differentiable iff \( \psi \circ g \circ \phi^{-1} : \R^n \to \R^m \) is differentiable on all coordinate patches. By hypothesis, we have that \( f \) and \( f \circ \pi \) are differentiable, so that by the previous statement \( f \circ \psi^{-1} \) and \( f \circ \pi \circ \phi^{-1} \) differentiable. It suffices to show that \( \psi \circ \pi \circ \phi^{-1} \) is differentiable — however, this is obvious since
As we know \( f \circ \pi \circ \phi^{-1} \) and \( f \circ \psi^{-1} \) are differentiable, the result follows.
$$\tag*{$\blacksquare$}$$
Exercise 3.1.B:
Show that a morphism of differentiable manifolds \( \pi : X \to Y \) with \( \pi(p) = q \) induces a morphism of stalks \( \pi^{ \sharp} : \OO_{Y, q} \to \OO_{X,p} \). Show that \( \pi^{\sharp} ( \mm_{Y,q} ) \subset \mm_{X, p} \). In other words, if you pull back a function that vanishes at \(q\), you get a function that vanishes at \( p \) — not a huge surprise. (In §6.3, we formalize this by saying that maps of differentiable manifolds are maps of locally ringed spaces.)
Proof:
This follows from Exercise 2.1.A and Exercise 2.2.I (and I suppose Zorn's lemma). Indeed, for every neighborhood \( V \) of \(q\), \( \pi^{-1}(V) \) is an open neighborhood of \(q\). Then by the definition of \( \OO_{X,p} \) as the colimit of open sets containing \( p \), there is a canonical map \( \iota_V : \OO_X{\pi^{-1}(V)} \to \OO_{X,p} \) that commutes with restrictions. Composing this with the map \( \OO_Y(V) \to \pi_* \OO_X(V) \ (=\OO_X(\pi^{-1}(V) ) ) \), we get a commuting cone for each open neighborhood \(V \ni q\), so by the universal property of the colimit there is an induced map \( \OO_{Y,q} \to \OO_{X,p} \) which we call \( \pi^{\sharp} \).
To see that \( \pi^\sharp ( \mm_{Y,q} ) \subset \mm_{X,p} \), notice from the above diagram (and the fact that differentiable functions form a sheaf) that if \( f_q \in \mm_{Y,q} \) (so that \( f_q = 0 \)) and \((\overline{f}, V)\) is a representative of \(f\) (such that the stalk \( \overline{f}_q = f_q \)) then by commutivity \( \pi_* \overline{f} = \overline{f} \circ \pi \Rightarrow ( \pi_* \overline{f} )_p = (\overline{f} \circ \pi)_p = 0 \Rightarrow \pi^\sharp (f_q) = 0
\). That is, \( (\pi^\sharp (f))_p \in \mm_{X,p} \).
$$\tag*{$\blacksquare$}$$
Section 3.2: The Underlying Set of Affine Schemes
Exercise 3.2.A:
Describe the set \( \spec k[\epsilon] / (\epsilon^2 ) \). The ring \( k[\epsilon] / (\epsilon^2 ) \) is called the \(\textbf{ring}\ \textbf{of}\ \textbf{dual}\ \textbf{numbers}\), and will turn out to be quite useful. You should think of \( \epsilon \) as a very small number, so small that its square is 0 (although it itself is not 0). It is a nonzero function whose value at all points is zero, thus giving our first example of functions not being determined by their values at points. We will discuss this phenomenon further in §3.2.11.
Describe the set \( \spec k[x]_{(x)} \)(see §1.3.3 for a discussion of localization). We will see this scheme again repeatedly, starting with §3.2.8 and Exercise 3.4.K. You might later think of it as a shred of a particularly nice "smooth curve".
Proof:
We will see in Exercise 3.2.J that \( \spec k[\epsilon] / (\epsilon^2) \) is in correspondence with the prime ideals \( (\epsilon^2) \subset I \subset k[\epsilon] \) — in particular, the only prime ideal is \( (\epsilon) \) so that \( \spec k[\epsilon] / (\epsilon^2) \) is a singleton. Now every "function" \( f \in k[\epsilon] / (\epsilon^2) \) is of the form \( f = a + b\epsilon \) for \( a, b \in k \) — "evaluating" this at \(\epsilon\) is equivalent to taking \( f \mod \epsilon \) which is simply the projection \( a + b\epsilon \mapsto a \). Indeed, this goes to show that we can have two functions when evaluated at all points (i.e. just \( \epsilon \)) are the same, but need not agree. Indeed, the ideal \( (\epsilon) \) is obviously maximal as well, so that \( k[\epsilon] / (\epsilon^2) \) is a local ring.
We will see in Exercise 3.2.K that the prime ideals of \( \spec k[x]_{(x)} \) are in correspondence with those prime ideals \( \pp \) that do not intersect \( k[x] - (x) \) (i.e primes contained in \( (x) \)). Thus, we get that \( \spec k[x]_{(x)} = \{ [(0)], [(x)] \} \) where \( [(x)]\) is closed as it is maximal.
$$\tag*{$\blacksquare$}$$
Exercise 3.2.B:
Show that for the last type of prime, of the form \((x^2 + ax + b)\), the quotient is always isomorphic to \( \C \).
Proof:
For ease of notation we will instead suppose our last quadratic is of the form \( (x^2 - 2ax + b) \). By completing the square, roots of \( (x^2 - 2ax + b) = (x - a)^2 + (b-a^2) \) correspond to roots of \( x^2 +1 \) via the map
$$
x \mapsto \frac{x - a}{\sqrt{b - a^2 }}
$$
Notice we need \( b - a^2 > 0 \) for \( x^2 - 2ax + b \) to be irreducible.
$$\tag*{$\blacksquare$}$$
Exercise 3.2.C:
Describe the set \( \A^1_\Q \). (This is harder to picture in a way analogous to \(\A^1_\R\). But the rough cartoon of points on a line, as in Figure 3.1, remains a reasonable sketch.)
Proof:
Similar to \( \A^1_\R \), we can consider \( \spec \Q[x] \) as \( \overline{\Q} \subset \C \) "folded" along the rational points; however, this does not in fact fold in the same way as the real line, since we now have algebraic elements such as \( \sqrt{2}, -\sqrt{2} \) which become identified / glued in \( \spec \Q[x] \) even though it lies between rational points. For this reason, \( \spec \Q[x] \) cannot be "nicely" folded, but the important part is to recognize conjugate algebraic elements are identified.
$$\tag*{$\blacksquare$}$$
Exercise 3.2.D:
If \(k\) is a field, show that \( \spec k[x]\) has infinitely many points. (Hint: Euclid’s proof of the infinitude of primes of \(\Z\).)
Proof:
The proof is by contradiction. First notice that since \( k \) is a field, \( k[x] \) is a PID so that each prime ideal is of the form \( (p(x)) \) for some prime element \( p(x) \in k[x] \). Now assume to the contrary that there are only finitely many prime ideals, generated by some \( p_1(x), p_2(x), \dots, p_n(x) \in k[x] \). Since every PID is a unique factorization domain, in which case prime elements and irreducible elements coincide up to a unit, then we may assume without loss of generality that \( p_1(x), p_2(x), \dots, p_n(x) \) are the only irreducible elements. Taking \( f(x) = 1 + \prod_{i=1}^n p_i(x) \) and again using the fact that \( k[x] \) is a UFD, \(f(x)\) may be uniquely decomposed into a product of irreducible (prime) elements:
$$
f(x) = q_1(x)^{i_1} \dots q_r(x)^{i_r}
$$
but each \(q(x)\) is of the form \( p_j(x) \) for some \( j \), so that there is some common factor of the form \( p_j(x) \) in
In other words, \( p_j(x) \) divides \(1\) which is impossible.
$$\tag*{$\blacksquare$}$$
Exercise 3.2.E:
Show that we have identified all the prime ideals of \(\C[x,y]\). Hint: Suppose \( \pp \) is a prime ideal that is not principal. Show you can find \(f(x, y), g(x, y) \in \pp\) with no common factor. By considering the Euclidean algorithm in the Euclidean domain \(\C(x)[y]\), show that you can find a nonzero \(h(x) \in ((f(x, y), g(x, y)) \subset \pp\). Using primality, show that one of the linear factors of \(h(x)\), say \((x − a)\), is in \(\pp\). Similarly show there is some \((y − b) \in \pp\).
Proof:
Suppose \( \pp \) is a prime ideal of \( \C[x,y] \) that is not principal. Since \( \pp \) is Noetherian, we may find a minimal set of generators \( f_1(x,y), \dots, f_n(x,y) \) of \( \pp \). Let \( h = \gcd(f_1, f_2) \) (since \( \C[x,y] \) is a UFD by Gauss' lemma, this is valid.) Then \( f_1 = h g_1 \), \( f_2 = h g_2 \) — clearly \( g_1 \) and \(g_2\) cannot have common factors, as otherwise \(h\) would not be the greatest common divisor. Now since \( \pp \) is prime and \( f_1 = hg_1 \in \pp \), we have \( h \in \pp \) or \( g_1 \in \pp \) — now if \( h \in \pp \), then we would have \( h, f_3, \dots, f_n \) is a minimal set of generators of \( \pp \), a contradiction. By repeating the argument for \( g_2 \), we may take \( f = g_1 \), \( g = g_2 \) as our polynomials in \( \pp \) with no common factor
Following the proof of Kenny Wong's answer on Mathematics StackExchange, if we consider the greatest common divisor \( \overline{h}(x,y) \) of \( f \) and \(g\) (notice that it may be different here, since \( \C(x)[y] \) and \( \C[x,y] \) are not the same UFD) we can allow the coefficient in \( \C(x) \) to absorb all non-trivial factors that are purely polynomials in \(x\):
So that all irreducible factors of \( c(x,y) \) are non-constant in \(y\). Since \( f, g \in ( \overline{h} ) \) (by definition of the GCD), we can find rational functions \( \frac{p(x, y)}{q(x)}, \frac{r(x, y)}{s(x)} \) with
Since we assumed \( c(x,y) \) has no factors purely in terms of \(x\), \( c(x,y) \) cannot divide \( a, b, q \) or \(s\), so it must be the case that \( c \) divides \(f\) and \(g\). Since \( f \) and \(g\) are coprime by the first paragraph, it must be the case that \( c \) is constant on \( \C^2 \). Thus, we can write \( \overline{h}(x, y) = \frac{a(x)}{b(x)} \) as a linear combination of \(f, g\) with coefficients in \( \C(x) \) — by multiplying out denominators, we get a linear combination of \( f, g \) (i.e. in the ideal \( (f, g) \subset \pp \) ) that is purely in terms of \(x\). Since \( \C \) is algebraically closed, we may decompose this into linear factors — using the fact that \( \pp \) is prime, one of the linear factors \( (x - a) \) is in \( \pp \). An identical argument shows the same for \( (y - b) \in \pp \) (by considering the Euclidean domain \( \C(y)[x] \)).
$$\tag*{$\blacksquare$}$$
Exercise 3.2.F:
Show that the Nullstellensatz 3.2.5 implies the Weak Nullstellensatz 3.2.4.
