Solutions to The Rising Sea (Vakil)

Proof:

Suppose to the contrary that \( \P^n_k \) is a reducible so that there exists some non-empty closed sets \( Y_1, Y_2 \) with \( \P^n_k = Y_1 \cup Y_2 \). Since \( \P^n_k \) is covered by the \( n+1 \) distinguished open sets \( D(x_i) \), which are isomorphic to \( \A^n_k \) and thus irreducible by Exercise 3.6.C, each \( D(x_i) \) is either disjoint from the \( Y_1 \) or is contained in \(Y_1\). If there exists some distinct \( D(x_i) \) contained in \( Y_1 \) and \( D(x_j) \) contained in \( Y_2 \), then \(\{ D(x_i), D(x_j)\} \) form a separation so that \( \P^n_k \) is disconnected — a contradiction. Thus, we must have either \( Y_1 = \P^n_k \) or \( Y_2 = \P^n_k \).

Proof:

Given a scheme \( (X, \OO_X) \), the map \( p \mapsto \overline{ \{ p \} } \) is still clearly well-defined using the Zariski topology and sends points to irreducible closed subsets (if it was not irreducible, we would be able to use an argument similar to that above to show that it is reducible in some affine set which would contradict Exercise 3.7.E).

Conversely, if \( Z \subset X \) is a closed irreducible subset, then for each open affine set \( U \subset X \) nontrivially intersecting \( Z \), we have \( U \cap Z \) is an irreducible closed subset in the subspace topology, in which case \( U \cap Z = \{ p \} \). To see that this holds on intersections, if \( U^\prime \) is another open set nontrivially intersecting \( Z \) then \( U^\prime \cap Z = \{ p^\prime \} \) must also be a point — if we suppose to the contrary that \( p \) and \( p^\prime \) are distinct, then \( Z \) is irreducible as \( Z \backslash U \cup \overline{\{ p \}} \). As the affine open sets form a base by Exercise 4.3.D, \( Z \) must then be a point.

Proof:

Following the hint, suppose \( X \) is a finite union of Noetherian subspaces \{ X = \bigcup_{i=1}^n U_i \} and

is a descending chain of closed subsets of \( X \). Then for each \( i \),

is a descending chain of closed subsets in \( U_i \) (w.r.t the subspace topology), so by the Noetherian condition there exists some \( n_i \) with \( Z_{n_i} \cap U_i = Z_{n_i + 1} \cap U_i \). Taking \( N = \max \{ n_i \} \) we have that

The result follows from Exercise 3.6.S since each \( A_i \) being Notherian implies each \( \spec A_i \) is a Noetherian topological space.

Proof:

The forward direction is trivial; by definition a scheme \( (X, \OO_X) \) is one such that for every point \( x \in X \) we can find an affine neighborhood \( U_x = \spec A_x \) (subscript does not denote localization, just index) — in other words, we may create an affine cover \( \{ \spec A_x \}_{x \in X} \). By the quasicompact condition, a finite number of these cover \( X \).

Conversely, suppose \( (X, \OO_X) \) is covered by finitely many affine open subschemes \( \{ (\spec A_i, \OO_{\spec A_i}) \}_{i=1}^n \) and \( \{ U_i \} \) is an arbitrary open cover of \( (X, \OO_X) \). By Exercise 3.6.G we know that \( (\spec A_i, \OO_{\spec A_i}) \) is quasicompact for each \( i \) so that there exist finitely many \( U_i \) covering \( (\spec A_i, \OO_{\spec A_i}) \). Since there are finitely many affines each covered by finitely many \( U_i \), there must then exist a finite subcover.

Proof:

By Exercise 3.6.H(b), closed subsets of quasicompact spaces are quasicompact, so it suffices to prove that \( X \) has a closed point. By the previous exercise, we know that \( X \) is the finite union of affines so that \( X = \bigcup_{i= 1}^n \spec A_i = \bigcup_{i=1}^n U_i \). Notice that in \( U_1 \), \( U_1 \cap (U_2 \cup \dots \cup U_n)^C \) is a closed subset. Assuming \( U_1 \not \subset U_j \) for any \( j > 1 \), \( U_1 \cap (U_2 \cup \dots \cup U_n)^C \) is a nonempty closed subset in \( U_1 \) corresponding to some prime ideal \( \pp \subset A_1 \). Since every prime ideal is contained in a maximal ideal by Zorn's lemma, \( U_1 \cap (U_2 \cup \dots \cup U_n)^C \) contains a closed point \( p \in U_1 \). It follows from construction that since \( \{ p \} \) is closed in \( (U_2 \cap \dots \cap U_n)^C \) which is closed in \(X\), \( \{ p \} \) is also closed in \(X\).

Proof:

For the forward direction suppose that \( X \) is quasi-separated and \( U_1 = \spec A_1\), \( U_2 = \spec A_2 \) are affine. Since both \( U_1 \) and \( U_2 \) are quasicompact By Exercise 3.6.G, we have that \( U_1 \cap U_2 \) is quasicompact (by definition of quasi-separated). By Exercise 5.1.D this is equivalent to \( X \) being the union of finitely many affine open subschemes.

Conversely, let \( W \) and \( V\) be quasicompact open subsets. Again using Exercise 5.1.D, this implies that \( W = \bigcup_{i=1}^n \spec A_i \) and \( V = \bigcup_{i=1}^m \spec B_i \). Then

By assumption, the intersection \( \spec A_i \cap \spec B_j \) is a finite union of affine open sets. Since \( i, j \) range over a finite index, \( W \cap V \) can be written as the union of finitely many open affine sets. By the converse of Exercise 5.1.D above, \( W \cap V \) is quasicompact.

Proof:

Let \( X = \spec R \); suppose \( W, V \) are quasicompact open subsets of \(X\) (which we know by now are precisely unions of affine open sets). If we suppose without loss of generality that \( W, V \) are also affine, then by Exercise 3.6.G \( W, V \) can both be written as the finite unions of distinguished open subsets \( W = \bigcup_{i=1}^n D(f_i) \), \( V = \bigcup_{j=1}^m D(g_j) \) (as distinguished open sets form the base of the Zariski topology on \( \spec R \)). Thus,

which is then a finite union of affine open subsets ( since \( D(f_i g_j) \) is isomorphic to \( \spec A_{f_ig_j} \) ) so that \( X \) is quasiseparated.

Proof:

For the forward direction, suppose \( X \) is quasicompact and quasiseparated. By quasicompactness and Exercise 5.1.D, we may write \( X = \bigcup_{i=1}^n U_i \) with \( U_i = \spec R_i \). On intersections \( U_i \cap U_j \), since affine sets are quasicompact and \( X \) is quasiseparated, \( U_i \cap U_j \) is thus quasicompact — using the exercise above a second time, this implies \( U_i \cap U_j\) is the intersection of affine open subsets.

Conversely, suppose that \( X \) can be covered by finitely many open subsets, any two of which have intersectiona also covered by a fininte number of affine opens. The first condition immediately gives us quasicompactness by Exercise 5.1.D (a third time) — thus we may assume \( X = \bigcup_{i=1}^n \spec R_i = \bigcup_{i=1}^n U_i \). By Exercise 5.1.F, it suffices to show that if \( W = \spec S_1 \), \( V = \spec S_2 \), then \( W \cap V \) has affine cover. By assumption we know this is true if \( W, V \) are in our open cover \(\{ U_i\}\). If \( W \) and \(V\) both lie in some \( U_i \) then we are done by the previous exercise. Otherwise, \( U_i \cap V \) and \( U_i \cap W \) are open sets in \( U_i \) and thus are finite unions of affine \( D(f_i) \) (via basis of the Zariski topology). Now the intersections \( U_i \cap U_j \) by assumption are the finite union of affine open sets; intersecting \( V \) and \( W \) with each affine open set in the intersection, the same logic (using the distinguished open sets from basis of Zariski topology) shows that \( W \cap V \) is the union of finitely many open distinguished subsets.

