Solutions to The Rising Sea (Vakil)

Proof:

Let \( \phi : \mm / \mm^2 \to \R\) denote our \( \R \)-linear map. Let \( \OO_p \to \R \) be the composition of \( \OO_p \to \mm \) ( given by \( f \maps to f - f(p) \) ) followed by the natural quotient, followed by \( \phi \). To show this satisfies the Leibniz rule, let \( f, g \in \OO_p \). Then under the first map we get \( fg \mapsto fg - f(p)g(p) \). Since \( (f - f(p))(g - g(p)) \in \mm^2 \), we obtain \( fg + f(p)g(p) \sim f(p)g + g(p)f \) in the quotient. Flipping \( f - f(p) \) to \( f(p) - f \) under the quotient (as both evaluate to 0)

so that by \( \R \)-linearity of \( \phi \) we have that the induced map \( \widetilde{\phi} : \OO_p \to \mm / \mm^2 \to \R \) satisfies

Proof:

Since \( \mm / \mm^2 \cong \mm A_\mm / (\mm A_\mm)^2 \), we may assume without loss of generality that \( (A, \mm) \) is a local ring and thus \( A/(f) \) is a local ring as well. Thus, the Zariski tangent space is of the form

Now we obtain an obvious surjection \( \mm / \mm^2 \to \mm / (\mm^2 + (f)) \) given by \( a \mapsto a + (f) \); passing to the dual gives an injection, such that the Zariski tangent space of \( A / (f) \) is isomorphic to the image of the the Zariski tangent space of \( A \) cut out by \( f \mod \mm^2 \).

Proof:

1. Recall that since \( \kappa(p) = \OO_{X,p} / \mm_p\) we have that \( T_{X, p} := \mm_p / \mm_p^2 \cong \mm_p \otimes_{\OO_{X,p}} \kappa(p) \) is a \( \kappa(p) \)-vector space in a natural way. Since \( Z \) is a closed subscheme, we may find an affine open cover \( U_i = \spec B_i \) of \( X \) such that \( Z \cap U_i \) is of the form \( \spec B_i / J_i \) for some ideal \( J_i \). Then at every point \( p = [\pp_i] \) of \(Z\) we are able to identify \( \OO_{Z, p} = ( B_i / J_i)_\pp \cong (B_i)_{\pp_i} / J_i (B_i)_{\pp_i} \). By applying the previous exercise, we may consider \( T_{Z,p} \) as the sub-\( \kappa(p) \) vector space cut out by \( J_i + \mm_p^2 \)

2. Part (a) (non-explicitly) shows that it suffices to check such a computation affine-locally. Thus, let \( Y = \spec A / I \) and \( Z = \spec A / J \). By the previous exercise, since \( Y \cap Z = \spec A / (I + J) \subset Y, Z\), we get inclusions of vector spaces

   $$ \begin{CD} T_{Y \cap Z, p} @>>> T_{Y, p} \\ @VVV @VVV \\ T_{Z, p} @>>> T_{X, p} \end{CD} $$

   It is easy to see that this gives a commuting diagram, but we also wish to show that it satisfies the universal property. Switching over to the dual, we have a diagram

   $$ \begin{CD} \mm / (I + J + \mm^2) @<<< \mm / (J + \mm^2) \\ @AAA @AA{\pi_J}A \\ \mm / (I + \mm^2) @<<{\pi_I}< \mm / \mm^2 \end{CD} $$

   If we let \( W \) be another \( \kappa(p) \)-vector space with \( \alpha : \mm / (I + \mm^2) \to W\) and \( \beta : \mm / (J + \mm^2) \to W \) morphisms such that \( \alpha \circ \pi_I = \beta \circ \pi_J \), then for any \( f \in I \) we have \( \alpha(\pi_I(f)) = \alpha(0) = 0 = \beta(\pi_J(f)) \). Similarly any \( f \in J \) maps to 0, so \( I + J \subset \ker ( \mm / \mm^2 \to W ) \). By the universal property of the quotient we obtain a unique map \( \mm / ( I + J + \mm^2) \to W \). Dualizing, this gives us a unique morphism \( W^\vee \to T_{Y \cap Z, p} \) so that \( T_{Y \cap Z, p} \cong T_{Y, p} \times_{ T_{X, p} } T_{Z, p} \cong T_{Y, p} \cap T_{Z, p}\) by Exercise 9.2.C