Proof:
Assume that every maximal ideal of \( k[x_1, \dots, x_n] \) has a residue field which is simply a finite extension of \(k\). Let \( \mm \) be a maximal ideal in \( k[x_1, \dots, x_n] \) so that by hypothesis \( k \to k[x_1, \dots, x_n] / \mm \) is a finite extension. Since \( k \) is algebraically closed, \( k[x_1, \dots, x_n] / \mm \) is an algebraic extension so that every \( a(x_1, \dots, x_n) + \mm \) is a root of some polynomial in \( k[x_1, \dots, x_r] \) — in particular, this gives us an isomorphism \( k[x_1, \dots, x_n] / \mm \cong k \). By looking at the coordinate functions, we have that \( x_1 + \mm \) maps to some \( a_i \) — thus, \( x_i - a_i \in \mm \). However, since \( ( x_1 - a_1, \dots, x_n - a_n ) \subset \mm\) is maximal (all polynomials may be decomposed into their linear factors in an algebraically closed field), the two must coincide.
$$\tag*{$\blacksquare$}$$
Exercise 3.2.G:
Any integral domain \(A\) which is a finite \(k\)-algebra (i.e., a \(k\)-algebra that is a finite-dimensional vector space over \(k\)) must be a field. Hint: for any nonzero \( x \in A \), show \(\times x: A \to A\) is an isomorphism. (Thus, in combination the Nullstellensatz 3.2.5, we see that prime ideals of \(k[x_1 , \dots , x_n ]\) with finite residue ring are the same as maximal ideals of \(k[x_1, \dots , x_n]\). This is worth remembering.)
Proof:
Fix \( a \neq 0 \) in \(A\) and consider the \(k \)-linear map \( \phi_a : A \to A \) given by left multiplication \( x \mapsto a\cdot x \). Since \(A\) is an integral domain, this map is necessarily injective ( if \(a \cdot x = 0 \) for some \( x\neq 0 \) then it would necessarily be a zero-divisor.) Since \(A \) is finite-dimensional (as a vector space), injective implies surjective (c.f. rank-nullity) so that \( x \mapsto a\cdot x \) is an isomorphism. Thus, \( a\) has an inverse.
$$\tag*{$\blacksquare$}$$
Exercise 3.2.H:
Describe the maximal ideal of \(\Q[x, y]\) corresponding to \( ( \sqrt{2}, \sqrt{2} ) \)
and \( ( - \sqrt{2}, - \sqrt{2} ) \). Describe the maximal ideal of \(\Q[x, y]\) corresponding to \( ( \sqrt{2}, - \sqrt{2} ) \)
and \( ( -\sqrt{2}, \sqrt{2} ) \). What are the residue fields in each case?
Proof:
We have that \( \mm_1 = (x^2 + y^2 - 4, x - y) \) is the maximal ideal in \( \spec \Q[x,y] \) corresponding to the points \( (\sqrt{2}, \sqrt{2}), (-\sqrt{2},-\sqrt{2} ) \) in \( \overline{Q^2} \).
Similarly, \( \mm_2 = (x^2 + y^2 - 4, x + y) \) is the maximal ideal in \( \spec \Q[x,y] \) corresponding to the points \( (\sqrt{2}, -\sqrt{2}), (-\sqrt{2},\sqrt{2} ) \).
In both cases, our reside field is simply \( \Q[\sqrt{2}] \), so that different points in \( \spec A \) can yield the same residue field.
$$\tag*{$\blacksquare$}$$
Exercise 3.2.I:
Consider the map of sets \( \phi : \C^2 \to \A_{\Q}^2 \) defined as follows. \( (z_1, z_2) \) is sent to the prime ideal of \( \Q[x,y]\) consisting of polynomials vanishing at \( (z_1, z_2) \).
What is the image of \( (\pi, \pi^2) \)
Show that \( \phi\) is surjective. (Warning: You will need some ideas we haven’t discussed in order to solve this. Once we define the Zariski topology on \( \A_\Q^2 \), you will be able to check that \(\phi\) is continuous, where we give \( \C^2 \) the classical topology. This example generalizes. For example, you may later be able to generalize this to arbitrary dimension.)
Proof:
Though we have not gotten to the topic yet, the vanishing set of polynomials vanising on a set \(S\) is often denoted \( V(S) \).
Notice we certainly have that \( (y - x^2) \subset V(\{\pi, \pi^2 \} ) \) since any multiple of \( y - x^2 \) evaluated at \((\pi, \pi^2) \) vanishes. To see the reverse inclusion, it is important to notice that since \( \pi \) is transcendental over \(\Q\) the minimal polynomial of \( \pi\) cannot be in terms of just one variable. By applying the standard division algorithm argument here in \(x\) (i.e. dividing some \(f \in V( \{ \pi, \pi^2 \} ) \) by \( y - x^2 \) and showing the remainder \(r(x)\) is zero), it follows from the previous sentence that since \( \pi \) is transcendental then any remainder in terms of one variable cannot vanish at \( \pi \).
We will see from the Strong Nullstellensatz that \( I(V( (x - z_1, y - z_2) )) = \sqrt{ (x - z_1, y - z_2) } \). Since \( (x - z_1, y - z_2) \) is prime (in fact maximal), it is equal to its radical so that \( I \) is an inverse to \( V \).
$$\tag*{$\blacksquare$}$$
Exercise 3.2.J:
Suppose \(A\) is a ring, and \(I\) an ideal of \(A\). Let \( \phi : A \to A/I \). Show that \( \phi^{-1} \) gives an inclusion-preserving bijection between prime ideals of \( A/I \) and prime ideals of \(A\) containing \( I \). Thus we can picture \( \spec A/I\) as a subset of \( \spec A \).
Proof:
For the forward direction, if \( I \subset J \subset A \) is a prime ideal, then as \( J \neq \emptyset \) we must have \( \phi(J) \neq \emptyset\). Since \( \phi \) is surjective, for any \( \overline{a} \in A/I \) there exists some \( a \in A \) with \( \phi(a) = \overline{a} \) — as \( \phi \) is a ring homomorphism we have
so that \( \phi(J)\) is indeed an ideal of \( A/I \). To see that it is prime, suppose \( \overline{x}, \overline{y} \in A/I \) with \( \overline{x}\overline{y} \in \phi(J) \). Again by surjectivity there exist \( x, y \in A \) with \( \phi(x) = \overline{x} \) and \( \phi(y) = \overline{y} \); since \( \phi \) is a ring homomorphism, \( \phi(xy) = \phi(x)\phi(y) = \overline{x}\overline{y} \) and thus \( xy \in J \). Since \( J \) is prime \( x \in J\) or \( y \in J \ \Rightarrow \overline{x} \in \phi(J) \) or \( \overline{y} \in \phi(J) \).
The reverse direction will be made clear by Exercise 3.2.M
$$\tag*{$\blacksquare$}$$
Exercise 3.2.K:
Suppose \(S\) is a multiplicative subset of \(A\). Describe an order-preserving bijection of the prime ideals of \( S^{-1}A \) with the prime ideals of \( A \) that don’t meet the multiplicative set \(S\).
Proof:
In the forward direction, suppose \( I \subset A \) is an ideal, and define
$$
\overline{I} := \{ \frac{x}{s} \mid x \in I,\, s \in S \}
$$
Notice that if \( \exists x' \in I \cap S \) then we have \( 1 \in \overline{I} \Rightarrow \overline{I} = S^{-1}A \). This gives us a map \( \psi \) of ideals in \(A\) to ideals in \( S^{-1}A \) via \( I \mapsto \overline{I} \) — we are more concerned in showing \( \psi \) induces a map on the spectrum of \( A \backslash S \). That is, if \( \pp \) is a prime ideal disjoint from \(S\), we wish to show that \( \psi(\pp) = \overline{\pp} \) is prime. Suppose \( \frac{z}{u} = \frac{x}{s}\frac{y}{t} \in \overline{\pp} \), so that \( z \in \pp \). Then \( \exists v \in S \) with
$$
v( stz - uxy ) = 0
$$
in \(A\). Since \( \pp \) is an ideal and \( z \in \pp \), we have \( vstz \in \pp \) so that \( uvxy \in \pp \). Now since \( u,v \in S\) and \( \pp \) is prime and disjoint from \(S\), we must have \( xy \in \pp \) — again using the fact that \(\pp \) is prime, \( x \in \pp \) or \( y \in \pp \Rightarrow \frac{x}{s} \in \overline{\pp}\) or \( \frac{y}{t} \in \overline{\pp} \).
In the reverse direction, we wish to show that if \( \overline{\pp} \) is a prime ideal in \( S^{-1}A \), then
$$
\phi^{-1}(\overline{\pp}) := \left\{ x \in A \mid \phi(x) = \frac{x}{1} \in \overline{\pp} \right\}
$$
is a prime ideal in \( A \) disjoint from \(S\). Now the fact that this set is disjoint from \(S\) is obvious from definition; if we suppose to the contrary \( t \in \phi^{-1}(\overline{\pp}) \cap S \), then \( \frac{t}{1} \in \overline{\pp} \Rightarrow \frac{1}{t} \frac{t}{1} \in \overline{\pp} \) (since \(\overline{\pp} \) is an ideal) and thus \( \overline{\pp} = S^{-1}A \) which is not prime (a contradiction). It suffices to show that \( \phi^{-1}(\overline{\pp}) \) is prime — but this is immediate from the fact that the preimage of a prime ideal under a ring map is prime ( Exercise 3.2.M).
It is easy to see from construction that both maps are order preserving.
$$\tag*{$\blacksquare$}$$
Exercise 3.2.L:
Show that these two rings are isomorphic. (You will see that \(y\) on the left goes to 0 on the right.)
Proof:
By localizing at \( x \), we are considering the points where \( x \) doesn't vanish. Since \( \C[x,y] \) is a domain, if \( xy = 0\) in \( (\C[x,y])_x \) then we necessarily have \( y = 0 \). In particular, since \( x \) is invertible in \( (\C[x,y])_x \) we have that \( (xy) = (y) \), giving us
If \(\phi: B \to A\) is a map of rings, and \( \pp \) is a prime ideal of \(A\), show that \( \phi^{-1}(\pp) \) is a prime ideal of \( B \).
Proof:
Notice \( \phi^{-1}(\pp) \) is non-empty, as \( 0 \in \pp \) and \( \phi(0) = 0 \). Similarly, \( \phi^{-1}(\pp) \) cannot be the whole ring, since if \( 1 \in \phi^{-1}(\pp) \) then \( \phi(1) = 1 \in \pp \) which would imply \( \pp = A \) (contradicting the fact that \( \pp \) prime). Now as \( \phi \) is a ring morphism, if \( x, y \in \phi^{-1}(\pp) \) then \( \phi(x), \phi(y) \in \pp \Rightarrow \phi(x + y) = \phi(x) + \phi(y) \in \pp\) — thus, \( x + y \in \phi^{-1}(\pp) \) so that \( \phi^{-1}(\pp) \) is an additive group. By a similar reasoning, for any \( s \in B \) and \( x \in \phi^{-1}( \pp) \) we have \( \phi(sx) = \phi(s)\phi(x) \in \pp \) since \( \pp \) is an ideal — thus \( sx \in \phi^{-1}(\pp) \).
$$\tag*{$\blacksquare$}$$
Exercise 3.2.N:
Let \(B\) be a ring.