Proof:

By definition a projective \(A\)-scheme is of the form \( X = \proj S_\bullet \) where \( S_\bullet \) is a finitely generated graded ring over \( A \). By Exercise 4.5.D this is equivalent to \( S_\bullet \) being generated over \( A = S_0 \) by a finite number of homogeneous elements \( x_0, \dots, x_n \) of positive degree. By Exercise 4.5.F and Exercise 4.5.G, the distinguished open sets \( D(x_1), \dots, D(x_n) \) form an affine open cover for \(X\) — thus, \(X\) is quasicompact. By Exercise 4.5.K we have that the intersection of two of these affine open sets is indeed affine. Since this in fact holds for any distinguished open sets (which form a base of our topology on \(X\)), we have that \(X\) is quasiseparated.

Proof:

As suggested in the hint, let \( X_1 \) and \( X_2 \) denote the two copies of \( \spec k[x_1, x_2, \dots] \) and let \( Y \) denote the space obtained by gluing the two affine schemes along the compliment of \( \mm \). By virtue of the quotient map, \( X_1 \) and \( X_2 \) are still open in \( Y \) and are affine (as they are isomorphic to \(X_i\) ). However, \( X_1 \cap X_2 = U \) which was shown not to be quasicompact in Exercise 3.6.G.

Proof:

For the forward direction, assume that \( X \) is reduced and suppose \( f_p \in \OO_{X, p} \) with \( f_p^n = 0 \) for some point \( p \in X \) and \( n \geq 1 \). If we let \( [ \widetilde{f}, U] \) be a local representative of \( f_p \) (i.e. \( \widetilde{f} \in \OO_{X}(U) \) with equivalent stalks ), then it must also be the case that \( \widetilde{f}^n = 0 \) on a possibly smaller open neighborhood \( V \subset U \) of \( p \). By assumption this implies that \( \widetilde{f} \vert_V = 0 \), and using Exercise 2.4.A, we know that the map \( \OO_X(V) \hookrightarrow \prod_{p \in V} \OO_{X, p} \) is injective so that \( \widetilde{f}_p = f_p = 0 \) in \( \OO_{X, p} \).

The converse direction is practically identical. IF we know that all stalks have no nonzero nilpotents and \( f \in \OO_X(U) \) satisfies \( f^n = 0 \) for some \( n \geq 1 \), then \( f^n_p = 0 \) for each \( p \in U \) which implies \( f_p = 0 \) for each \( p \). Again using the injectivity of the map \( f \mapsto \prod_{p \in U} f_p \), this tells us that \( f = 0 \).

Lastly, if \( f = g \) agree at all points \( P \in X \) then on any affine open subset \( U = \spec A \) we have \( f - g \vert_U \in \pp \) for all \( \pp \in \spec A \). Thus \( f - g \vert_U \in \mathfrak{N}(A) = 0 \) since \( A = \OO_X(U)\) has no nonzero nilpotents — since the open affines cover our scheme \(X\) by definition, \( f = g \).

Proof:

Suppose there is some nonempty open set \( U \subset \spec A \) and \( f \in \OO_{\spec A}(U) \) with \( f^n = 0 \) — since distinguished open subsets form a base for the Zariski topology on \( \spec A \), we may assume without loss of generality that \( U = D(g) \) for some nonzero \( g \in A \) (which we also require to be nilpotent by Exercise 3.5.F). By Exercise 4.3.B \( \OO_X(U) = \spec A_f \). Then \( f = \frac{s}{g^m} \) for some \( s \in A \), and since \( f^n = 0 \) there exists some \( k \geq \) with \( g^k s^0 = 0 \) in \(A\). In particular, either \( gs \) is nilpotent in \(A\) (which is impossible) or \( s = 0 \) in which case \( f = 0 \) in \( A_f = \OO_{\spec A}(U) \).

To see that \( \A^n_k \) is reduced simply follows from the fact that \( k[x_1, \dots, x_n] \) is a domain and thus has no nilpotent elements. Since \( \P^n_k \) is covered by \( n+1 \) copies of \( \A^n_k \), it is also necessarily reduced.

Proof:

Similar to Exercise 3.2.L, if \( f(x,y) \in ( k[x, y] / (y^2 , xy ) )_x \) satisfies \( f^n = 0 \) then by lifting to \( k[x, y]_x \) there is some representative \( \widetilde{f} \) with \( \widetilde{f}^n = a_1 y^2 + a_2 xy \). However, since \( x \) is invertible in \( k[x, y]_x \), we must have \( y \mid \widetilde{f} \) so that \( f(x, y) = 0 \) in \( k[x, y] / (y^2, xy) \). To see that the only point of \( \spec k[x, y] / (y^2, xy) \) with reduced stalk is the origin, we must have that for \( (y^2, xy) \subset \pp \subset k[x, y] \) prime \( y \in \pp \) so that our point geometrically lies on the \( x \)-axis. If instead \( x - a \mid \pp \) for some \( a \neq 0 \), then \( ay \) would vanish on the stalk \( \OO_{X, \pp} \) so that \( y = 0 \) and thus there are no nilpotents. Therefore, we only get a non-reduced stalk at the point \( \pp = (x, y) \).

Proof:

By Exercise 5.2.A above, it suffices to check on the level of stalks. Thus, the only nontrivial direction is \( \OO_{X, \pp} \) reduced for closed \( \pp \ in X \ \ \Rightarrow \OO_{X, \pp} \) reduced at all points \( \pp \). Following the hint, suppose \( \qq \in X \) is not a closed point — by Exercise 5.1.E earlier, we know that \( \overline{\{ \qq \}} \) contains a closed point \( \pp \). By Exercise 3.6.N, we know that for any affine neighborhood \( U = \spec A \) of \( \pp \) we must also have \( \qq \in U \). In particular \( U \cap \overline{\{ \qq \}} = V(\qq) \) and since \( \pp \in V(\qq) \) we have \( \qq \subset \pp \). From definitions, \( A_\qq = \OO_{X, \qq} \) is the localization of \( A_\pp = \OO_{X, \pp} \) at \( \qq A_{\pp} \) — since \( A_\pp \) is reduced, all of its localizations are reduced (to see this, notice that if \(S\) is a multiplicatively closed subset, then \( (\frac{f}{s})^n = 0 \) implies \( \exists s^\prime \) with \( s^\prime f^n = 0 \) in \( A_\pp \) so that \( s^\prime f \) is nilpotent in \( A_\pp \)).

Proof:

Since \(X\) is quasicompact, it may be covered by finitely many affine sets \( \spec A_1, \dots, \spec A_n \). Then on each \( \spec A_i \), we know that \( f \) vanishes at all points so that by Exercise 3.5.F, \( f \in \mathfrak{N}(A_i) \). Thus, there exists some integer \( k_i \geq 0 \) such that \( f^{k_i} = 0 \) — by taking \( N = k_1 \cdot \dots\cdot k_n \), we have that \( f^N \) vanishes on \( \OO_X(U) \) for every open set \( U \subset X \) (since each open set must intersect some affine set in the cover).