3. Recall that the union of \( \spec A / I \) and \( \spec A / J\) may be given by \( \spec A / ( I \cap J) \). Since we have natural surjections \( \mm / (I \cap J) + \mm^2 \to \mm / I + \mm^2 \) and \( \mm / (I \cap J) + \mm^2 \to \mm / J + \mm^2 \), we obtain a surjection \( \mm / (I \cap J) + \mm^2 \to \mm / I + \mm^2 \oplus \mm / J + \mm^2 \). Dualizing, this gives us an injection \( T_{Y, p} \oplus T_{Z, p} \hookrightarrow T_{Y \cup Z, p} \).

4. Following the hint, consider the union of the \( x \)-axis \( \spec k[x, y] / (y) \) and the parabola \( \spec k[x, y] / (y - x^2) \). Individually, the cotangent spaces of both of these affine schemes at the origin are equal to \( (x) / (x^2) \) (e.g. \( (x, y) / (y - x^2) \cong (x) \) and \( (x^2, xy, y^2) / (y - x^2) \cong (x^2, x^3) \) ) which is one dimensional. However, the intersection of the two is given by \( \spec k[x, y] / (y(y - x^2)) \). It is easy to see that since \(y\) does not vanish in this case we obtain a 2-dimensional tangent space.

Proof:

Following the solution of 12.1.4, we know that \( \spec A = \spec k[w, x, y, z] / (wz - xy) \) has dimension 3 by Krull's Principal Ideal Theorem since \( (wz - xy) \) is irreducible and not a zerodivisor in \( k[x, y, z, w] \) (and thus has codimension 1 which is equivalent to \( \dim k[w, x ,y ,z] - 1 \) for varieties in \( \A^4 \)). Now \( A / (x, z) \cong k[w, y] \) clearly has dimension 2, which tells us that \( (x, y) \) is codimension 1 as codimension is the difference of dimension for varieties.

To show it is not principle, we again let \( \mm = (w, x, y, z) \) be the maximal prime corresponding to the origin, and compute

(notice one generator gets dropped) so that the Zariski tangent space \( \mm / \mm^2 \) has dimension 4. However, \( \mm / (x, z) = (w, y) \) and \( \mm^2 / (x, z) = (w^2, wy, y^2) \) so that the Zariski tangent space is dimension 2. Since the Zariski tangent space of \( V(f) \subset \spec A \) of a prime ideal \( (f) \) must be at least 3 dimensional by Exercise 12.1.B, we conclude that \( (x, z) \) cannot be principle.

Proof:

This basically follows from the hint, but recall that the primes ideals in \( A_\mm \) (where \( \mm = (w, x, y, z) \) is the maximal ideal corresponding to the origin) correspond to the prime ideals contained in \( \( x, y, z, w \) \). As \( (x, z) \subset (w, x, y, z) \) from the previous exercise is a prime ideal of codimension 1 that is not principal, so that \( A_\mm \) cannot be a UFD by Lemma 11.1.6.

Proof:

1. By a proof similar to Lemma 4.2.3 in Qing Liu's Algebraic Geometry & Arithmetic Curves, by taking \( \mm = (x, y, z) \) and \( I = (xy, xz, yz) \) we have that \( (I / \mm I)_\mm \cong I / \mm I \) (more generally this will hold for finitely-generated modules). It suffices to show that \( I / \mm I \) is a 3-dimensional vector space over \( k \) (as \(k[x, y, z] \) is a graded vector space, this will tell us the minimal number of homogeneous generators). However, it is easy to see that \( xy, xz, yz \) are \( k \)-linearly independent over \( k \): if \( axy + bxz + cyz \in \mm I \) for \( a, b, c \in k\), then as \( \mm I \) is generated by degree 3 elements and each of \( axy, bxz, cyz \) are degree \( 2 \) we must simply have \( a = b = c = 0 \).