Suppose \(I \subset B\) is an ideal. Show that the map \(\spec B/I \to \spec B\) is the inclusion of §3.2.7.
Suppose \( S \subset B\) is a multiplicative set. Show that the map \(\spec S^{-1} B \to \spec B\) is the inclusion of §3.2.8.
Consider the map of complex manifolds sending \( \C \to \C \) via \( x \mapsto y = x^2 \). We interpret the "source" \( \CC \) as the "x-line", and the "target" \(\C\) the "y-line". You can picture it as the projection of the parabola \(y = x^2\) in the \(xy\)-plane to the \(y\)-axis (see Figure 3.6). Interpret the corresponding map of rings as given by \(\C[y] → \C[x]\) by \( y \mapsto x^2 \). Verify that the preimage (the fiber) above the point \( a \in \C \) is the point(s) \(\pm \sqrt{a} \in \C\), using the definition given above. (A more sophisticated version of this example appears in Example 9.3.3. Warning: the roles of x and y are swapped there, in order to picture double covers in a certain way.)
Proof:
The point \( a \in \C \) corresponds to the ideal \( (x - a) \in \spec \C \). If we let \( \phi : \C[y] \to \C[x] \) be the map \( y \mapsto x^2 \), then the image of \( (x^2 - a) \) under the induced map \( \spec \C[x] \to \spec \C[y] \) is \( \phi^{-1}( (x^2 - a) ) = (y - \sqrt{a})(y + \sqrt{a}) \)
$$\tag*{$\blacksquare$}$$
Exercise 3.2.P:
Suppose \(k\) is a field, and \(f_1,\dots,f_n \in k[x_1,\dots,x_m]\) are given. Let \(\phi: k[y_1,\dots ,y_n] \to k[x_1,\dots,x_m]\) be the ring morphism defined by \( y_i \mapsto f_i \).
Show that \( \phi \) induces a map of sets \( \spec k[x_1,\dots,x_m]/I \to \spec k[y_1,\dots,y_n]/J \) for any ideals \(I \subset k[x_1,\dots,x_m]\) and \(J \subset k[y_1,...,y_n]\) such that \(\phi(J) \subset I\). (You may wish to consider the case \(I = 0\) and \(J = 0\) first. In fact, part (a) has nothing to do with \(k\)-algebras; you may wish to prove the statement when the rings \(k[x_1 , \dots , x_m ]\) and \(k[y_1, \dots , y_n]\) are replaced by general rings \(A\) and \(B\).)
Show that the map of part (a) sends a point \( (a_1, \dots, a_m) \in k^m \) (or more precisely, \( [(x_1 - a_1, \dots, x_m - a_m)] \in \spec k[x_1, \dots, x_m] \) ) to
$$
(f_1(a_1, \dots, a_m), \dots, f_n(a_1, \dots, a_m)) \in k^n
$$
Proof:
Notice there is a clear induced ring map on the quotients: if we let \( \pi : k[x_1, \dots, x_m] \to k[x_1, \dots, x_m] / I \) denote the canonical quotient map, then since \( \phi(J) \subset I \) and \( I = \ker \pi \), we have \( J \subset \ker ( \pi \circ \phi ) \). By the first isomorphism theorem, there is an induced map \( \overline{\phi \circ \pi} : k[y_1, \dots, y_n] / \ker (\pi \circ \phi ) \to k[x_1, \dots, x_n] / I \), which by the universal property of the kernel extends to a map \( \phi : k[y_1, \dots, y_n] / J \to k[x_1, \dots, x_n] / I \). The rest follows from Exercise 3.2.M.
This is a generalization of Exercise 3.2.O. Skipping ahead a bit in notation to 3.7, let \( I(S) \subset k[y_1, \dots, y_n] \) denote the ideal of polynomials vanishing on the set \(S\); in particular, for any \( b_1, \dots, b_n \in k \)
As \( k \) is a field (and thus \( k[y_1, \dots, y_{n-1}] \) a Euclidean domain ), the division algorithm allows us to show \( g(b_1, \dots, b_{n-1}, b_n) = 0 \Leftrightarrow (y_n - b_n) \mid g\). Consequently, one has
The set \( I((b_1, \dots, b_n)) \) may also be seen as the kernel of the evaluation morphism \( k[y_1, \dots, y_n] \to k\) at \( (b_1, \dots, b_n) \); by the first isomorphism theorem, we have that \( k[y_1, \dots, y_n] / I((b_1, \dots, b_n)) \cong k \) so that \( I((b_1, \dots, b_n)) \) is necessarily maximal.
As \( \phi \) is the unique \(k\)-algebra homomorphism sending \( y_i \) to \( f_i \), we have that elements of \( k \) are fixed by \( \phi \) (since \( 1 \mapsto 1 \) and it is \(k\)-linear). Thus
vanishes when evaluated at \( a_1, \dots, a_m \in k \), so it is contained in \( I((a_1, \dots, a_m)) = ( x_1 - a_1, \dots, x_m - a_m ) \) by the same reasoning as above. Thus,
However, since \( (y_1 - b_1, \dots, y_n - b_n) \) is maximal, the two ideals are indeed equal. Thus, under the induced map \( \spec k[x_1, \dots, x_m] \to \spec k[y_1, \dots, y_n] \), \( \pp = (x_1 - a_1, \dots, x_m - a_m) \) is sent to \( \phi^{-1}( \pp ) = (y_1 - f_1(a_1, \dots, a_m), \dots, y_n - f_n(a_1, \dots, a_m) ) \).
$$\tag*{$\blacksquare$}$$
Exercise 3.2.Q:
Consider the map of sets \( \pi : \A^n_\Z \to \spec \Z \), given by the ring map \(\Z \to \Z[x_1 , \dots , x_n ]\). If \(\pp\) is prime, describe a bijection between the fiber \(\pi^{-1}([(\pp)]) \) and \(\A^n_{\F_p}\). (You won’t need to describe either set! Which is good because you can’t.) This exercise may give you a sense of how to picture maps (see Figure 3.7), and in particular why you can think of \( \A^n_\Z \) as an "\(\A^n \)-bundle" over \(\spec \Z\).(Can you interpret the fiber over \([(0)]\) as \( \A^n_k \) for some field \( k \)?)
Proof:
Our ring map \( \Z \to \Z[x_1, \dots, x_n] \) is simply the inclusion (which we will denote \(\iota\)), so that \( \pi : \A^n_\Z \to \spec \Z \) is given by \( \iota^{-1}( ( x_1 - r_1, \dots, x_n - r_n ) ) \). Thus if \( (p) \in \spec \Z \) and \( \pi(I) = \iota^{-1}(I) = (p) \), then by definition of the inclusion we have \( I \supseteq (p) \). Thus, the fibre \( \pi^{-1}( (p) ) \) consists of all prime ideals \( I \subset \Z[x_1, \dots, x_n] \) containing \( (p) \) — however, by Exercise 3.2.J this is precisely \( \spec \Z / (p) [x_1, \dots, x_n] = \spec \mathbb{F}_p [x_1, \dots, x_n] = \A^{n}_{\mathbb{F}_p} \).
Now if \( \iota^{-1}(I) = (0) \) for some prime ideal \( I \subset \Z[x_1, \dots, x_n] \), then by taking \( S = \Z - \{ 0\} \) we have that the prime ideals that map to \( (0) \) under \( \pi \) are precisely those which do not intersect our multiplicative subset \( S \). By Exercise 3.2.K, this is precisely \( \spec S^{-1}\Z [x_1, \dots, x_n] = \spec \Q[x_1, \dots, x_n] = \A^n_\Q \).
$$\tag*{$\blacksquare$}$$
Exercise 3.2.R:
Ring elements that have a power that is 0 are called \(\textbf{nilpotents}\).
Show that if \(I\) is an ideal of nilpotents, then the inclusion \(\spec B/I \to \spec B\) of Exercise 3.2.J is a bijection. Thus nilpotents don’t affect the underlying set. (We will soon see in §3.4.5 that they won’t affect the topology either — the difference will be in the structure sheaf.)
Show that the nilpotents of a ring B form an ideal. This ideal is called the \(\textbf{nilradical}\), and is denoted \(\mathfrak{N} = \mathfrak{N}(B)\).
Proof:
Every prime ideal \( \pp\) already contains all nilpotents — to see this, let \( x \) be a nilpotent such that \( \exists n > 0 \) with \( x^n = 0 \). Since every ideal contains \( 0 \), we have \( x^n \in \pp \). Since \( \pp \) is prime, we may deduce \( x \in \pp \). By Exercise 3.2.J, the prime ideals of \( \spec B/I \) are in bijection with the prime ideals containing \( I \) — by the first two sentences, this is necessarily every \( \pp \in \spec B \).
This is a common abstract algebra exercise that should hopefully have been seen before coming to algebraic geometry — however, it is required that the ring \(B\) be commutative (an easy counterexample is \( B = M_{2 \times 2}(\Z) \)). Following the standard proof, let \( N \) denote the set of nilpotent elements. We know \( N \neq \emptyset \) since \( 0 \in N \). Now if \( x, y \in N \), there exist some \( m, n > 0 \) with \( x^m = y^n = 0 \). Then
$$
(x + y)^{m+n} = \sum_{i=0}^{m + n} {m + n \choose i} x^{m + n - i} y ^i
$$
Notice that for each \( 0 \leq i \leq m+n \), one either has \( i \geq m \) or \( m + n - i \geq n \), so that \( x^{m+n-i}y^i = 0 \) for each \(i\) and thus the sum vanishes \( \Rightarrow x + y \in N \). In addition, if \( r \in B \) is arbitrary, then \( (rx)^m = r^m x^m = r^m 0 = 0 \) (here we needed that \( B \) is commutative).
$$\tag*{$\blacksquare$}$$
Exercise 3.2.S:
If you don’t know this theorem, then look it up, or better yet, prove it yourself. (Hint: Use the fact that any proper ideal of \(A\) is contained in a maximal ideal, which requires Zorn’s Lemma. Possible further hint: Suppose \(x \notin \mathfrak{N}(A)\). We wish to show that there is a prime ideal not containing \(x\). Show that \(A_x\) is not the 0-ring, by showing that \(1 \neq 0\).)
Proof:
As we have already seen that the nilradical is contained in the intersection of all primes, it suffices to show that the intersection of all primes is contained in the nilradical. Proceeding by contrapositive, suppose \( x \in A \) is not nilpotent and let \( \mathscr S \) denote the collection of ideals \( I \) not containing \( x^n \) for any \( n \). Since \( x \) is not nilpotent, \( (0) \in \mathscr S \). By Zorn's lemma, there exists a maximal element \( \mm \) (i.e. maximal ideal) of \( \mathscr S\) which is necessarily prime — thus, \( \mm \) is a prime ideal that does not contain \(x\).
$$\tag*{$\blacksquare$}$$
Exercise 3.2.T:
Suppose we have a polynomial \(f(x) \in k[x]\). Instead, we work in \(k[x, \epsilon]/(\epsilon^2 )\). What then is \(f(x + \epsilon)\)? (Do a couple of examples, then prove the pattern you observe.) This is a hint that nilpotents will be important in defining differential information (Chapter 21).