To see that this need not hold in the non-quasicompact case, we follow the hint and consider \( X = \coprod_{n} \spec k[\epsilon] / (\epsilon^n) \) with the linear term \( f = \epsilon \). Then for each \( n \geq 0 \), \( f^n \) only vanishes on the first \( n \) affine sets, but clearly does not vanish on

Proof:

For the forward direction suppose that \( X \) is integral, and let \( f \neq 0 \). Then if \( f^n = 0 \) for some \( n > 1 \) then \( f\) and \( f^{n - 1} \) would be zero divisors — a contradiction. To see that \( X \) is irreducible, suppose to the contrary that there exists some \( X = A \cup B \) for closed \( A \), \( B \). Taking \( U = X \backslash A \), \( V = X \backslash B \) we have that \( U, V \) are disjoint open subsets. By a similar argument to Exercise 3.6.A we have a natural isomorphism \( \OO_X(U \cup V) \cong \OO_X(U) \times \OO_X(V) \) — then we can take some sections \( (s, 0), (0, t) \in \OO_X(U) \times \OO_X(V) \) which would necessarily be zero divisors, a contradiction.

In the reverse direction, suppose that \( X \) is irreducible and reduced. Then the restriction to any open subset \( U \) is also necessarily reduced and irreducible (the latter is not proved in earlier exercises, but if \(U\) were to be reducible, then there exist proper closed subsets \( A, B \) in the subspace topology so that \( A = U \cap A^\prime \), \( B = U \cap B^\prime\) and thus \( X \backslash U \cup A^\prime \) and \( X \backslash U \cup B^\prime \) would make \( X \) reducible ), so it suffices to check on the level of affine open sets. By assumption \( A = \OO_X(U) \) is reduced so that \( \mathfrak{N}(A) = (0) \) — by definition, this implies that \( (0)\) is prime and thus an integral domain.

Proof:

The forward direction is trivial since if \( \spec A \) is integral then \( \OO_X(U) \) is a domain for all \( U \) open and thus \( A = \OO_X(X) \) is a domain.

For the reverse direction, suppose \( A \) is an integral domain, and let \( U \subset \spec A \) be an open subset. Since the distinguished open subsets form a basis for the topology, we may assume without loss of generality that \( U = D(f) \) for some nonzero \( f \in A \). From our construction of the structure sheaf, we know that \( \OO_X(D(f)) = A_f \). Since \( A \) is an integral domain then \( A_f \) must also be an integral domain since \( \frac{s}{f^n} \cdot \frac{t}{f^m} = 0 \) in \(A_f\) if and only if \( f^k \cdot st = 0 \) in \( A \) for some \( k \geq 0\). Since \( f \neq 0 \), it must be the case that either \( s = 0 \) or \( t = 0 \).

Proof:

Since \(X\) is irreducible, \( \spec A \) is also necissarily irreducible and \( \eta \) is the unique generic point so that \( \eta\vert_{\spec A} = (0) \). Then

Proof:

We know from the previous exercise that the generic point \( \eta \in X \) is unique. We may assume without loss of generality that \( U = \spec A \) is affine and \( V = D(f) \) for some \( f \in A \). Then we know that \( \spec A \to \spec A_f \) is necessarily injective since \( g = 0 \) in \(A\) implies \( \frac{g}{f^r} = 0 \) in \( A_f \) by an argument similar to Exercise 5.2.G above.

By a similar argument, we know the map \( A \to \textrm{Frac}\,(A) \) is injective, so it follows that whenever \(X\) is an integral scheme the map \( \OO_X(\spec A) \to \OO_{X, \eta} = K(A) \) is injective.

Proof:

Suppose \( X \) is a locally Noetherian scheme, such that \( X \) can be covered by \( \{ U_i = \spec A_i \} \) with each \( A_i \) Noetherian — in addition, let \( V = \spec B \) be an affine open set. Now \(V\) can be covered by finitely many (since affines are quasicompact) \( V \cap U_i \), which we may assume without loss of generality are distinguished opens. But then the \( V \cap U_i = \spec B_{f_i} \) are also necessarily Noetherian, as \(B\) is Noetherian. By Exercise 3.6.T, any open subset of \( U_i \cap B_{f_i} \) must then be quasicompact — by taking another affine open set \( W\), this implies \(V \cap W\) is quasicompact.

Proof:

Let \(X\) be a Noetherian scheme and suppose to the contrary that \(X\) has infinitely many irreducible components. Then we can construct an infinite ascending chain of closed subsets by adding each irreducible component, and thus construct an infinite descending chain of open subsets which covers \(X\) and has no finite subcover.

To see that the latter statement is true, we know by Exercise 3.6.Q that every connected component \( X_i \) is the union of irreducible components — since the closure of an irreducible set Exercise 3.6.B(b) is irreducible, irreducible components are closed. If \( X_i \) did not have a finite subcover, \( X \) would have an infinite cover with no finite subcover so that \( X \) is not quasicompact — a contradiction.

Proof:

For the forward direction, suppose that \( X \) is integral. We know that since \( \OO_{X}(U) \) is integral for every open set \( U \subset X \) then each \( \OO_{X, p} \) must be integral by construction. We require that the empty set is not irreducible, as otherwise schemes would not have a unique decomposition into irreducible components — by Exercise 5.2.F it must be the case that integral schemes are nonempty as well. Moreover, \( X \) is connected by Exercise 3.6.D.

For the converse, we follow the hint as suggested: suppose that \( X \) is nonempty and connected with integral stalks. By Exercise 5.2.A and Exercise 5.2.F we know that \(X\) is reduced and thus it suffices to show that \( X\) is irreducible. Since \(X\) is Noetherian and connected, it has a single connected component which is the finite union of irreducible components \( \{Y_i\} \) by the previous exercise — if we suppose without loss of generality \( X = Y_1 \cup Y_2 \), then if \( Y_1 \) and \(Y_2\) do not intersect we could write \( X = X \backslash Y_1 \cup X \backslash Y_2 \) contradicting the fact that \( X \) is connected. However, if \( Y_1, Y_2 \) intersect at some point \( p \), then \( \OO_{X, p} \) cannot be an integral domain as (affine locally on some \( \spec A \)) each irreducible component may be cut out by finitely many polynomials \( f_1, \dots, f_r \in A\) (by the Noetherian condition), so by multiplying generators from the two sets we would obtain zero divisors.

Proof:

1. From our prior definition of quasiprojectiveness, \(X\) is a quasicompact open subset of \( \proj S_\bullet \), where \( S_\bullet \) is a finitely generated (graded) algebra over \( A = S_0 \). By quasicompactness, we need only show that \(X\) satisfies the locally finite type condition, i.e. each affine open \( \spec B_i \subset X \) is such that \( B_i \) is a finitely-generated \(A\)-algebra. We may assume without loss of generality that our affine open sets \( \spec B_i \) are distinguished opens and thus of the form \( D(f) = \spec (S_f)_0 \) for some homogeneous \(f\). However, since \( S_\bullet \) is a finitely generated \( S_0 \)-algebra, \( (S_f)_0 \) must be as well by Proposition 5.3.3.

   By looking at the preceding paragraph in the text, we know that if \( S_0 = A \) is Noetherian, then any finitely-generated algebra over \(A\) is also Noetherian so that by definition \(X\) is a Noetherian scheme. The last statement follows from Exercise 5.3.B

2. The argument is effectively the same as part (a) above, however we are now dropping the assumption that our subscheme is quasicompact. Thus, it again follows by proposition 5.3.3 that each affine distinguished open \( D(f) = \spec (S_f)_0 \) satisfies \( (S_f)_0 \) is a finitely-generated \( S_0 \)-algebra, so that \( U \) is indeed a subscheme of finite-type. Now if we assume \( A \) is Noetherian, since \( S_\bullet \) is a finitely generated (graded) \(A\)-algebra it must be Noetherian as well — in particular it is also a Noetherian ring. Again using 5.3.3 (this time part (a)), each distinguished open subset \( D(f) = \spec (S_f)_0 \subset \proj S_\bullet \) is also necessarily Noetherian. By Exercise 3.6.T, any open subset of a Noetherian topological space is quasi-compact so that each \( U \cap D(f) \) is quasicompact. Since there are only finitely many distinguished open sets by the \( \proj \) constrution from the previous chapter, this implies that \( U \) is indeed quasicompact.