2. Letting \( \iota : \spec k[x, y, z] / I \hookrightarrow \spec k[x, y, z]\) denote our closed embedding of the coordinate axes, recall that this is a regular embedding if at all \( p = [\pp] \) we have the kernel of \( k[x,y,z]_\pp \twoheadrightarrow (k[x,y,z]/I)_\pp \) (i.e. the ideal of \( V(I) \)) is generated by a regular sequence (of length \( \operatorname{height}(I) \)). Now since \( V(I) \) is codimension 2, this would tell us that \( \iota \) is a regular embedding if there exist \( f, g \in k[x,y,z]_\pp \) such that \( I_\pp = (f, g) \) and \( (f, g) \) is a regular sequence. By taking \( \pp = (x, y, z) \), part (a) already tells us that the first requirement alone cannot hold.

Proof:

It is a standard and easy fact to show that the derivative at a point is a \( k \)-linear map; since translation itself is an invertible linear map, we may assume without loss of generality that our closed \( k \)-valued point \( p \) is simply the origin \( \mm = (x_1, \dots, x_n) \). Letting \( I = (f_1, \dots, f_r) \), it then follows (using the notation in the preceding paragraph in the text) that \( \mathfrak{n} = \mm / I \). Let \( dx_i \) denote the equivalence class \( x_i + \mm^2 \) so that\( \mm / \mm^2 \) is the \( k \)-vector space generated by the \( dx_i \). Similar to Exercise 12.1.B we know that \( \mathfrak{n} / \mathfrak{n}^2 = \mm / ( I + \mm^2 ) \). Now for any \( g \in \mm \), we may write

for some \( h \in \mm^2 \). Using the formal derivative defined above, it is easy to see that the \( a_i = \frac{\partial g}{\partial x_i}(p) \). As the image of \( g \) in \( \mm / \mm^2 \) is simply \( \sum_{i=1}^n a_i \, dx_i \) and the kernel of the map \( \mm / \mm^2 \to \mathfrak{n} / \mathfrak{n}^2 \) is generated by the images of the \( f_i \) in \( \mm / \mm^2 \), we obtain that the Zariski cotangent space is locally generated by the image of the \( \sum_{i=1}^n \frac{ \partial f_i }{ \partial x_j } (p) dx_j \) in \( \mathfrak{n} / \mathfrak{n}^2 \) proving the claim.

Proof:

1. This simply follows by virtue of our "formal partial derivatives" described in the problem statement of the previous exercise as \(A\)-linear maps. Indeed, since \( g \in I \) we have that there exist some \( a_1, \dots, a_r \in A \) such that

   $$ g = a_1 f_1 + \dots + a_n f_n $$

   Then for any \( 1 \leq i \leq n \) we obtain

   $$ \frac{\partial g}{\partial x_i} = a_1 \frac{ \partial f_1 }{\partial x_i } + \dots + a_n \frac{ \partial f_n }{\partial x_n} $$

   But then by writing

   $$ J^\prime = \begin{pmatrix} \frac{\partial f_1}{\partial x_1}(p) & \dots & \frac{\partial f_r}{\partial x_1}(p) & \frac{ \partial g }{\partial x_1} \\ \vdots & \ddots & \vdots \\ \\frac{\partial f_1}{\partial x_n}(p) & \dots & \frac{\partial f_r}{\partial x_n}(p) & \frac{\partial g }{\partial x_n} \end{pmatrix} $$

   we see that the last column is always a linear combination of the first \( n \) columns so that the rank is unchanged. In fact, if we know that some \( a_j \neq 0 \) is a unit, then we may always rewrite \( f_j \) in terms of \( f_1, \dots, f_{j-1}, g, f_{j+1}, \dots, f_r \) so that by substituting we obtain the same corank. Therefore, the corank does not depend on the choice of generators.

2. A simple computation shows that the resulting Jacobian \( J^\prime \) obtained after appending \( g \) is

   $$ J^\prime = \begin{pmatrix} \frac{\partial f_1}{\partial x_1}(p) & \dots & \frac{\partial f_r}{\partial x_1}(p) & \frac{ -\partial q }{\partial x_1} \\ \vdots & \ddots & \vdots \\ \\frac{\partial f_1}{\partial x_n}(p) & \dots & \frac{\partial f_r}{\partial x_n}(p) & \frac{-\partial q }{\partial x_n} \\ 0 & \dots & 0 & 1 \end{pmatrix} $$