Proof:
Assume \( d = \textrm{deg}\,f > 1 \) (the base case is somewhat trivial) with \( f(x) = \sum_{k=0}^d a_k x^k \). Then by the binomial formula
For fixed \( k \), we get two non-zero terms at \( i = 0 \) and \( i = 1\). In the first case, we are left with \( a_kx^k \) and in the second case we are left with \( a_k { k \choose 1 } x^{k-1}\epsilon = a_k k x^{k-1} \epsilon \). Thus,
There are no exercises in this section, just commentary.
Section 3.4: The Underlying Topological Space of Affine Schemes
Exercise 3.4.A:
Check that the \(x\)-axis is contained in \( V(xy, yz) \). (The \(x\)-axis is defined by \( y = z = 0 \), and the \(y\)-axis and \(z\)-axis are defined analogously.)
Proof:
Notice \( (xy, yz) \subset (y, z) \) since if \( f = a xy + byz \) for some \( a, b \in \C[x, y, z] \), then clearly \( f \) is a linear combination of \( y \) and \(z\) with coefficients in \( \C[x,y,z] \). By definition of the vanishing set, it follows that \( [ (y, z) ] \in V(xy, yz) \).
$$\tag*{$\blacksquare$}$$
Exercise 3.4.B:
Show that if \( (S) \) is the ideal generated by \(S\), then \( V(S) = V((S)) \)
Proof:
Since \( S \subset (S) \) it is easy to see that if \( ( S) \subset \pp \) then \( S \subset \pp \) by transitivity — thus, \( V((S)) \subseteq V(S) \). In the reverse direction, suppose that \( [\pp] \in V(S) \), such that \( S \subset \pp \). Since \( \pp \) is an ideal, any \( f \in (S) \) which is a linear combination of elements in \(S\) must also be a linear combination of elements in \( \pp \) and thus in \( \pp \).
$$\tag*{$\blacksquare$}$$
Exercise 3.4.C:
Show that \( \emptyset \) and \( \spec A \) are both open subsets of \( A \)
If \( I_i \) is a collection of ideals (as \(i\) runs over some index set), show that \( \bigcap_i V(I_i) = V( \sum_i I_i ) \). Hence the union of any collection of open sets is open.
Show that \( V(I_1) \cup V(I_2) = V( I_1I_2 ) \). Hence the intersection of any finite number of open sets is open.
Proof:
Taking compliments, it suffices to show that the two sets are closed. But this is immediate since \( \emptyset = V(1) = V((1)) = V(A) \) (as no prime ideal can contain the entire ring) and \( \spec A = V(0) \).
\( [\pp] \in \bigcap_i V(I_i) \) iff \( I_i \subset \pp \) for each \( i\) — since \( \pp \) is an ideal and \( 0 \) is an element of each \( I_i \) (needed for reverse direction), this is true iff \( \sum_i I_i \subset \pp \) iff \( [\pp] \in V( \sum_i I_i ) \)
Since \( I_1I_2 \subset I_1 \) and \( I_1I_2 \subset I_2 \), we have \( V(I_1) \cup V(I_2) \subseteq V(I_1I_2) \). Now if \( [\pp] \in V(I_1I_2) \), we have \( I_1I_2 \subset \pp \). Now suppose \( I_1 \not\subset \pp \) so that there exists some \( f \in I_1 \) with \( f \notin \pp \) — then for each \( g \in I_2 \), \( fg \in I_1I_2 \subset \pp \). Since \( \pp \) is prime, \( fg \in \pp \) and \( f \notin \pp \) we must have \( g \in \pp \) so that \( I_2 \subset \pp \Rightarrow V(I_1I_2) \subseteq V(I_2) \subseteq V(I_1) \cup V(I_2) \).
$$\tag*{$\blacksquare$}$$
Exercise 3.4.D:
If \( I \subset A \) is an ideal, then define its \( \textbf{radical} \) by
$$
\sqrt{I} = \{ r \in A \mid r^n \in I \ \textrm{for}\ \textrm{some}\ n \in \Z^{> 0} \}
$$
For example, the nilradical \(\mathfrak{N}\) (§3.2.R) is \( \sqrt{(0)} \). Show that \(\sqrt{I}\) is an ideal (cf. Exercise 3.2.R(b)). Show that \( V( \sqrt{I}) = V(I) \). We say an ideal is radical if it equals its
own radical. Show that \(\sqrt{\sqrt{I}} = \sqrt{I}\), and that prime ideals are radical.
Proof:
Since \( \emptyset \neq I \subset \sqrt{I} \), \( \sqrt{I} \) is clearly nonempty. Proving that it is closed under addition looks the exact same as the proof above for the nilradical — that is, if \( x, y \in \sqrt{I} \) then there exist \( m, n > 0 \) with \(x^m, y^n \in I \). Then
As before, for each \( i \) we either have \( i \geq m \) or \( m + n - i \geq n \) so that the term is always contained in the ideal \( I \). Therefore, \( x + y \in \sqrt{I} \). By a similar line of logic, for any \( r \in A \) we have \( (rx)^m = r^mx^m \in r^m I \subset I \) — again, we require \(A\) to be commutative here. This proves that the radical is an ideal
To see that \( V(\sqrt{I}) = V(I) \), notice that forward inclusion follows from the fact that \( I \subset \sqrt{I} \). However, before we show reverse inclusion we wish to prove two facts:
\( I \subset J \Rightarrow \sqrt{I} \subset \sqrt{J} \)
Prime ideals are radical.
The first fact is obvious from the definitions, since if \( x^m \in I \) and \( I \subset J \) then \( x^m \in J \). Moreover, the converse is true whenever \( J \) is a prime ideal. To see that the second statement is true, suppose \( \pp \) is a prime ideal and \( x \in \sqrt{\pp} \). Then there exists some \( r > 0 \) with \( x^r \in \pp \) — however, since \( \pp \) is prime this implies that \( x \in \pp \) so that \( \sqrt{\pp} \subseteq \pp \Rightarrow \pp = \sqrt{\pp}\).
We are now ready to show that \( V(I) \subset V( \sqrt{I} ) \) — suppose \( [\pp] \in \spec A \) is a prime ideal with \( I \subset \pp \). By the first statement this implies \( \sqrt{I} \subset \sqrt{\pp} \); by the second statement \( \sqrt{\pp} = \pp \) so we have \( \sqrt{I} \subset \pp \) and thus \( [\pp] \in V( \sqrt{I} ) \).
Lastly, we wish to show that \( \sqrt{\sqrt{I}} = \sqrt{I} \) — reverse inclusion is obvious since every ideal is contained in its radical. For forward inclusion, suppose \( x \in \sqrt{\sqrt{I}} \). Then there exists some \( a > 0 \) with \( x^a \in \sqrt{I} \) — but then there exists some \( b > 0 \) with \( (x^a)^b \in I \). Hence, \( x^{ab} \in I \) so that \( x \in \sqrt{I} \).
$$\tag*{$\blacksquare$}$$
Exercise 3.4.E:
If \( I_1, \dots, I_n \) are ideals of a ring \(A\), show that \( \sqrt{ \bigcap_{i=1}^n I_i } = \bigcap_{i=1}^n \sqrt{I_i} \). We will use this property repeatedly without referring back to this exercise.
Proof:
By induction, it suffices to show that \( \sqrt{I \cap J} = \sqrt{I} \cap \sqrt{J} \) for two ideals \( I, J \subset A \). For forward inclusion, notice that since \( I \cap J \subset I, J \) we have \( \sqrt{I \cap J} \subset \sqrt{I}, \sqrt{J} \) so that \( \sqrt{I \cap J} \subset \sqrt{I} \cap \sqrt{J} \).
In the reverse direction, suppose that \( x \in \sqrt{I} \cap \sqrt{J} \). Then there exist some \( m, n > 0 \) with \( x^m \in I \) and \( x^n \in J \). Taking \( l = \lcm( m, n ) \), it follows that \( x^l \in I, J \) so that \( x^l \in I \cap J \). Thus, \( x \in \sqrt{I \cap J} \).
$$\tag*{$\blacksquare$}$$
Exercise 3.4.F:
Show that \( \sqrt{I} \) is the intersection of all prime ideals containing \( I \). (Hint: Use Theorem 3.2.12 on an appropriate ring.)
Proof:
Recall from Exercise 3.2.J that if we let \( \phi : A \to A/I \) denote the natural projection, then \( \spec A/I \) is the set of all prime ideals containing \( I \). In addition, \( \phi( \sqrt{I} ) = \mathfrak{N} (A/I) \) (since \( I \) is the zero element in the ring \(A/I\)). By Theorem 3.2.12, \( \mathfrak{N}(A/I) \) is the intersection of all prime ideals in \( A/I \), which by our first sentence is precisely the prime ideals in \(A\) containing \(I\).
$$\tag*{$\blacksquare$}$$
Exercise 3.4.G:
Describe the topological space \( \A^1_k \) (cf. Exercise 3.2.D). (Notice that the strange new point \([(0)]\) is "near every other point" — every neighborhood of every point contains \([(0)]\). This is typical of these new points, see Easy Exercise 3.6.N.)
Solution:
This is no different from the preceeding paragraph in the text. Regardless of whether \( k \) is finite or infinite, \( \A^1_k = \spec k[x] \) has infinitely many points. The closed points are those cut out by irreducible polynomials, which necessarily have finitely many roots. However, in the cofinite topology every point is closed (as its compliment is obviously the compliment of a finite set) — this is not true for \( \A^1_k \) since the point \( [ (0) ] \) is not closed. In other words, you cannot find a finite set of polynomials \( S \subset k[x] \) with \( (0) \) being the only prime ideal containing \(S\) (since then all prime ideals would contain \(S\)). Thus, \( \A^1_k \) is slightly coarser than the cofinite topology.
$$\tag*{$\blacksquare$}$$
Exercise 3.4.H:
By showing that closed sets pull back to closed sets, show that \(\pi\) is a continuous map. Interpret \( \spec \) as a contravariant functor \( Rings \to Top \).
Solution:
Let \( I \subset B \) be an ideal, and consider the set \( \pi^{-1}( V(I) ) \) in \( \spec A \). If \( \mathfrak{a} \in \pi^{-1} (V(I))\), then \( \pi(\mathfrak{a}) = \phi^{-1}(\mathfrak{a}) \in V(I) \) (and by Exercise 3.2.M we know that this is a prime ideal). Thus, \( \pp = \phi^{-1}(\mathfrak{a}) \) corresponds to a prime ideal containing \( I \), so that \( \mathfrak{a} = \phi(\pp) \) is a prime ideal containin \( \phi(I) \). Thus, \( \pi^{-1}(V(I)) = V( \phi(I) ) \).
$$\tag*{$\blacksquare$}$$
Exercise 3.4.I:
Suppose \( I, S \subset B \) are an ideal and a multiplicative subset, respectively.
Show that \( \spec B/I \) is naturally a closed subset of \( \spec B \). If \( S = \{ 1, f, f^2, \dots \} \) \( f \in B \). Show that \( \spec S^{-1}B \) is naturally an open subset of \( \spec B \). Show that for arbitrary \(S\), \(\spec S^{-1} B\) need not be open or closed. (Hint: \(\spec \Q \subset \spec \Z\), or possibly Figure 3.5.)
Show that the Zariski topology on \(\spec B/I\) (resp. \(\spec S^{−1}B\)) is the subspace topology induced by inclusion in \(\spec B\). (Hint: compare closed subsets.)