   Modifying the previous example of Exercise 3.6.G to the projective case, take \( S_\bullet = k[x_0, x_1, \dots] \) and let \( U \) by the open subscheme \( \proj S_\bullet - V( (x_0, x_1, \dots) ) \) of the projective scheme \( \proj S_\bullet \). Then \( U \) cannot be quasicompact, as it would then be quasicompact on the restriction to affine open sets contradicting the statement of that exercise.

Proof:

1. We indeed know that \( k[x_0, \dots, x_n] \) is a finitely generated \( k \)-algebra, and thus \( k[x_0, \dots, x_n] / I \) is as well. Thus, it suffices to show that \( k[x_0, \dots, x_n] \) is reduced — that is, it has no nonzero nilpotents. However, an element \( f \in k[x_0, \dots, x_n] / I\) is nilpotent if and only if \( f^n \in I \) for some \( n \geq 0 \) (which is precisely the definition of \( \sqrt{I} \)).

2. By a similar line of logic, it suffices to show that \( X = \proj k[x_0, \dots, x_n] / I \) is a reduced scheme (which requires more than showing the graded ring \( S_\bullet = k[x_0, \dots, x_n] / I\) is reduced now since projective space is not affine). Though we will not prove it here, a useful lemma is that reducedness can be checked on open affine sets; that is, \( X \) is reduced iff \( \OO_X(U) \) is a reduced ring for all \( U \subset X \) affine open. In fact, \( X \) is reduced iff there exists an affine open cover \( X = \bigcup_{i} U_i \) with each \( \Gamma(U_i, \OO_X) \) reduced (see [Stacks, tag 01OJ]). By taking our standard open cover \( D(x_i) = \spec ((S_\bullet)_{x_i})_0 \), we know that \( (S_\bullet)_{x_i} \cong k[x_0, \dots, x_i^\pm, \dots, x_n] / I_{x_i} \). Now \( I \subset k[x_0, \dots, x_n] \) radical implies \( I_{x_i} \subset k[x_0, \dots, x_i^\pm, \dots, x_n] \) radical, since if \( \frac{f}{x_i^k} \) is such that \( \frac{f^n}{x^{kn}} \in I_{x_i} \) for some \( n \), then multiplying by large enough \( x_i \) gives us \( f^n \in I \) and thus \( f \in I \Rightarrow \frac{f}{x_i^k} \in I_{x_i}\). By part (a) above, it follows that the affine subscheme \( \spec ((S_\bullet)_{x_i})_0 \) does not have any nilpotent elements, so that \( \proj k[x_0, \dots, x_n] / I \) is covered by reduced \( D(x_i) \) and therefore also reduced.

Proof:

For the forward direction, suppose that \( X \) is a locally finite-type \( k \)-scheme with affine cover \( U_i = \spec B_i \) (where the \( B_i\) are finitely-generated \(k \)-algebras) and \( p \in X \) is a closed point. If \( p \in U_i \) for some \(i\), then locally we may write \( p = [\mm] \) where \( \mm \) is a maximal ideal in \( B_i\). Then we have that the stalk \( \OO_{X, p} \) is precisely \( (B_i)_{\mm} \) and thus our residue field \( \kappa(p) \) is precisely \( \kappa(p) = (B_i)_{\mm} / \mm (B_i)_{\mm} = \textrm{Frac}( B_i / \mm) \). Then as \( B_i \) is a finitely-generated \(k\)-algebra, this implies that \( \kappa(p) \) is a finite extension by Zariski's lemma.

Conversely, if \( p \in X \) is such that \( \kappa(p) \) is a finite-extension of \(k \), then for any affine open \(U_i = \spec B_i \) containing \( p \) (such that \( p = [\pp] \) for some prime ideal \( \pp \subset B_i \) ) we have \( B_i / \pp \) is a finite \( k \)-algebra. By Exercise 3.2.G this implies that \( B_i / \pp \) is in fact a field so that \( \pp \) is maximal and thus \( p \) is a closed point.

Note that denseness of our closed points may be checked affine locally, as our closed points are dense if and only if every open set \( V \subset X\) contains a closed point. However, as affine opens \( U_i = \spec B_i \) cover \(X \), if we showed that \( U_i \cap V \) contains a closed point then we're done. Thus, the result follows from Exercise 3.6.J(a)

Proof:

To start off we simply define \( X_{an} \) as the set of closed points of \(X\) with the subspace topology. If we first consider the case that \(X = \spec R\) is a reduced finite-type affine \(\C\)-scheme, then \( R \) is a finitely generated \(\C\)-algebra. Since every maximal ideal \( \mm \subset R \) is the kernel of a unique \( \C \)-algebra homomorphism, there is a natural bijection

(most of this follows from argument that \(\spec \) is a contravariant functor and further the category of affine \(k\)-varities is equivalent to the category of reduced finite \(k\)-algebras). Now in the more general case that \( X \) is just a reduced finite-type \( \C \)-scheme, if we have open subsets \( U = \spec R \) and \( V = \spec S \) then our closed point \( p \) corresponds to maximal ideals in \( R, S\) which are precisely the kernels of some \( \C \)-algebra homomorphisms \( \phi : R \to \C \) and \( \phi^\prime : S \to \C \) respectively. Then by universal property of the kernel, we may find a unique map \( \psi : R \to S\) with \( \psi(\ker \phi) = \ker \phi^\prime \).

Proof:

Following the hint, we know that by the base identity axiom our map \( A \hookrightarrow \prod A_{f_i} \) is injective (as mentioned in equation 4.1.3.1 in the text). Consider

If we indeed had (for some fixed \(j\)) \( I_{i, j} = I_{i, j+1} \) for all \( i \), then the map \( q^\prime \) would be an isomorphism. Since our horizontal maps are injective, this would imply that the map \( q \) is an isomorphism (we know it is surjective as it is the quotient map arising from the third isomorphism theorem). Thus, we must have \( I_j = I_{j+1} \) which is absurd.

Proof:

From the universal property of quotients, it is easy to see that \( K(S^{-1}A) = K(A) \); moreover, since \( A \) is an integral domain \( S^{-1}A \) is as well. Following the hint, suppose \(x \in K(A)\) is integral over \( S^{-1}A \), satisfying some equation

where our \( a_i = \frac{b_i}{s_i} \). If we take \( s = \prod_i s_i \in S\) and multiply the above equation by \(s^n \) then we get

In other words, \( sx \in K(A) \) satisfies a monic polynomial in \( A[x] \) --- however, since \( A \) is integrally closed, this implies that \( sx \in A \) which itself implies that \( x \in S^{-1}A \).

Proof:

For the forward direction, suppose that \( X \) is a Noetherian normal scheme. By normality, we know that all stalks are integral domains and thus \(X\) is reduced. Furthermore, by Exercise 5.3.C we know that \(X\) has finitely man irreducible components — if any two of them meet at a point \( p\) then the stalk \( \OO_{X, p} \) is not an integral domain. Thus, the irreducible components must necessarily be disjoint, and since each component is irreducible + reduced, it is therefore integral.