   Since the matrix rank equals the column rank, adding one column can only increase the rank by at most one (and precisely one when the new column is independent of the others). But it is obvious that the last column must be independent from the first \( r \) columns since the last entry is \( 1 \) and all other columns have \( 0\) as their last entry giving us \( \textrm{rank}(J^\prime) = \textrm{rank}(J) + 1\). Since \( J^\prime \) is now a map to \( k^{n+1} \) we obtain \( \textrm{corank}(J^\prime) = (n+1) - \textrm{rank}(J) = \textrm{corank}(J) \)

Proof:

As the tangent space is local by construction, suppose that \( \spec A \hookrightarrow X \) is an affine neighborhood of \( p\). As any morphism \( \spec k[\epsilon] / (\epsilon^2) \to \spec A \) clearly induces a morphism \( \spec k[\epsilon] / (\epsilon^2) \to X \), we may assume without loss of generality that \( X = \spec A \) with \( A = k[x_1, \dots, x_n] / I \) for \( I \) prime.

For the forward direction, suppose \( p \in \spec A \) is a closed point with residue field \( k \) so that there is an isomorphism \( \OO_{X, p}/ \mm_p \cong k \) — thus, every element of \( \OO_{X, p} \) may be uniquely written as \( z + a \) for \( z \in \mm_{p} \) and \( \alpha \in k \). Then for any linear map \( f : \mm_p / \mm_p^2 \to k \), we uniquely define \( \phi_f : \OO_{X, p} \to k[\epsilon] / (\epsilon^2) \) via

giving us in turn a morphism \( \spec k[\epsilon] / (\epsilon^2) \to \spec \OO_{X, p} \). From Exercise 6.3.J we know there is a canonical morphism \( \spec \OO_{X, p} \to \spec A \), so by composing we get a unique morphism \( \spec k[\epsilon] / (\epsilon^2) \to \spec A \).

Conversely, let \( f : \spec k[\epsilon] / (\epsilon^2) \to \spec A \) be given. Setting \( p = f((\epsilon)) \), the map \( A \to k[\epsilon] / (\epsilon^2) \) induces a map of residue fields \( \kappa(p) \to k \) which is necessarily an isomorphism (as maps of fields are always injective). By definition, the stalk map \( f^\sharp_p \) must factor through the quotient giving a map \( \mm_p / \mm_p^2 \to k \) as an element of \( T_{X,p} \).

Proof:

Notice that in both cases, our point corresponds to ( [(2, x)] \) when considered as a prime ideal of the polynomial ring \( \Z[x] \). This is indeed a maximal ideal since \( \Z[x] / (2, x) \cong \mathbb{F}_2 \) is a field, so we have \( \mm / \mm^2 = (2, x) / (4, 2x, x^2) \). In the first ring where \( x^2 = -4 \), we simply have \( (2, x) / (4, 2x) \) is a 2-dimensional \( \mathbb{F}_2 \)-vector space given by \( \{ 0, 2, x, x+2 \} \). In the latter ring where \( x^2 = -2 \), we obtain \( (2, x) / (4, 2x, -2) = (2, x) / (2, 2x) \cong \mathbb{F}_2 \) so that the tangent space is now 1-dimensional.

Proof:

The forward direction is obvious, as if \( k \) is a field then its unique maximal prime is \( \mm = (0) \) and thus \( \mm^2 = \mm \) implies the cotangent space is trivial, i.e. \( \dim \mm / \mm^2 = 0\).

Conversely, if \( (A, \mm) \) is dimension \( 0 \) then every prime ideal is maximal. However since \( (A, \mm) \) is a local ring it has a unique maximal ideal which is necessarily \( (0) \).

Proof:

Since the Zariski tangent space is a local construction, we may reduce to some affine open neighborhood \( \spec A \) of \( p \). As \( p \) is a closed point, it corresponds to some maximal ideal \( \mm \) so that \( \OO_{X, p} = A_\mm \) and \( T_{X, p} = (\mm / \mm^2)^\vee \). By assumption, we have that \( \dim A_\mm = n = \dim \mm / \mm^2 \). By Exercise 12.1.B, we know that the tangent space of \( A_\mm / (f) \) is \( \mm / (f + \mm^2) \). From this, it is immediate that if \( f \in \mm^2 \) then \( \mm / (f + \mm^2) = \mm / \mm^2 \) so the local ring \( A_m / (f) \) is no longer regular as \( \dim A_m / (f) = n - 1 < \dim \mm / \mm^2 \). Conversely, Exercise 12.1.B tells us that the dimension of the tangent space \( \mm / (f + \mm^2) \) is one less.