Solution:
Recall by Exercise 3.2.J \( \spec B / I \) corresponds to the prime ideals containing \( I \) — this is precisely \( V(I) \). Similarly, \( \spec S^{-1} B \) corresponds to all prime ideals not intersecting \( S \) — that is all prime ideals \( \pp \) with \( f \notin \pp \). From the following section, we see that this is precisely the distinguished set \( D(f) \).
Now consider \( \spec \Q \subset \spec \Z \) — in this case since \( \Q \) is the fraction field of \( \Z \), we have that \( S = \Z - (0) \). Thus, the prime ideals of \( \spec \Q \) are those which do not intersect \( \Z - (0) \), i.e. \(0\) (notice that this aligns with the fact that \(\Q\) is a field). In particular, the generic point \( (0) \) is neither open nor closed.
As before, the prime ideals of \( \spec B/ I \) correspond to prime ideals containing \( \pp \supset I \) — thus, if \( V(S) = \{ \pp \in \spec B / I \mid S \subset \pp \} \) is a closed subset, then each \( \pp \in V(S) \) corresponds to a prime ideal containing \( (S) + I \). Since each \( (S) + I \) is always contains \(I\) and \( (S) \), this is similar to intersecting with \( V(I) \) since \( V( (S) ) \cap V(I) = V((S) \cup I) \). The proof for \( \spec S^{-1} B \) is similar.
$$\tag*{$\blacksquare$}$$
Exercise 3.4.J:
Suppose \( I \subset B \) is an ideal. Show that \( f\) vanishes on \( V(I) \) if and only if \( f \in \sqrt{I} \) (i.e., \( f^n \in I \) for some \( n \geq 1 \)). (Hint: Exercise 3.4.F. If you are stuck, you will get another hint when you see Exercise 3.5.E.)
Solution:
For the forward direction, suppose that \(f\) vanishes on \( V(I) \) — that is, for each \( \pp \in V(I) \) with \( \pp \supset I \) we have \( f(\pp) = 0 \Leftrightarrow f \in \pp \). However, by Exercise 3.4.F \( \sqrt{I} \) is the intersection of all prime ideals containing \(I\), so that we have \( f \in \sqrt{I} \).
The reverse argument is identical — if \( f \in \sqrt{I} \) then \(f\) is contained in all prime ideals containing \( I \). Thus, for every \( \pp \in V(I)\), \( f \in \pp \Leftrightarrow f(\pp) = 0 \).
$$\tag*{$\blacksquare$}$$
Exercise 3.4.K:
Describe the topological space Spec \( k[x]_{(x)} \).
Solution:
By Exercise 3.2.K we have that the prime ideals of \( k[x]_{(x)} \) are precisely those primes ideals not intersecting \( k[x] - (x) \) — that is, the primes \( \pp \subseteq (x) \) (i.e. polynomials with constant terms). However, the only prime ideals in this case are \((x)\) and the prime ideals of \(k\) — since \( k \) is a field, the only prime ideals are \( (0) \) and \( (x) \), corresponding to the generic point and the unique closed point (respectively). In particular, the generic point \( (0)\) is open since \( D(x) = (0) \) and \( (x) \) is closed since it is maximal.
$$\tag*{$\blacksquare$}$$
Section 3.5: A Base of the Zariski Topology on \(\spec A\): Distinguished Open Sets
Exercise 3.5.A:
Show that the distinguished open sets form a base for the (Zariski) topology. (Hint: Given a subset \(S \subset A\), show that the complement of \(V(S)\) is \(\bigcup_{f \in S} V(f)\).)
Solution:
First we wish to show that \( V(S) = \bigcap_{f \in S} V(f) \) for \( S \subset A \) — however, this follows from Exercise 3.4.C(b) since
$$
\langle S \rangle = \sum_{ f \in S } (f)
$$
Now for any \( f \in A \), we have that \( D(f) \) is simply the compliment of \( V(f) \) in \( \spec A \) since \( f \in \pp \Leftrightarrow (f) \subset \pp \). Therefore, letting \( X = \spec A \) we have
$$
X \backslash V(S) = X \backslash \bigcap_{f \in S} V(f) = \bigcup_{f \in S} X \backslash V(f) = \bigcup_{f \in S} D(f)
$$
Thus, for any open set \( U \), \( X \backslash U \) is a closed set and therefore of the form \( V(S') \) for some \( S' \subset A \). In this case, \( U = \bigcup_{f \in S'} D(f) \) so that every open set is a union of distinguished open sets — consequently, the distinguished open sets form a base.
$$\tag*{$\blacksquare$}$$
Exercise 3.5.B:
Suppose \(f_i \in A\) as \( i \) runs over some index set \(J\). Show that \(\bigcup_{i \in J} D(f_i) = \spec A\) if and only if \( ( \{f_i \}_{i \in J}) = A\), or equivalently and very usefully, if there are \(a_i\ (i \in J)\), all but finitely many \(0\), such that \( \sum_{i \in J} a_i f_i = 1 \). (One of the directions will use the fact that any proper ideal of \(A\) is contained in some maximal ideal.)
Solution:
The reverse direction is trivialized by Exercise 3.4.C(b) — indeed, if \( \sum_{i \in J} (f_i) = A\), then \( V(A) = \emptyset = V(\sum_{i \in J} (f_i) ) = \bigcap_{i \in J} V(f_i) \). By the previous exercise, we have for \( X = \spec A \)
$$
X = X \backslash \emptyset = X \backslash \bigcap_{i \in J} V(f_i) = \bigcup_{i \in J} D(f_i)
$$
For the forward direction, we proceed by contrapositive — suppose that \( \sum_{i \in J} (f_i) \) is a proper ideal in \( A \). Following the hint, this tells us that \( \sum_{i \in J} (f_i) \) is contained in some proper ideal \( \mm \); in other words, \( [\mm] \in V( \sum_{i \in J} (f_i) ) = \bigcap_{i \in J} V(f_i) \) or equivalently \( [\mm] \notin \bigcup_{ i \in J } D(f_i) \).
$$\tag*{$\blacksquare$}$$
Exercise 3.5.C:
Show that if \( \spec A \) is an infinite union of distinguished open sets \( \bigcup_{j \in J} D(f_j) \), then in fact it is a union of a finite number of these, i.e. there is a finite subset \( J' \) so that \( \spec A = \bigcup_{j \in J'} D(f_j) \). (Hint: Exercise 3.5.B).
Solution:
This is a direct application of the previous exercise; by hypothesis, since \( \bigcup_{j \in J} D(f_j) = \spec A \) then by the previous exercise \( \sum_{i \in J} (f_i) = A \). Using the equivalent condition, there exist \( a_i \in A \) with \( \sum_{i \in J} a_i f_i = 1 \) with all but finitely many of the \( a_i = 0 \). If we let \( J_0 \) be the (finite) subset of indices such that \( a_j \neq 0 \), then we still have \( \sum_{j \in J_0} a_j f_j = 1 \) so that \( \sum_{j \in J_0} (f_j) = A \). Using the reverse direction of the above exercise, it follows that \( \bigcup_{j \in J_0} D(f_j) = \spec A \).
$$\tag*{$\blacksquare$}$$
Exercise 3.5.D:
Show that \( D(f) \cap D(g) = D(fg) \).
Solution:
This simply follows from the fact that \( D(f) = \spec A \backslash V(f) \) and Exercise 3.4.C(c) which tells us \( V(f) \cup V(g) = V(fg) \).
$$\tag*{$\blacksquare$}$$
Exercise 3.5.E:
Show that \( D(f) \subset D(g) \) if and only if \( f^n \in (g) \) for some \( n \geq 1 \), if and only if \( g \) is an invertible element of \(A_f\).
Solution:
For the first equivalence, by taking compliments we have that \( D(f) \subset D(g) \) if and only if \( V(g) \subset V(f) \) if and only if all prime ideals containing \( g \) also contain \( f \) if and only if \( f \) is contained in the intersection of all prime ideals containing \( g \) (i.e. \( f \in \bigcap_{\pp \supset g} \pp \) ). By Exercise 3.4.F, this is the same as saying \( f \in \sqrt{(g)} \).
To show that this is equivalent to the second condition, we again have \( D(f) \subset D(g) \) if and only if \( V(g) \subset V(f) \). By the paragraph above, this implies that \( f^n \in (g) \) so that there exists some \( a \in A \) with \( f^n = a g \). But then \( \frac{a}{f^n} \) is inverse to \( g \). Conversely, if \( \frac{g}{1} \) has some inverse \( \frac{b}{f^n} \) in \(A_f\) then there exists some \( f^k \in S = (f) \) with
$$
f^k(bg - f^n) = 0
$$
or \( f^{n+k} = (bf^k) g \in (g) \) so that \( f \in \sqrt{(g)} \Leftrightarrow V(g) \subset V(f) \).
$$\tag*{$\blacksquare$}$$
Exercise 3.5.F:
Show that \( D(f) = \emptyset \) if and only if \( f \in \mathfrak{N} \)
Solution:
\( D(f) = \emptyset \) if and only if for every prime ideal \( \pp \in \spec A \), \( f \in \pp \) if and only if \( f \in \bigcap_{\pp \subset A} \pp = \mathfrak{N} \) (where the last equality follows from Exercise 3.2.S).
If \( A = A_1 \times A_2 \times \dots \times A_n \), describe a homeomorphism \( \spec A_1 \coprod \spec A_2 \coprod \dots \coprod \spec A_n \to \spec A \) for which each \( \spec A_i \) is mapped onto a distinguished open subset \( D(f_i) \) of \( \spec A \). Thus, \( \spec \prod_{i=1}^n A_i = \coprod_{i=1}^n \spec A_i \) as topological spaces. (Hint: reduce to \(n=2 \) for convenience. Let \( f_1 = (1, 0) \) and \( f_2 = (0, 1) \).)
Solution:
First let \( \pi_i : A \to A_i \) denote the canonical surjection onto the \( i^{th} \) component. This induces a morphism of topological spaces \( \phi_i : \spec A_i \to \spec A \) for each \( i \) — by the universal property of the coproduct, this factors through as a map \( \phi : \spec A_1 \coprod \dots \coprod \spec A_n \to \spec A \). Alternatively, this argument may be circumvented by using the fact that a contravariant functor sends products to coproducts and vice-versa.
We next wish to show that the image of \( \phi_i \) is the distinguished open set \( D(f_i) \), where \( f_i \) is the element of \(A\) with \( 1 \) in the \(i^{th}\) component and \( 0 \) everywhere else. To see the equivalence, if \( P \in \im \phi_i \) there exists some prime ideal \( Q \in \spec A_i \) with \( \phi_i(Q) = \pi_{i}^{-1}(Q) = P \) or \( \pi_i(P) = Q\). As \( Q \subset A_i \) is a proper ideal, \(\pi_i(f_i) = 1 \notin Q \Leftrightarrow f_i \notin P \) which is equivalent to \( P \in D(f_i) \).
$$\tag*{$\blacksquare$}$$
Exercise 3.6.B:
Show that in an irreducible topological space, any nonempty open set is dense. (For this reason, you will see that unlike in the classical topology, in the Zariski topology, nonempty open sets are all "huge".)