Conversely, suppose that \(X\) is a finite disjoint union of integral Noetherian normal schemes. Since the finite union of Noetherian schemes is itself Noetherian, and each stalk \( \OO_{X,p} \) above some \( p \) lies inside an irreducible component that is normal by assumption, each stalk \( \OO_{X, p} \) is integrally closed.

Proof:

Since \( A \) is a domain, for every maximal ideal \( \mm \) the map \( A \to A_\mm \) is injective so that \( A \subseteq \cap_{\mm} A_\mm \). Conversely, fix some \( s \in K(A) \) such that \( s \in A_\mm \) for every maximal ideal \( \mm \). If we suppose to the contrary that \( s \notin A\), then the ideal of denominators \( I = \{ x \in A \mid xs \in A \} \) is clearly not the whole ring as \( 1 \notin I \). By Zorn's lemma, \( I \subseteq \mm \) for some maximal ideal \( \mm \) — if \( s = \frac s 1 \in A_\mm \) then there exists some \( a \in A \) and \( t, t^\prime \notin \mm \) with \( t^\prime (t s - a) = 0 \) or \( (tt^\prime) s = t^\prime a \in A \). Thus, \( (tt^\prime) \in I \subseteq \mm \) which is a contradiction.

Proof:

If we denote \( a = \frac w y = \frac x z \) and \( I_a \) the ideal of denominators, we clearly have that \( (y, z) \subset I_a \) since \( y a = w \in A \) and \(z a = x \in A \). Conversely, fix some \( r \in A \) such that \( r a \in A \). Then there exists some \( b \in A \) such that either \( \frac{rw}{y} = \frac{b}{1} \in A \) or \( \frac{rx}{z} = \frac{b}{1} \in A \). Since \( A \) is an integral domain, we do not need to scale by some factor in our multiplicative subset, and thus we either have \( rw = by \) or \( rx = bz \). In other words, \( rw \in (y) \) or \( rx \in (z) \). Since both are prime ideals, \( w \) does not divide \( y \), and \( x \) does not divide \( z \), it must be the case that \( r \in (x, y) \).

Proof:

Let \( A \) be a UFD and \( S \subset A \) be a multiplicative subset not containing \( 0 \) (so that \( S^{-1}A \) nontrivial ). Fix some \( \frac{a}{s} \in S^{-1}A \) with \( a \in A \) and \( s \in S \). Since \( A \) is a UFD, we may write \( a = u p_1^{r_1}\dots p_n^{r_n} \) for some unit \( u \in A \) and irreducibles \( p_i \in A \). Now some of the \( p_i \) may become units in \( S^{-1}A \):

Thus, we first wish to show that any \( p_i \) that does not become a unit remains irreducible in \( S^{-1}A \) (from there it remains to show such a decomposition is unique). To see this, suppose \( p_i \) does not become a unit under localization and \( \frac{p_i}{1} = \frac{q_1}{s_1} \frac{q_2}{s_2} \). Then as \( A \) is a domain and \( 0 \notin S \), this implies \( s_1s_2 p_i = q_1q_2 \) in \(A\) — however since \( A \) is a UFD, it must be the case that \( p_i \mid q_1 \) or \( p_i \mid q_2 \). Without loss of generality, if \( q_1 = t p_i \), then

so that \( \frac{q_2}{s_2} \) is a unit.

It remains to show that this decomposition is unique; if we also have \( \frac a s = \frac{u^\prime}{ t^\prime } \frac{ q^{l_1} }{ t_1^{l_1} }\dots \frac{ q^{l_n} }{ t_n^{l_n} } \), notice we may absorb the \( t_i \) into our unit \( \frac{u^\prime}{t^\prime} \) as they are invertible. Thus,

in \( A \) as a decomposition of irreducibles. Since \( A \) is a UFD, we ultimately get that the two decompositions are equal.

Proof:

Let \( A \) be a UFD, \( f = x^n + c_{n-1}x^{n-1} + \dots + c_0 \in A[x] \), and suppose \( \alpha = \frac{a}{b} \in \textrm{Frac}(A) \) is a root of \( f \) — we may assume without loss of generality that \( a, b \) do not share common irreducible factors. Similar to Exercise 5.4.A, if we multiply \( f \) by \( b^n \) we see that

Since \( b \) divides \( c_{n-1} b a^{n-1} + \dots c_n b^n \), it must divide \( a^n \) as well; since \( a \) and \( b \) do not share common factors, this implies that \( b \) is a unit; thus \( \frac{a}{b} = ab^{-1} \in A \).

Proof:

We know that \( \Z \) and \( k[x_1, \dots, x_n] \) are both UFDs (for the latter one may require Gauss' lemma), so by the preceding exercise it follows that \( \spec \Z \) and \( \A_k^n = \spec k[x_1, \dots, x_n] \) are factorial (and thus normal). To see that \( \P_k^n \) is also normal, we instead prove the stronger statement that it is factorial and use the preceding exercise. As each \( p \in \P^n_k \) must lie in some affine distinguished open set \( D_+(x_i) \cong \A_k^n \); thus the stalk, \( \OO_{\P^n_k, p} \) is in fact isomorphic to the stalks over \( A_k^n \) so that \( \P_k^n \)is normal.

Proof:

1. Since \(f\) has no repeated prime factors, \( z^2 - f \) is irreducible in \( A[x] \) and thus \( (z^2 - f) \) is prime (as \( A \) UFD \( \Rightarrow A[x] \) UFD \( \Rightarrow A[x] \) PID); therefore we have that \( B\) is indeed an integral domain. Following the hint, suppose \( F(T) \in B(T) \) is a monic polynomial with root \( \alpha \in K(B) \) — by the hint, we may assume without loss of generality that \( F(T) \in A[T] \) in fact. Since \( K(B) = K(A)[z] / (z^2 - f) \), we may write \( \alpha = g + hz \) for some \( g,h \in K(A) \); since \( \alpha \notin K(A) \) we know that the minimal polynomial of \( \alpha \) must be a quadratic. Since \( (\alpha - g)^2 = (hz)^2 = h^2f \), the minimal polynomial \( Q(T) \) of \( \alpha \) can be written \( Q(T) = T^2 -2gT + (g^2 - h^2f) \) so that \( F(T) = Q(T)P(T) \) in \( K(A)[T] \); by Gauss' lemma we may assume this is the factorization over \( A[T] \) so that \( 2g, g^2 - h^2f \in A \). Since \( 2 \) is invertible, \( g \in A \). If we write \( h = \frac{s}{t} \) for \( s, t \in A\), then \( s^2 f = t^2 h^2f\) — since \( t^2 \) does not divide \( s^2 \) it must be the case that \( t^2 \mid f \). Since \( f \) has no repeated prime factors by assumption, \( t^2 \) (and thus \(t\)) must be a unit so that \( h \in A \Rightarrow \alpha \in A \).

2. Take \( f = p^{2n} f^\prime \) with \( p \) irreducible not dividing \( f^\prime \). Then for any \( g, s \in A \), \( \alpha = g + \frac{s}{p^n} \in K(B) \) satisfies \( Q(T) = T^2 - 2gT + (g^2 - h^2 f^\prime) \) but \( \alpha \notin B \).

Proof:

1. Taking \( \alpha \) in the fraction field of \( \Z[x]/(x^2 - n) \) we have that \( \alpha = a + i\beta \) is a root of \( \chi(X) = X^2 - 2aX - (a^2 - nb^2) \). The discriminant of this polynomial is \( 4na^2 \) --- since \(n \) is square-free by hypothesis (assuming \( n = 0 \)). Replacing \( \chi(X) \) with \( \chi(X) \chi(\overline{X}) \), we may assume without loss of generality (by Gauss' lemma) that \( \chi(X) \in \Z[X] \) --- the statement is trivial for \( b = 0 \) since we may make \( \chi(X) \) a perfect square, so we must have \( b = p/q \) with \( q^2 \) dividing 4. If \( q = 1 \) then \( \alpha \in \Z[x] / (x^2 - n) \) so that our scheme is normal; otherwise, if \(q = 2\) then \(q / 2\) is a square mod-4 which is a contradiction.