Proof:

Since the Zariski tangent space is a local construction, we may shrink to a sufficiently small neighborhood \( \spec A \) of \( p \) such that \( D = V(f) \) for some non-zerodivisor \( f \in A \). If we let \( \mm \) be the maximal ideal such that \( p = [\mm] \) then by the previous exercise we have \( f \in \mm \backslash \mm^2 \). As \( p \) is a regular point of \(D\), \( \dim A / (f) = \dim A - 1 \) and \( \dim \mm / ( f + \mm^2 ) = \dim \mm / \mm^2 - 1 \) by Exercise 12.1.B, the claim follows.

Proof:

By Exercise 12.1.G we have that the Zariski cotangent space at \( p \) is given by the cokernel of the Jacobian matrix

Thus, the dimension of the Zariski cotangent space is the dimension of the cokernel which is simply the corank. Since \(X\) is regular at \( p \) iff the Zariski cotangent space is dimension \( d \), the claim follows.

Proof:

By the previous exercise, we simply evaluate the corank of the Jacobian matrix at any closed point \( p \in X \) — in this case, it is the \( n \times 1 \) matrix

When this is not the zero vector, it spans a 1 dimensional subspace so that the corank is \( \dim X - 1 = \dim V(f) \); that is to say the point \( p\) is regular. Thus, the only time the corank of \( J_p \) is strictly larger than \( \dim V(f) = \dim X - 1 \) is when the rank is \( 0 \) which holds iff all entries are 0.

Proof:

1. To see that \( \A^n_k \) is rather trivial: it may be interpreted as \(\spec k[x_1, \dots, x_n] / (1)\) where all partial derivatives of the constant 1 vanish. Thus, the Jacobian matrix is either a \(n \times 0 \) matrix which isn't well-defined, or it is the \( n \times 1 \) zero vector. In the latter case, this clearly has rank 0 and thus corank \( n = \dim \A^n_k \).

   For the second example, we consider the standard affine cover \( D(x), D(y), D(z) \). On \( D(x) \), our homogeneous polynomial is of the form \( y^2z - 1 + z^2 = 0 \), which has Jacobian

   $$ J_x = \begin{pmatrix} 2yx \\ y^2 + 2z \end{pmatrix} $$

   which is nonzero except at \( y = z = 0 \) which does not lie on the curve. However, when \( \textrm{char}\, k = 2 \) the Jacobian vanishes at points \( [1 : 0 : z] \) with \( z^2 = 1 \) (such that the curve is singular at those points).

   On the affine open set \( D(y) \), our curve is of the form \( z - x^3 - xz^2 = 0 \). In this case, the Jacobian is of the form

   $$ J_y = \begin{pmatrix} -3x^2 - z^2 \\ 1 - 2xz \end{pmatrix} $$

   which remains smooth even in characteristic 3 (since we must have \( z = 0 \) which would make the bottom component nonzero). Lastly, on the affine open subset \( D(z) \) we have the curve \( y^2 - x^3 - x = 0 \) (notice this is almost the nodal cubic which would be singular even in characteristic 0). In this case, we get

   $$ J_z = \begin{pmatrix} -3x^2 - 1 \\ 2y \end{pmatrix} $$

   which always has rank 1 (in \( \textrm{char}\,k \neq 2 \)) on the curve, since \( y = 0 \) implies either \( x = 0 \) or \( x^2 = -1 \) (plugging either into the top entry of the Jacobian gives something nonzero). However, in \( \textrm{char}\,k = 2 \) we have that any point \( [x, y, 1]\) with \( x^2 = 1 \) will be singular.

2. If

   $$ \frac{\partial f}{\partial x_1} = \dots = \frac{\partial f}{\partial x_n} = 0 $$

   has no solution in \( \overline{k} \) then it certainly doesnt obtain any solution in \( k \) — in particular the rank doesn't drop to 0 so that \( V(f) \) remains smooth.