If \(X\) is a topological space, and \(Z\) (with the subspace topology) is an irreducible subset, then the closure \(\overline{Z}\) in \(X\) is irreducible as well.
Solution:
Let \( U \subseteq X\) be a nonempty open set, and suppose to the contrary that \( \overline{U} \neq X \). Then \( \overline{U}, X \backslash U \) are proper closed subsets (the latter since \(U\) nonempty) with \( X = \overline{U} \cup X \backslash U \), contradicting the fact that \( X \) is irreducible.
Suppose \( \overline{Z} = A \cup B \) where \( A, B \) nonempty closed subsets of \( \overline{Z} \) (and thus closed in \(X\)). Then \( Z \cap A \) and \( Z \cap B \) are closed subsets in \(Z\) by the subspace topology; since \( Z \) is irreducible, we have \( Z = Z \cap A \) or \( Z = Z \cap B \). However, this is equivalent to \( Z \subset A \) or \( Z \subset B\) — by taking closures in \(X\), this tells us \( \overline{Z} \subseteq A \subseteq \overline{Z} \) or \( \overline{Z} \subseteq B \subseteq \overline{Z} \) from which the result follows.
$$\tag*{$\blacksquare$}$$
Exercise 3.6.C:
If \(A\) is an integral domain, show that \( \spec A \) is irreducible. (Hint: pay attention to the generic point \([(0)]\).) We will generalize this in Exercise 3.7.F.
Solution:
By our exercises from §3.4 we know that every closed set of \( \spec A \) is of the form \( V(I) \) for some set \(I \subset A\). If we suppose \( \spec A = V(I) \cup V(J) \), then as \( A \) integral domain \( \Rightarrow (0) \) prime \( \Rightarrow (0) \in V(I) \) or \( (0) \in V(J) \). However, this is equivalent to \( I \subseteq (0) \), \( J \subseteq (0) \) — if we assume without loss of generality \( I = (0) \), then we necessarily have \( V(I) = V(0) = \spec A \).
$$\tag*{$\blacksquare$}$$
Exercise 3.6.D:
Show that an irreducible topological space is connected.
Solution:
We proceed by contrapositive; suppose \( X \) is disconnected, so that there exist \( U, V \) nonempty open subsets with \( U \cap V = \emptyset \) and \( X = U \cup V \). Then \( V \subset X \backslash U \) and \( U \subset X \backslash V \) are non-empty proper closed subsets with \( X = (X \backslash U) \cup (X \backslash V) \) so that \( X \) is reducible.
$$\tag*{$\blacksquare$}$$
Exercise 3.6.E:
Give (with proof!) an example of a ring \(A\) where \( \spec A \) is connected but reducible. (Possible hint: a picture may help. The symbol "\( \times \)" has two "pieces" yet is connected.)
Solution:
Take \( X = \spec k[x,y] / (xy) \) as the two coordinate axes intersecting at the origin. This is clearly connected as a topological space; however, from Exercise 3.4.C(c) we have that \( V(x) \cup V(y) = V(xy) = V(0) = X \).
$$\tag*{$\blacksquare$}$$
Exercise 3.6.F:
Suppose \(I = (wz−xy, wy−x^2 , xz−y^2 ) \subset k[w, x, y, z]\). Show that \(\spec k[w, x, y, z]/I\) is irreducible by showing that \(k[w, x, y, z]/I\) is an integral domain. (This is hard, so here is one of several possible hints: Show that \(k[w, x, y, z]/I\) is isomorphic to the subring of \(k[a, b]\) generated by monomials of degree divisible by 3. There are other approaches as well, some of which we will see later. This is an example of a hard question: how do you tell if an ideal is prime?) We will later see this as the cone over the twisted cubic curve (the twisted cubic curve is defined in Exercise 8.2.A, and is a special case of a Veronese embedding, §8.2.6).
Note that the generators of the ideal of part (a) may be rewritten as the equations ensuring that
$$
\textrm{rank} \begin{pmatrix} w & x & y \\ x & y & z \end{pmatrix} \leq 1
$$
i.e., as the determinants of the \( 2 \times 2 \) submatrices. Generalize part (a) to the ideal of rank one \(2 \times n\) matrices. This notion will correspond to the cone (§8.2.12) over the degree \(n\) rational normal curve (Exercise 8.2.J).
Solution:
Using the Veronese embedding mentioned, it should be clear that the ring morphism \( \phi : k[x, y, z, w] \to k[a, b] \) that we want is \( x \mapsto a^2b, y \mapsto ab^3, z \mapsto a^3\) and \( w \mapsto b^3 \) so that \( I \subset \ker \phi \). By using a standard division algorithm argument (likely what makes this problem difficult) any polynomial \( f \in \ker \phi \) can be decomposed as
by first choosing some \( i \in I \). Then by considering exponents modulo 3, under the mapping of \( \phi \), the only way that the degree of exponents in \(a\) and \(b\) can cancel is when \( g = h = k = 0 \) so that \( \ker \phi = I \). By the first isomorphism theorem, it suffices to ensure that \( \im \phi \) is the subring of homogeneous polynomials of degree divisible by 3 — however, this is fairly immediate from the construction of \( \phi \) since each of the coordinates \( x, y, z, w \) gets mapped to a degree 3 monomial, so that any homogeneous polynomial of degree \(d\) in \( k[x,y,z,w] \) is a degree \(3d\) polynomial in \( k[a,b] \).
There isn't much to state here; the degree \( n \) rational curve is the variety whose ideal is generated by the determinants of \(2 \times 2\) submatrices of
Show that \( \spec A \) is quasicompact (Hint: Exercise 3.5.C).
(Less Important) Show that in general \( \spec A \) can have nonquasicompact open sets. Possible hint: let \(A = k[x_1,x_2,x_3,\dots] \) and \(\mm = (x_1,x_2,\dots) \subset A\), and consider the complement of \(V(\mm)\). This example will be useful to construct other "counterexamples" later, e.g., Exercises 7.1.C and 5.1.J. In Exercise 3.6.T, we will see that such weird behavior doesn’t happen for "suitably nice" (Noetherian) rings.
Solution:
Suppose \(\{ U_i \}\) is an open cover of \( \spec A \) — since the distinguished open sets form a base for the Zariski topology, we may decompose \(\{ U_i \}\) into a cover of distinguished open sets. By Exercise 3.5.C, there is some finite subcover of distinguished open subsets. By possible re-combining distinguished open sets, we obtain a finite subcover.
As the hint suggests, consider the (non-Noetherian) coordinate ring in infinitely many variables \( A = k[x_1, x_2, x_3, \dots] \) and let \( \mm = (x_1, x_2, \dots) \). The compliment of \( V(\mm) \) is clearly an open subset of \( \spec A \) by construction of the Zariski topology; by definition this prevariety \(X\) is
$$
X = \spec A \backslash V(\mm) = \{ [\pp] \in \spec A \mid \pp \subseteq \mm \}
$$
It is fairly easy to see that \( X = \bigcup_{i=1}^\infty D(x_i) \); if this had any finite subcover by finitely many linear forms \(\{ x_{i_1}, \dots, x_{i_n} \}\), then we would have
$$
\spec A \backslash X = V(\mm) = \bigcap_{j=1}^n V( x_{i_j} )
$$
so that \( \mm \subseteq (x_{i_1}, \dots, x_{i_n}) \), contradicting the fact that \( k[x_1, \dots] \) is a free \(k\)-algebra.
$$\tag*{$\blacksquare$}$$
Exercise 3.6.H:
If \(X\) is a topological space that is a finite union of quasicompact spaces, show that \(X\) is quasicompact.
Show that every closed subset of a quasicompact topological space is quasicompact.
Solution:
Suppose \( X = \bigcup_{i=1}^n X_i \) where each \( X_i \) is quasi-compact, and let \( \{ U_j \} \) be an open cover of \(X\). Since \( X_i \subseteq X \), we have that \( \{ U_j \} \) is also necessarily an open cover of \( X_i \). Since \( X_i \) is quasicompact, there exists some finite subcover \( \{ U_{i,1}, \dots, U_{i, r_n} \} \). Taking the union of finitely many finite subcovers will again be a finite collection — it also must necessarily cover all of \(X\) since the \( X_i \) cover \(X\).
The proof is the same as that in standard topology — if \( Y \subset X \) is a closed subset and \( \{ U_i \} \) is an open cover of \( Y \), then \( \{ U_i \} \cup \{ X \backslash Y \} \) is an open cover of \( X \) by open sets (since the compliment of an open set is necessarily open). Then \( \{ U_i \} \cup \{ X \backslash Y\} \) admits a finite subcover, so that only finitely many of the \( U_i \) necessarily cover \(Y\).
$$\tag*{$\blacksquare$}$$
Exercise 3.6.I:
Show that the closed points of \(\spec A\) correspond to the maximal ideals. (In particular, non-empty affine schemes have closed points, as nonzero rings have maximal ideals, §0.3.)
Solution:
\( [\pp] \) is a closed point in \( \spec A \) if and only if \( \{ \pp \} = V( \pp ) = \{ \mathfrak{q} \in \spec A \mid \pp \subset \mathfrak{q} \} \) if and only if the only prime ideal containing \( \pp \) is \( \pp \) itself if and only if \( \pp \) is maximal.
$$\tag*{$\blacksquare$}$$
Exercise 3.6.J:
Suppose that \( k \) is a field and \( A \) is a finitely-generated \( k \)-algebra. Show that the closed points of \( \spec A \) are dense by showing that if \( f \in A \) and \( D(f) \) is a non-empty (distinguished) open subset of \( \spec A \), then \( D(f) \) contains a closed point of \( \spec A \). Hint: note that \(A_f\) is also a finitely generated \(k\)-algebra. Use the Nullstellensatz 3.2.5 to recognize closed points of \(\spec\) of a finitely generated \(k\)-algebra \(B\) as those for which the residue field is a finite extension of \(k\). Apply this to both \(B = A\) and \(B = A_f\).
Show that if \(A\) is a \(k\)-algebra that is not finitely generated the closed points need not be dense. (Hint: Exercise 3.4.K.)
Solution:
As the hint suggests, let \( f \in A \) and consider \( D(f) = \spec A_f \). Now any ideal \( I \subset A_f \) is contained in a maximal ideal \( \mm \subset A_f \) by Zorn's lemma — thus, \( [\mm] \in D(f) \). To see that \( [\mm] \) is indeed a closed point, since \( A_f \) is also a finitely-generated \( k\)-algebra and \( \mm \) is maximal in \( A_f \), \( k(\mm) \) is a finite extension of \( k \). Using the Nullstellensatz in the reverse direction, this implies that \( \mm \) is maximal in \( A \).
By taking \( A = k[x]_{(x)} \), we know from Exercise 3.4.K that \( \spec A \) has a unique closed point \( (x) \) (as \(A \) is a local ring) and a generic point \( (0) \) — in particular, \( D(x) = \{ [\pp] \mid x \notin \pp\} = \{ (0) \}\). Thus, \( D(x) \) does not contain the closed point \( (x) \).
$$\tag*{$\blacksquare$}$$
Exercise 3.6.K:
Suppose \( k \) is an algebraically closed field and \( A = k[x_1, \dots, x_n] / I \) is a finitely-generated \( k \)-algebra with \( \mathfrak{N}(A) = \{ 0 \}\) (so the discussion of §3.2.13 applies). Consider the set \( X = \spec A \) as a subset of \( \A^n_k \). The space \( \A^n_k \) contains the "classical" points \( k^n \). Show that functions on \(X\) are determined by their values on the closed points (by the weak Nullstellensatz 3.2.4, the "classical" points \(k^n \cap \spec A\) of \(\spec A\)). Hint: if \(f\) and \(g\) are different functions on \(X\), then \(f − g\) is nowhere zero on an open subset of \(X\). Use Exercise 3.6.J(a).