2. By induction, it can easily be shown that \( x_1^2 + \dots + x^2_k \) has no repeated prime factors, so that by Exercise 5.4.H the scheme is normal.

3. We may write $$ wx - yz = \Big( ( \frac{w + z}{2} )^2 + ( \frac{y - x}{2} )^2 \Big) + \Big( (\frac{x + y}{2})^2 + ( \frac{w - z}{2} )^2\Big) $$ so that by part (b) the scheme is normal.

Proof:

1. This is a standard result from linear algebra that any quadratic form \(q(x_1, \dots, x_n)\) may be represented as a matrix operation

   $$ q(x_1, \dots, x_n) = x^TQx $$

   for some symmetric matrix \(T\). Since symmetric matrices are diagonalizable, we may consider the orthogonal matrix \(B\) whose columns consist of the eigenvectors of \(q\) such that \( B^T QB \) is a diagonal matrix whose entries are the eigenvalues of \(q\).

2. Since a change of basis matrix is invertible, the rank of \(Q\) (using the notation from above) is independent of basis.

Proof:

From Exercise 5.4.I(a) we know that since \( -5 \equiv 3 \mod 4 \) the ring \( \Z[\sqrt{-5}] = \Z[x] / (x^2 + 5) \) is integrally closed. However, \( 2 \times 3 = (1 + \sqrt{-5})(1 - \sqrt{-5}) \) shows that it is not a UFD.

Proof:

1. By Exercises 5.4.I(c) the scheme \( \spec A \) is normal and thus \(A\) is integrally closed.

2. To see that our monomials \( w, x, y \) and \(z\) are irreducible, suppose to the contrary that \( w = fg \). Since \( w \in A_1 \), we must have that \( \deg f + \deg g = 1 \) so that either \( f \) or \(g\) is a constant and thus unit. Since \( wz = xy \), \( A \) cannot be a UFD.

Proof:

As they are equivalent definitions, a field extension is an algebraic extension if and only if it is integral extension. Now since \( A \otimes_k l \) is integral, its fraction field \( K(A \otimes_k l) \) is (by the universal property) the smallest field containing \( A \otimes_k l \). However, we note that \( K(A) \otimes_k l \) must also be a field since it is a finite-dimensional \( K(A)\) vector space which is an integral domain (field extensions are faithfully flat) — by Exercise 3.2.G it must be a field. Clearly, it contains a(n isomorphic) copy of \( A \otimes_k l \) since we need only consider elements of the form \( \frac{a}{1} \otimes s_l \) so that \( K(A \otimes_k l) \subset K(A) \otimes_k l \). As suggested in the hint, we also have \( A \otimes_k l \subset K(A) \otimes_k l \subset K(A \otimes_k l) \) where the last map is the inclusion \( \frac{a}{b} \otimes s_l \mapsto \frac{a \otimes s_l}{b \otimes 1} \), giving us equality.

Again since field extensions are faithfully flat, the map \( A \to A \otimes_k l \) is an injection so that we get a commutative diagram of injections

Where the last injection follows from \(K(A) \otimes_k l = K(A \otimes_k l) \). To see that \( (A \otimes_k l) \cap K(A) = A \), notice that for the reverse direction that \( A \subset A \otimes_k l \) and \( A \subset K(A) \). Next, suppose that \(t = \sum a_i \otimes s_i \in (A \otimes_k l) \cap K(A) \). Since \( \sum_i a_i \otimes s_i \in K(A) \), we can multiply by some \( b \in A \) such that \( b \cdot ( \sum_i a_i \otimes s_i) = (\sum_i ba_i \otimes s_i) \in A\). In particular, we may write \( b\cdot t = ba^\prime \otimes 1_l \) so that \( t \in A \).

Lastly, suppose that \( P[T] \in A[T] \) is monic and has a root \( \alpha \in K(A) \). Then \( \alpha \otimes 1_l \) is a root in \( K(A \otimes_k l) \) of \( P[T] \) — however, since \( A \otimes_k l \) is integrally closed by assumption, this implies that \( \alpha \otimes 1 \in A \otimes_k l \cap K(A)\) so that \( \alpha \in A \) by the previous paragraph.

Proof:

By exercise 3.5.B we know that since \( 1 \in \big( \frac{x^2}{x^2 + y^2} \big) + \big( \frac{y^2}{x^2 + y^2} \big) \), \( D( \frac{x^2}{x^2 + y^2} ) \cup D( \frac{y^2}{x^2 + y^2} ) = A\). There is an obvious isomorphism \( \Q[t]_{t^2 + 1} \to A_{ \frac{x^2}{x^2 + y^2} } \) given by \( t \mapsto \frac{x}{y} \) that is well defined since

A similar case can be made for \(A_{ \frac{y^2}{x^2 + y^2} }\). By Exercise 5.4.E it follows that both \( A_{ \frac{x^2}{x^2 + y^2} } \) and \(A_{ \frac{y^2}{x^2 + y^2} }\) are UFDs, so that by remark 5.4.5 in Vakil, \( D( \frac{x^2}{x^2 + y^2} ) \) and \( D( \frac{y^2}{x^2 + y^2} ) \) are factorial. However, as \( A \) is not a UFD as shown by

\( \spec A \) cannot be factorial.

Proof:

Since any function \( f \in k[x, y] / (y^2, xy) \) can be expressed as \( f = ay + \sum_{i = 0}^n b_i x^i \), we consider three cases:

• \( a = b_i = 0 \) for all \( i \geq 0 \): in this instance, \(\textrm{Supp}\,f = \emptyset\)

• \( a \neq 0 \) but all \(b_i = 0\): In this instance the nilpotent element \( (x, y) \) is the only prime at which \(f\) does not vanish so that \( \textrm{Supp}\,f \) is the origin.

• Otherwise, \( f \) only vanishes at \( (0) \) or \((y)\), where the later is a nilpotent so that the support is the entire space.

Proof:

By exercises 5.2.F and 5.2.G \( \spec A\) is an integral scheme and thus irreducible — in other words, \( \spec A \) only has one irreducible component so that the associated point is the generic point \( [0] \).

Proof:

Since \( \spec A \) is Noetherian, it has finitely many irreducible components \( X_i \). If \( D(f_i) \) intersects \( X_i \), then this intersection is an open subset of \( X_i \) and must contain the generic point \( \{ x_i \} \). Thus \( \textrm{Supp} (f) \) is a closed subset containing the \( x_i \) and so contains all of \(X\) — hence it suffices to show that \( \textrm{Supp}\,f = \overline{ D(f) } \).

For forward inculsion, suppose that \( D(f) \subset \textrm{Supp}\,f = V( \textrm{Ann}(f) ) \). Then for \( [\pp] \in D(f) \) we have that \( f \in \pp \). Thus, for any \( g \in \textrm{Ann}(f) \), \( fg = 0 \in \pp \) so that \( g \in \pp \) which implies \( \textrm{Ann}(f) \subset \pp \Leftrightarrow [\pp] \in V(\textrm{Ann}(f)) \). Since the later is closed, \( \overline{ D(f) } \subset \textrm{Supp}\,f \). For the reverse direction, notice that since \(A\) is reduced \( z \in \textrm{Supp}\,f \) if and only if \( fA_z \neq 0 \). In order to show that \( z \in \overline{D(f)} \) it suffices to show that every neighborhood of \( z \) intersects \( D(f) \). Suppose to the contrary there is some neighborhood \( D(g) \) that does not intersect \( D(f) \) so that \( D(fg) = \emptyset \) which implies \( fg \) is nilpotent. However, as \(A\) is reduced, this implies that \( fg = 0 \) or in other words \( f = 0 \) in \(A_z\), a contradiction.