Proof:

It suffices to check this affine locally since smoothness is checked on an affine cover. Thus, let \( I = (f_1, \dots, f_r)\), \( B = k[x_1, \dots, x_n] / I \) and \( X = \spec B\) where the Jacobian matrix \( J \) has corank \(d\) at all points \( p \in X \). Since field extension are faithfully flat, \( \otimes_k \ell \) is an exact functor — applying it to the short exact sequence

yields the exact sequence

which proves what we already know from the affince construction of the fibred product \( X \times_{\spec k} \spec \ell = \spec B \otimes_k \ell \). More importantly, applying \( \otimes_k \ell \) to the exact sequence

gives the exact sequence

Since dimension behaves well under base by field extensions by Exercise 11.2.J we have that \( \dim \mm_\ell / \mm_\ell^2 = X_\ell \).

Proof:

Let \( I \) denote the ideal generated by the \( (n - d) \)-minors of the Jacobian matrix \( J\). Since the Jacobian has rank \( n - d\) at every closed point (i.e. maximal ideal), \( I \) is contained in the intersection of all maximal ideals which is the Jacobson radical — a helpful commutative algebra fact tells us that this is the nilradical for polynomial rings in particular so that \( V(I) = \emptyset \). Thus, at every point the Jacobian matrix must have rank at least \( n - d \), i.e. the corank is \( \leq d \).

For the reverse direction, suppose there is some (non-closed) point \( [\pp] \in X \) with Jacobian \( J_\pp \) such that there is a \( (n- d + 1) \)-minor that does not vanish. Since a polynomial not vanishing is an open condition and this open set is non-empty, the denseness of closed points implies that there must also be some closed point in this set — a contradiction.

Proof:

By the previous exercise, \(X\) is smooth if and only if it is smooth at closed points, i.e. iff the Jacobian is corank \(d\) at all closed points. By Exercise 12.2.D, this is equivalent to regularity at closed points.

(N.B. we are implicitly using the fact that regularity does not depend on choice of affine cover since it is a stalk-local condition).

Proof:

If \( p \in Z_1 \cap Z_2 \) for closed irreducible components \( Z_i \) corresponding affince locally (say on \( \spec A \)) to minimal ideals \( \pp_1, \pp_2 \subset A \), then we could choose some \( f_1 \in \pp_1 \backslash \pp_2 \) and \( f_2 \in \pp_2 \backslash \pp_1 \) such that \( f_1f_2 = 0 \) in \( \OO_{X, p} \) contradicting the theorem above that regular rings are domains.

Proof:

For the forward direction, if Exercise 5.3.C tells us that a Noetherian scheme \(X\) that is connected with all stalks integral domains is itself and integral domain (and thus irreducible).

Conversely, Exercise 3.6.D tells us irreducible implies connected.

Proof:

1. As regular rings are Noetherian by definition, \( I \) must be finitely generated. Therefore, by an argument similar to Exercise 12.1.B, we may assume the Zariski cotangent space of \( A / I \) is cut out by \( \mm / (I + \mm^2) \). Therefore, the natural surjection \( A \to A / I \) induces a surjective morphism of \( k \) vector spaces

   $$ \mm / \mm^2 \to \mm / (I + \mm^2) $$

   where the domain has dimension \( n \) and the codomain has dimension \( d \) by assumption. By the rank-nullity theorem, the kernel of this map is dimension \( r \) as a \( k \)-vector space. It is fairly easy to see that the kernel is of the form \( I / (I \cap \mm^2) \) — therefore, we may choose some \( f_1, \dots, f_r \in I \) so that their images in the quotient span \( I / (I \cap \mm^2) \). In fact, by an argument similar to Exercise 11.3.I we should be able to choose the \(f_i\) in such a way that sucessive quotients \( A / (f_i) \) each reduce the dimension by 1. Since the quotient of local rings is always local (the unique maximal ideal is always preserved by the Correspondence theorem), we have that \( A / (f_1, \dots, f_r) \) and \( A / I \) are both local rings (and hence integral domains by the theorem above) of dimension \(d\).

   To prove the last claim, let \( \phi : A \to B \) be a surjection of domains with the same Krull dimension. Since \( B \cong A / \ker \phi \), we may without loss of generality work in \(A\). Since the prime ideals of \( A / \ker \phi \) correspond to the prime ideals of \(A\) containing \( \ker \phi \), consider a maximal chain of primes in \( A / \ker \phi \):

   $$ (\ker \phi \ \subset)\ \pp_0 \subset \dots \subset \pp_d = \mm $$

   However, since \( \dim A = \dim A / \ker \phi \), this is also a maximal chain of primes in \( A \). Since \( A \) is a domain, the unique minimal prime is \( (0) \) so that \( \ker \phi = (0) \) and thus we have an isomorphism.