Solution:
Suppose that \( f, g \in A \) are two functions that agree on all closed points — that is, for every maximal ideal \( \mm \subset A \), we have \( f = g \mod \mm \). If we suppose to the contrary that \( f \neq g \), then by the preceding exercise \( D(f - g) \neq \emptyset \Rightarrow D(f - g) \) must contain some maximal point \( \mm \) (as the set of closed points are dense). That is to say, there exists some \( \mm \) with \( f \neq g \mod \mm \) contradicting our assumption. Thus, \( D(f - g) = \emptyset \). By Exercise 3.5.F, this implies that \( f - g \in \mathfrak{N}(A) = \{ 0\} \) so that \( f = g \).
$$\tag*{$\blacksquare$}$$
Exercise 3.6.L:
If \( X = \spec A \), show that \( [\mathfrak{q}] \) is a specialization of \( [\pp] \) if and only if \( \pp \subset \mathfrak{q} \). Hence, show that \( V( \pp ) = \overline{ \{ [\pp] \} } \).
Solution:
This is simply moving around definitions — for the latter statement, the closure of \( \pp \) is the smallest closed subset in \( \spec A \) containing \( \pp \). However the Zariski topology is precisely the coarsest topology in which the closed subsets are \( V(-) \). Since \( \pp \) is prime, \( \sqrt{\pp} = \pp \) so that \( V(\pp) \) is indeed the smallest closed subset containing \( \pp \).
Consequently, \( [\mathfrak{q}] \) is a specialization of \( [\pp] \) if and only if \( \mathfrak{q} \in \overline{ \{ [\pp] \} } = V( \pp ) \) if and only if \( \pp \subset \mathfrak{q} \).
$$\tag*{$\blacksquare$}$$
Exercise 3.6.M:
Verify that \( [(y - x^2)] \in \A^2 \) is a generic point of \( V(y - x^2) \).
Solution:
Again, this is unwinding definitions. It suffies to show that every \( \mathfrak{q} \in V(y - x^2) \) is a specialization of \( \pp = [(y - x^2)] \). But this is obvious since \( \mathfrak{q} \in V(y - x^2) \) iff \( (y - x^2) \subseteq \mathfrak{q} \), so that we may apply the above exercise.
$$\tag*{$\blacksquare$}$$
Exercise 3.6.N:
Suppose \(\pp\) is a generic point for the closed subset \(K\). Show that it is "near every point \(\mathfrak{q}\) of \(K\)" (every neighborhood of \(\mathfrak{q} \) contains \(\pp\)), and "not near any point \(r\) not in \(K\)" (there is a neighborhood of \(r\) not containing \(\pp\)). (This idea was mentioned in §3.2.3 Example 1 and in Exercise 3.4.G.)
Solution:
Let \( [\mathfrak{q}] \in K \) and \( U \) be an open neighborhood of \( \mathfrak{q} \) — since distinguished open sets form a base for the Zariski topology, we may assume without loss of generality that \( U = D(f) \) so that \( f \notin \mathfrak{q} \). Then \( K \backslash U \) is a closed subset of \( K \) — however, since the closure of a point is the smallest closed subset containing that point, and \( \overline{ \{\pp\} } = K \), it cannot be the case that \( \pp \in K \backslash U \). In other words, we must also have that \( \pp \in U \).
The second statement is obvious since \( K \) is closed and \( r \notin K \Rightarrow r \in X \backslash K\) which is an open subset that does not contain \( \pp \).
$$\tag*{$\blacksquare$}$$
Exercise 3.6.O:
Show that every point \(x\) of a topological space \(X\) is contained in an irreducible component of \(X\). Hint: Zorn’s Lemma. More precisely, consider the partially ordered set \(S\) of irreducible closed subsets of \(X\) containing \(x\). Show that there exists a maximal totally ordered subset \(\{Z_\alpha\}\) of \(S\). Show that \(\bigcup Z_\alpha\) is irreducible.
Solution:
Let \( x \in X \) and consider the collection of all irreducible closed subsets containing \(x\) which we will label \( \mathfrak{I}_x \). First notice that \( \{ x \} \) is an irreducible set since if \( \{ x \} = A \cup B \) then by the pigeon-hole property \(x\) must be an element of either \( A \) or \(B\) so that \( A = \{x\} \). By Exercise 3.6.B, \( \overline{ \{ x \} } \) must be irreducible as well, so that \( \mathfrak{I}_x \) is non-empty — in addition, \( \mathfrak{I}_x \) is certainly bounded above by \( X \) and is a poset under set containment. Then by Zorn's lemma, there exists a maximal irreducibe closed \( Z_\alpha \) containing \(x\) — we use this notation instead of Vakil's concept of a "totally ordered subset" — it is easier for the moment to consider a maximal element.
$$\tag*{$\blacksquare$}$$
Exercise 3.6.P:
Show that \( \A^2_\C \) is a Noetherian topological space: any decreasing sequence of closed subsets of \( \A^2_\C = \spec \C [x, y] \) must eventually stabilize. Note that it can take arbitrarily long to stabilize. (The closed subsets of \( \A^2_\C \) were de- scribed in §3.4.3.) Show that \( \C^2 \) with the classical topology is not a Noetherian topological space.
Solution:
By §3.4.3 we know that the closed subsets of \( \A^2_\C \) are the closed points (corresponding to maximal ideals), a finite number of irreducible curves (corresponding to height 1 prime ideals, which are thus of Krull dimension 1), and the entire space (the closure of the generic point). Thus, any chain of closed subsets in \( \A^2_\C \) is is necessarily bounded below by a closed point and bounded above by the entire plane, so that the length of the chain is determined by the irreducible curves. However, since there are only finitely many, the chain must stabilize.
To see that \( \C^2 \) in the classical topology is not Noetherian, consider the chain of closed balls \( \overline{B_R(0)} \) — this clearly does not stabilize.
$$\tag*{$\blacksquare$}$$
Exercise 3.6.Q:
Show that every connected component of a topological space \(X\) is the union of irreducible components of \(X\). Show that any subset of \(X\) that is simultaneously open and closed must be the union of some of the connected components of \(X\). If \(X\) is a Noetherian topological space, show that the union of any subset of the connected components of \(X\) is always open and closed in \(X\). (In particular, connected components of Noetherian topological spaces are always open, which is not true for more general topological spaces, see Remark 3.6.13.)
Solution:
Suppose \( Y \) is a connected component of \(X\) and \( x \in Y \). By Exercise 3.6.O, \( x \) is contained in some maximal irreducible component \( Z_\alpha \). Notice that \(Z_\alpha\) irreducible implies its connected, since if \( Z_\alpha = U \cup V \) for \( U, V \) open, disjoint, then \( Z_\alpha = Z_\alpha \backslash U \cup Z_\alpha \backslash V \) — a contradiction. Since \( Y \cap Z_\alpha \neq \emptyset \), it is connected. By virtue of \( Y \) a connected component, this implies that \( Y = Y \cup Z_\alpha \) so that \( Z_\alpha \subset Y \). Since \( Z_\alpha \) is closed, and the compliment is the remainder of the closed irreducible components, \( Z_\alpha \) is clopen and thus \( Z_\alpha = Y \).
$$\tag*{$\blacksquare$}$$
Exercise 3.6.R:
Show that a ring \(A\) is Noetherian if and only if every ideal of \(A\) is finitely generated.
Solution:
For the forward direction, suppose \( I \subseteq A \). Since \( I \) nonempty, take \( f_1 \in I \) — if \( I = (f_1) \) we are done. Otherwise, there exists some \( f_2 \in I \backslash (f_1) \) so that \( (f_1) \subsetneq (f_1, f_2) \). Proceeding in this fashion, we may construct an ascending chain
which must terminate by the Noetherian hypothesis; therefore \( I \) is finitely generated.
Conversely, suppose
$$
I_1 \subsetneq I_2 \subsetneq \dots
$$
is an ascending chain of ideals, and define \( I = \bigcup I_n \). This is also necessarily an ideal since if \( x \in I_r \), \( y \in I_s \), the two are necessarily contained in the larger of the two ideals. By hypothesis, \( I = (f_1, \dots, f_n) \). For each \( f_k \), there exists some \( N_k \) such that \( f_i \in I_{N_k} \) — taking the maximum of all \( N_k \), we get that the chain must eventually terminate.
$$\tag*{$\blacksquare$}$$
Exercise 3.6.S:
If \(A\) is Noetherian, show that \(\spec A\) is a Noetherian topological space. Describe a ring \(A\) such that \(\spec A\) is not a Noetherian topological space. (Aside: if \(\spec A\) is a Noetherian topological space, \(A\) need not be Noetherian. One example is \(A = k[x_1,x_2,x_3, \dots]/(x_1,x_2,x_3, \dots)\). Then \(\spec A\) has one point, so is Noetherian. But \(A\) is not Noetherian as \( (x_1) \subsetneq (x_1, x_2) \subsetneq (x_1, x_2, x_3) \subsetneq \dots \) in \(A\).)
Solution:
Suppose
$$
Z_1 \supset Z_2 \supset \dots
$$
is a descending chain in \( \spec A \) of closed subsets, so that \( Z_i = V(I_i) \) for some \( I_i \subset A \) — since \( V(S) = V((S)) \) we may assume without loss of generality that each \( I_i \) is an ideal. Moreover, by Exercise 3.5.E, we have that \( V(I_i) \subset V(I_j) \) implies \( \sqrt{I_j} \subset \sqrt{I_i} \). Thus, we get an ascending chain of ideals in \( A \) which by hypothesis must terminate.
To see an example of when \( \spec A \) is not a Noetherian topological space, we consider the non-Noetherian ring \(A = k[x_1, x_2, \dots] \).
$$\tag*{$\blacksquare$}$$
Exercise 3.6.T:
Show that every open subset of a Noetherian topological space is quasicompact. Hence if \(A\) is Noetherian, every open subset of \( \spec A\) is quasicompact.
Solution:
Notice that a descending chain of closed subsets corresponds to an ascending chain of open subsets. Suppose \( \{ U_i \} \) is an open cover of a Noetherian topological space \(X\), and consider the collection \( \mathcal{F} \) consisting of finite unions of open sets in \( \{ U_i \} \). Since \( X \) is Noetherian, this collection has a maximal element \( U_{i_1} \cup \dots \cup U_{i_N} \) — notice that this must necessarily cover \(X\), since if \( x \in X\) is in the compliment of our cover, since the original \( \{ U_i \} \) covered \(X\) we could find another \( U_j \ni x \) so that \( U_{i_1} \cup \dots \cup U_{i_N} \subsetneq U_{i_1} \cup \dots \cup U_{i_N} \cup U_j \) contradicting maximality.
$$\tag*{$\blacksquare$}$$
Exercise 3.6.U:
Show that if \( M \) is a Noetherian \(A\)-module, then any submodule of \(M\) is a finitely generated \(A\)-module.