Proof:

As described in the hint, if we let \( S = \{ \pp \in \textrm{Supp}\,m \mid \pp \ \textrm{associated}\ \textrm{point} \} \) then clearly \( S \subset \textrm{Supp}\,m \) by definition — following the hint, since \( \textrm{Supp}\,m \) is closed, \( \overline{S} \subset \textrm{Supp}(m) \).

For the reverse direction, since \( A\) is Noetherian \( \textrm{Supp}(m) \) is a finite union of generic points of irreducible components (by property (B)) so that \( \textrm{Supp}(m) = \bigcup_{i=1}^r X_i \) where each \( X_i \) has generic point \( x_i \). But then by (A) these must also be associated points so that \( \textrm{Supp}(m) \subset \overline{S} \).

Proof:

Let \( \bigcup_i \{ \mathfrak{a}_i \} \) denote the locus of points where \( A_\pp \) has non-reduced stalk and let \( \bigcup_i \mathfrak{b}_i \) denote associated points whose stalks are non-reduced.

Suppose \( f \in A_\mathfrak{a} \) is nilpotent. Then on the generic point of its support \( \eta \), it must still be nilpotent. Conversely, if \( \pp \in \mathfrak{b} \) so that \( \widetilde{f} = \frac{f}{g} \in A_\mathfrak{b} \) for some \( g \notin \mathfrak{b} \), then by definition there exists some \(c \notin \mathfrak{b}\) such that \( acf \) is nilpotent.

Proof:

\( \pp \in \textrm{Supp}\,m \) if and only if \( rm \neq 0 \) for all \( r \notin \pp \) iff \( \pp \supset \textrm{Ann}(m) \) iff \( \pp \in V(\textrm{Ann}(m)) \). Now letting \(S\) be our multiplicatively closed subset, if \( \pp \in \spec S^{-1}A \) is an associated prime of some \( m \in M \), then for all \( r \notin \pp \) and \( s \in S \) we have that \( rs \notin \pp \) (as \( \pp \) is prime) so that \( rsm \neq 0 \). That is to say, for any \( s \in S \) we have \( \pp \in \textrm{Supp} \frac{m}{s}\) so that \( \pp \) is an associated prime of \( S^{-1}M \).

Conversely, let \( \pp \in \textrm{Supp}\,\frac{m}{s} \) is an associated point of \( S^{-1}M \). By taking \( s = 1 \), we can see that \( m_\pp \neq 0 \) as desired.

Proof:

By property (ii) we know that since \(X\) is locally Noetherian, each of the irreducible components of the support of some \( m \in \OO_X(U) \) is the union of closures of some subset of associated points. Thus, if we suppose that \( f \in \OO_X(U) \) has zero stalk on each \( \OO_{X, p} \) for \( p\) associated in \( U \) then it suffices to show \( f = 0 \). However, since the support of \(f\) must be a subset of the closure of the associated points the support must be empty so that \( f = 0 \).

Proof:

2. By Exercise 5.5.C, if we can find an embedded point then the scheme is non-reduced. As \( [y - x^2] \subset [(x - 1, y - 1))] \) the scheme is indeed non-reduced.

3. Yes - \( x + y - 2 \in [ (y - x^2) ] \subset [(x - 1, y - 1)] \) and clearly \( y - x^2 \in [y - x^2] \) so they are both zero divisors by (C). However, \( x + y - 3 \) is not a zero divisor as it is not contained in any associated points.

Proof:

Following the hint, let \( \overline{g} \in A\) be a zero divisor, and suppose \( g \in k[x_1, \dots, x_n] \) is such that \( \pi(g) = \overline{g} \) where \( \pi : A \to A/(f) \) is the natural projection. In particular, if \( \pp = \textrm{ass}(g) \) is the associated prime ideal, then \( \qq = \pi^{-1}(\pp) = \{ h \in k[x_1, \dots, x_n] \mid gh \in (f) \} \). Without loss of generality, we may assume \( g \) is irreducible and thus \(\overline{g}\) is also --- since \( \overline{g} \) is a zero-divisor, there exist \( e, h \) such that \( gh = ef \). We wish to show that \( (g) \) is minimal among primes contained in \( (f) \), as this would imply there cannot be an embedded point contained in an associated point. We also may assume without loss of generality that \( e, h \) are not divisible by \(f\), as \(g\) is irreducible and a zero-divisor.

Suppose to the contrary that there exists some \((f) \subset \pp \subset (g)\). Then for any irreducible \( y \in \pp \) we must have \( y \in \gg \) so that \( y = gr \) for some \(r \in k[x_1, \dots, x_n]\). Since \( y \) irreducible, we have that \(r\) is a unit i.e. \(r \in k\) so that \( (y) = (g) \) and thus \( \pp = (g) \). Therefore, \( (\overline{g}) \) must be generic.

Proof:

Suppose that \( xy \in I \) with \(y \notin I\) and \( A \supset I = \textrm{Ann}(m) \) for some \( m \in M \). Then by definition there exists some \( s \in A \) with \( sxy = 0 \) in \(M\). Since \( y \notin \textrm{Ann}(m) \), we have that \( sy \neq 0 \) so that \( I \subset (x, I) \subset \textrm{Ann}\,M\), which contradicts the fact that \( I\) is maximal with respect to annihilators. Thus, \( y \in I\) so that \( I \) is prime and thus an associated prime.

Proof:

The forward direction is obvious; thus, we wish to show \( m = 0 \) in each \(M_\pp\) implies that \(m = 0\). By contrapositive, if we assume that \( m \neq 0 \) then by Zorn's lemma we can find an ideal \( \pp \) maximal among ideals containing \( \textrm{Ann}(m) \), and by Exercise 5.5.J we know that \(\pp\) is prime and thus associated. By definition of \( M_\pp \), for any \( x \notin \pp \) we have \( x \notin \textrm{Ann}(m) \) so \( xm \neq 0 \) and thus \( \frac{m}{1} \neq 0 \) in \( M_\pp \).

Proof:

Let \( f : M^\prime \to M \) denote the injection; if \( \pp^\prime \) is an associated prime of \( M^\prime \) then by definition there exists some nonzero \( m^\prime \in M \) such that \( \pp^\prime = \{ a \in A \mid am^\prime = 0 \} \). Since \( f \) is injective, \( f(m^\prime) \neq 0 \) and \( f(\pp) = \{ a \in A \mid af(m^\prime) = f(am^\prime) = 0 \} \) so that \( f(\pp) \) is an associated prime.

Next, suppose that \(g : M \to M^{\prime\prime}\) denotes our surjection and let \( \pp \in \textrm{Ass}\, M \) so that there exists some \( m \neq 0 \) with \( \pp = \{ a \in A \mid am = 0 \} \). We consider two cases: if \( A.m \cap \textrm{Im}(f) \neq 0 \), then we can find some nonzero \( x \in \textrm{Im}(f) \cap A.m \) so that \( f(m^\prime) = x = am \) for some \( m^\prime \in M^\prime \) and \(a \in A\). Similar to the paragraph above, since \( f \) injective we have that \( \pp \) must also be the annihilator for \( m^\prime \) so that \( \pp \in \textrm{Ass}\,M^\prime \). However, if \( A.m \cap \textrm{Im}(f) = 0 \), then for any \(a \in \pp \) we have that \( am = 0 \Longrightarrow ag(m) = g(am) = 0 \) so that \( a \in \textrm{Ann}\,(g(m)) \). Conversely, if \( a \in g(m) \) then \( am \in \ker g = \textrm{Im}f \) so we would have \( am = 0 \). Thus, \( \pp = \textrm{Ann}(m) = \textrm{Ann}(g(m)) \).