Solution:
The proof is effectively the same as the proof of Exercise 3.6.R — if \( N\) is a submodule of \( M \), then \( N \) must be non-empty as an abelian group. Thus, we may fix some \( f_1 \in M \). If \( \langle f_1 \rangle \neq M \), then we can pick some \( f_2 \) and consider \( \langle f_1, f_2 \rangle \). Proceeding inductively, this yields an ascending chain
$$
\langle f_1 \rangle \subsetneq \langle f_1, f_2 \rangle \subsetneq \dots \subset N \subseteq M
$$
Since \( M \) is Noetherian, this chain must terminate at some \( k \in N \). Notice that we must have \( N = \langle f_1, \dots, f_k \rangle \) since otherwise we could find a \( g \in N \backslash \langle f_1, \dots, f_k \rangle \) with \( \langle f_1, \dots, f_k \rangle \subsetneq \langle f_1, \dots, f_k, g \rangle \) contradicting our assumption.
$$\tag*{$\blacksquare$}$$
Exercise 3.6.V:
If \( 0 \to M^\prime \to M \to M^{\prime \prime} \to 0 \) is exact, show that \( M^\prime \) and \( M^{ \prime \prime } \) are Noetherian if and only if \( M \) is Noetherian. (Hint: Given an ascending chain in \(M\), we get two simultaneous ascending chains in \(M′\) and \(M^{\prime\prime}\). Possible further hint: prove that if
$$
\require{amscd}
\begin{CD}
M^\prime @>>> M @>\phi>> M^{\prime\prime}
\end{CD} $$
is exact,and \(N \subset N^\prime \subset M\), and \(N \cap M′ = N′ \cap M′\) and \(\phi(N) = \phi(N′)\),then
N = N′.)
Solution:
For ease of notation suppose we label our first map \( M' \to M \) as \(\alpha\), and the second map as \( \beta \).
The forward direction is fairly easy — suppose \( A_1 \subsetneq A_2 \subsetneq \dots \) is an ascending chain in \( M^{\prime \prime} \). Then \( \beta^{-1}(A_1) \subsetneq \beta^{-1} (A_2) \subsetneq \dots \) is an ascending chain of submodules of \(M\), and thus must terminate at some \( k\). Since \( \beta \) is surjective by exactness, \( \beta^{-1}(A_k) = \beta^{-1}(A_{k+1}) \Rightarrow A_k = A_{k+1} \). Similarly, if \( B_1 \subsetneq B_2 \subsetneq \dots \) is an ascending chain in \( M' \), then \( \alpha(B_1) \subset \alpha(B_2) \subset \dots\) is an ascending chain of submodules of \(M\), which thus terminates at some \( k \). Since \( \alpha \) is injective \( \alpha(B_k) = \alpha(B_{k+1}) \) yields \( B_k = B_{k+1} \).
In the reverse direction, suppose that \( C_1 \subsetneq C_2 \subsetneq \dots \) is an ascending chain in \( M \). Taking the image under \( \beta \), then since \( M^{\prime \prime} \) is Noetherian, there exists some \( N_1 \geq 1 \) with \( \beta( C_k ) = \beta(C_{k+1}) \) for \( k \geq N_1 \). Similarly, \( \alpha^{-1}( C_1 ) \subsetneq \alpha^{-1}(C_2) \subsetneq \dots \) is an ascending chain in \( M^\prime \) so there exists some \( N_2 \geq 1 \) with \( \alpha^{-1}(C_k) = \alpha^{-1}(C_{k+1}) \) for \( k \geq N_2 \). Take \( N = \max\{N_1, N_2 \}\) and fix some \( x \in C_{N+1} \) so that \( \beta(x) \in \beta(C_{N+1}) = \beta(C_N) \). Then there exists some \( y \in C_N \) with \( \beta(y) = \beta(x) \) or \( y - x \in \ker \beta = \im \alpha \). Then there exists some \( z \in M^\prime \) with \( \alpha(z) = x - y \in C_{N+1} \) implying \( z \in \alpha^{-1}(C_{N+1}) = \alpha^{-1}(C_N) \). Thus, \( \alpha(z) = x - y \in C_N \Rightarrow x \in C_N \) so that \( C_{N} = C_{N+1} \).
$$\tag*{$\blacksquare$}$$
Exercise 3.6.W:
Show that if \(A\) is a Noetherian ring, then \(A^{\oplus n}\) is a Noetherian \(A\)-module.
Solution:
By the previous exercise, this follows from finite induction. For the trivial case, notice we get an exact sequence
$$
\require{amscd}
\begin{CD}
0 @>>> A @>>> A \oplus A @>>> A @>>> 0 \\
@. @. @. \\
@. a @>>> (a,0) \\
@. @. \\
@. @. (a,b) @>>> b
\end{CD}
$$
Since \( A \) is obviously Noetherian as a module over itself, \( A \oplus A \) must be a Noetherian \(A\)-module as well. For the inductive step, if we suppose \( A^{\oplus n} \) is Noetherian, then we get a similar sequence is exact:
Show that if \(A\) is a Noetherian ring and \(M\) is a finitely generated \(A\)-module, then \(M\) is a Noetherian module. Hence by Exercise 3.6.U, any submodule of a finitely generated module over a Noetherian ring is finitely generated.
Solution:
$$\tag*{$\blacksquare$}$$
Section 3.7: The Function \( I(\cdot) \) Taking Subsets of \( \spec A \) to ideals of \(A\)
Exercise 3.7.A:
Let \( A = k[x,y] \). If \( S = \{ [(y)], [(x, y-1)] \} \) (see Figure 3.10), then \(I(S)\) consists of those polynomials vanishing on the \(y\)-axis, and at the point \((0, 1)\). Give generators for this ideal.
Solution:
Using the definition \( I(S) = \bigcap_{ [\pp] \in S } \pp \subset A \), we see that
Suppose \(S \subset \A^3_\C\) is the union of the three axes. Give generators for the ideal \(I(S)\). Be sure to prove it! We will see in Exercise 12.1.F that this ideal is not generated by less than three elements.
Solution:
If \( P \) is a point on one of the coordinate axes, say the \(x\)-axis, then its other two coordinates must vanish. Thus, any function vanishing at \( P\) must be divisible by \( y \) or \(z\), so that \( I( x-\textrm{axis} ) = (y, z) \). Then
Show that \( V(I(S)) = \overline{S} \). Hence, \( V(I(S)) = S \) for a closed set \(S\).
Solution:
Since \( \overline{S} \) is closed, \( \overline{S} = V( J ) \) for some ideal \( J \), and \( S \subset V(J) \Rightarrow I(V(J)) \subset I(S) \) (here we are using Theorem 3.7.1 which may not have been intended.) Then using the following exercise,
However, since the topological closure of \(S\) is the smallest closed subset containing \(S\) we also get \( \overline{S} \subset V(I(S)) \) as desired.
$$\tag*{$\blacksquare$}$$
Exercise 3.7.D:
Prove that if \( J \subset A \) is an ideal, then \( I(V(J)) = \sqrt{J} \). (Huge hint: Exercise 3.4.J).
Solution:
This follows from our definition of \( I( Y ) \) as the set of functions \(f \in A \) vanishing on \(Y\), and Exercise 3.4.J where we proved a function \(f \in A\) vanishes on \( V(J) \) if and only if \( f \in \sqrt{J} \). There is nothing else to prove here.
$$\tag*{$\blacksquare$}$$
Exercise 3.7.E:
Show that \(V(\cdot )\) and \(I(\cdot )\) give a bijection between irreducible closed subsets of \(\spec A\) and prime ideals of \(A\). From this conclude that in \(\spec A\) there is a bijection between points of \(\spec A\) and irreducible closed subsets of \(\spec A\) (where a point determines an irreducible closed subset by taking the closure). Hence each irreducible closed subset of \(\spec A\) has precisely one generic point — any irreducible closed subset \(Z\) can be written uniquely as \(\overline{\{z\}}\).
Solution:
Suppose \( Z \) is an irreducible closed subset; by construction of the Zariski topology we may write \( V(I) \). By Exercise 3.4.B we may assume without loss of generality that \( I \subset A \) is an ideal. Suppose to the contrary that \( I \) is not prime; then we can find some \( fg \in I \) with \( f, g \notin I \). Then \( V(I, f) \cup V(I, g) = V(I) \) implying \( V(I) \) is irreducible — a contradiction.
Conversely, suppose that \( \pp \) is a prime ideal. Then \( V(\pp) = \overline{\{ \pp\}} \) is the closure of a singleton and thus irreducible.
$$\tag*{$\blacksquare$}$$
Exercise 3.7.F:
A prime ideal of a ring \(A\) is a minimal prime ideal (or more simply, minimal prime) if it is minimal with respect to inclusion. (For example, the only minimal prime of \(k[x, y]\) is \((0)\).) If \(A\) is any ring, show that the irreducible components of \(\spec A\) are in bijection with the minimal prime ideals of \(A\). In particular, \(\spec A\) is irreducible if and only if \(A\) has only one minimal prime ideal; this generalizes Exercise 3.6.C.
Solution:
This follows from the previous exercise and the definition of an irreducible component as a maximal irreducible subset. Indeed, if \( Z \) is an irreducible component, then as an irreducible closed subset \( Z \) corresponds to some prime ideal \( \pp \). If we suppose to the contrary that there exists some other prime \( \qq \subset \pp \), then by the previous exercise (and the fact that \( V(\cdot) \) is inclusion-reversing) we get that \( Z \subset V(\qq) \) with \( V(\qq) \) irreducible and closed — a contradiction! The reverse direction is identical.
$$\tag*{$\blacksquare$}$$
Exercise 3.7.F:
A prime ideal of a ring \(A\) is a minimal prime ideal (or more simply, minimal prime) if it is minimal with respect to inclusion. (For example, the only minimal prime of \(k[x, y]\) is \((0)\).) If \(A\) is any ring, show that the irreducible components of \(\spec A\) are in bijection with the minimal prime ideals of \(A\). In particular, \(\spec A\) is irreducible if and only if \(A\) has only one minimal prime ideal; this generalizes Exercise 3.6.C.
Solution:
This follows from the previous exercise and the definition of an irreducible component as a maximal irreducible subset. Indeed, if \( Z \) is an irreducible component, then as an irreducible closed subset \( Z \) corresponds to some prime ideal \( \pp \). If we suppose to the contrary that there exists some other prime \( \qq \subset \pp \), then by the previous exercise (and the fact that \( V(\cdot) \) is inclusion-reversing) we get that \( Z \subset V(\qq) \) with \( V(\qq) \) irreducible and closed — a contradiction! The reverse direction is identical.
$$\tag*{$\blacksquare$}$$
Exercise 3.7.G:
What are the minimal prime ideals of \(k[x, y]/(xy)\) (where \(k\) is a field)?
Solution:
We know that the minimal prime ideals correspond to the irreducible components; as \( V(xy) \) is simply the union of the coordinate axes, it is easy to see that the irreducible components are \( V(x) \) and \( V(y) \). Thus, the minimal primes are precisely \( (\overline{x}), (\overline{y}) \) (where we use the superscript to remind us of the quotient).