Proof:

1. Since \( M \neq 0 \) we know that \( \textrm{Ass}(M) \neq \emptyset \) (by Exercise 5.5.J) so we can fix some \( \pp_1 \in \textrm{Ass}(M) \). By the equivalent definition of (D) (see 5.5.9), \(M\) has a submodule \( M_1 \) isomorphic to \( A / \pp_1 \). Since \( M_0 = (0) \) we have that \( M_1 / M_0 \cong A / \pp_i \) as desired.

   If \( M / M_1 = 0 \) we are done --- otherwise we may aaply the same argument to \( M / M_1 \) and find some \( \pp_2 \in \textrm{Ass}(M / M_1) \) so that \( M / M_1 \) has a submodule \( \overline{M}_2 \) isomorphic to \( A / \pp_2 \). By letting \( \pi : M \to M / M_1 \) denote the natural projection, we define \( M_2 := \pi^{-1}( \overline{M}_2 ) \) so that \( M_2 / M_1 \cong A / \pp_2\) as desired. As \(A\) is Noetherian and \(M\) is finitely generated, \(M\) is Noetherian by Exercise 3.6.X and thus this process ends after finiely many steps.

2. For each \(0 \leq i \leq n\) (using the enumeration of submodules from the problem statement), consider the short exact sequence \( 0 \to M_i \to M_{i+1} \to A / \pp_i \to 0 \). By the previous exercise, we know that \( \textrm{Ass}(M_{i+1}) \subset \textrm{Ass}(M_i) \cup \textrm{Ass}(A / \pp_i) \) and thus \( \textrm{Ass}(M) \subset \bigcup_{i=1}^n \textrm{Ass}(A / \pp_i) \).

3. Following the hint, suppose that some \( \pp_i\) does not contain any associated prime. Let \( m \in M \) be arbitrary — then as \( A \) is Noetherian, there is an associated prime \( \qq \) maximal among annihilators of of elements of \(M\) that contain \( \textrm{Ann}(m) \). Since \( \pp_i \) does not contain \( \qq \) by assumption, we can pick some \( r \in \qq / \pp_i \) so that \( rm = 0 \). As \( m \in M \) was arbitrary, this implies that \( M_{\pp_i} = 0 \). Letting \( M_{i+1} \) and \( M_i \) be the submodules such that \( M_{i+1} / M_i \cong A / \pp_i \) we have that $$ 0 = (M_{i+1})_{\pp_i} / (M_i)_{\pp_i} = (M_{i+1} / M_i)_{\pp_i} \cong (A / \pp_i)_{\pp_i} \neq 0 $$ a contradiction.

Proof:

1. Following the hint (as well as Eisenbud's proof of Theorem 3.1), \( \pp \in \textrm{Ass}_A\,M \cap \spec S^{-1} A \). Then there exists an injection \( R / \pp \hookrightarrow M\) --- as localization is exact, this gives us an injection \( S^{-1} R / \pp S^{-1}R \hookrightarrow S^{-1} M \). Since \( \pp \in \spec S^{-1} A \), this implies that \( \pp S^{-1} R \) is a prime ideal such that \( S^{-1} M \) has a submodule isomorphic to \( S^{-1} R / \pp S^{-1} R \) --- that is to say \( \pp S^{-1} R \in \textrm{Ass}\, S^{-1} M \)

2. Following the hint, suppose \( \qq \in \textrm{Ass}_{S^{-1}A} S^{-1} M \) so that by Exercise 3.2.K we can find some \( \pp \in \spec A \) with \( \pp \cap S = \emptyset \) such that \( \qq = \pp S^{-1}A \). Since \( \qq \) is an associated prime, we again get an injection \( \phi : S^{-1} A / \pp S^{-1} A \to S^{-1} M \) — since \( A \) is Noetherian, \( \pp \) is finitely presented (and thus finitely generated) so that by Exercise 1.6.G \( \phi \) corresponds to some \( \widetilde{\phi} \in S^{-1} \Hom_A( A / \pp, M) \) which we will write as \( \widetilde{\phi} = \psi / s \) for \( \psi \in \Hom_A (A / \pp, M) \). Since \( s \) is a nonzerodivisor on \( R / \pp \) (as \( \pp \cap S = \emptyset\) ) we have that \( \psi \) is still an injection as desired.

Proof:

Handling the reverse direction first, let \( \{ \pp_1, \dots, \pp_n \} \) be a finite set of irreducible points (by property (B) of course) and for each \(i\) let \( \pp_i = \textrm{Ann}(m_i) \) for some \( m_i \in M \). Since embedded primes do not contribute to \( \bigcup_i \pp_i \), we may assume without loss of generality that each \(\pp_i \) is minimal.

We claim that the support of \( m = \sum m_i \) is precisely \( \overline{ \bigcup_i \pp_i } \). Notice that by definition of \( \pp_i = \textrm{Ann}(m_i) \) we have that \( m_i \neq 0 \) in \( M_{\pp_i}\). Additionally, by minimality of each \( \pp_j \) we can find some \( r_j \in \pp_j \backslash \pp_i \) so that \( m_j = 0 \) in \( M_{\pp_i} \). Thus \(m = m_i \neq 0 \) in \( M_{\pp_i} \). Now for any \( \qq \in \overline{\{\pp_i\}} \) by Exercise 3.6.L we have that \( \pp_i \subset \qq \) so that \( m \neq 0 \) in \( M_\qq \) --- therefore \( \bigcup_{i} \overline{\pp_i} \subset \textrm{Supp}(m) \).

Conversely, if \( \qq \notin \bigcup_i \overline{ \pp_i } \) then we can find \( r_i \in \pp_i \backslash \qq \) for all \( i \) and thus \( m = 0 \) in \( M_\qq \). Thus \( \textrm{Supp}(m) = \bigcup_{i} \overline{\pp_i} \).

To show that the support of any \( m \in M \) is the closure of a subset of associated points, we refer to part (a) of the next exercise.

Proof:

1. This is immediate from the fact that if \( \pp \) is an associated prime, then \( \pp = \textrm{Ann}(m) \) for some \(m \in M \), so affine-locally we have $$ \textrm{Supp}(m) = V(\textrm{Ann}(m)) = \overline{ \{\pp\} } $$ By looking at an affine open subset \( U = \spec R \) of some irreducible component \( X_i \), \( \overline{ \{ \pp \} } = U \) implies \( \overline{ \{\pp\} } = X_i \).

2. This was shown in the previous exercise.

Proof:

Let \( X \) be an integral scheme and take \( M = \OO_X\). By Exercise 5.2.F an integral scheme is both irreducible and reduced. As \(X\) is irreducible, it has a unique generic point \( \eta \) so our associated point must simply be the generic point of \(X\). In other words, \( \textrm{Ass}(X) = \textrm{Ass}_{\OO_X}(\OO_X) = \{ \eta \} \) — (B) follows immediatedly. Property (C) must then also hold since the only zerodivisor of \( \OO_X \) for \( X \) reduced is \( f = 0 \).

Proof:

The associated primes here are simply \( \pp_1 = (y - x^2) \), \( \pp_2 = (x - 1, y - 1) \) and \( (x - 2, y - 2) \) so property (B) is clearly satisfied.