2. This immediately follows from part (a) above since closed embeddings are locally of the form \( \spec A / I \hookrightarrow \spec A \) so that the stalk of the ideal sheaf at every point is generated by a regular sequence.

Proof:

As pointed out by David Benjamin Lim in this Stack Exchange answer, the example of Caution 12.2.11 can be used to show that the diagonal is, in fact, not a regular embedding.

Proof:

In \( k[x] \), all of our prime ideals are of the form \( (f) \) for \( f\) irreducible. Since every stalk is of the form \( \OO_{X, p} = (k[x])_{(f(x))} \) and \( \mm_p / \mm_p^2 = \mm \otimes_{\OO_{X,p}} \kappa(p) \), we get that every Zariski tangent space is simply the ideal of \( (f(x)) \) in \( \kappa(p) \). As this is principal, we obtain a 1 dimensional subspace.

By Exercise 3.2.E we know that the points of \( \A^2_{\overline{k}} \) are the maximal ideals / closed points, the irreducible curves, and the generic point. As we have already seen that smoothness implies regularity at closed points, it suffices to check regularity at the points \( \pp = (f(x,y)) \) for \( f(x, y) \) irreducible. But then using Exercise 12.1.B, we can see that by taking a general closed point \( [\mm] \) with \( f \in \mm \backslash \mm^2 \), the Zariski tangent space of \( \C[x, y] / (f) \) will have dimension 1 (since we chose a closed point we may simply use the Jacobi criterion). Thus, the dimension of the Zariski cotangent space corresponding to \( \pp \) has dimension 2.

Since regularity can be checked affine locally, this proves \( \P^1_k \) and \( \P^2_{\overline{k}} \) are regular.

Proof:

The first claim follows from Exercise 12.2.F(b) since singularity of a point may be checked in an affine neighborhood. To prove the Euler test, we must use the fact that \( f\) is necessarily a homogeneous polynomial so that the degree of each component is the same — in fact, if we show the result for monomials then the result will follow. Thus, suppose \( f = x_0^{\alpha_0}\dots x_n^{\alpha_n} \) is a monomial of degree \( d = \sum_i \alpha_i \). Since

by construction, we have that

as desired. Thus, if all the partials vanish and \( d \nmid \textrm{char}\,k \) we have that \( f = 0 \) as well.

Proof:

We showed smoothness for \( \textrm{char}\,k \neq 2 \) in Exercise 12.2.F(a). Referring to the stronger fact / theorem mentioned in Exercise 12.2.I, this implies regularity at closed points and thus everywhere since \( k = \overline{k} \). To see that the curve is irreducible, we use Eisenstein's criterion to see that there is no prime \( p \) dividing 1.

Proof:

By Exercise 5.4.J we may write our quadric of the form

for some symmetric \( M \). Up to a linear change of basis we may diagonalize this matrix, so that we may assume without loss of generality that

where \( r \leq n \) is the rank. Since \( \textrm{char}\,k \neq 2 \), we may use the Euler criterion, so if \( r = n \) then each \( \frac{\partial f}{\partial x_i} = 2a_i x_i = 0 \) implies our point is \( [0 , \dots, 0] \) which is not in \( \P^n \) — that is to say that if it is maximal rank then it is regular. Conversely, if \( r \leq n -1 \) then \( [0, \dots, 0, 1] \) is a singular point.

Proof:

Following the hint, we know by Euler's criterion that the Fermat curve is singular whenever all coordinates vanish which does not occur in \( \P^2 \) — that is to say its a regular curve.

Proof:

1. We assume by plane curve each closed subvariety is embedded in \( \A^2_k \). Thus using the Jacobi Criterion from Exercise 12.2.F, we see that \( V(y^2 - x^2 - x^3) \) is singular when \( x = y = 0 \) (we necessarily have \( y = 0 \) in order for \( \frac{\partial f}{\partial y} \) to vanish, and \( x = 0\) or \( x = -2/3 \) for \( \frac{\partial f}{\partial x} \) to vanish — however \( (-2/3, 0)\) does not lie on the curve ).

2. An even easier computation still results in the origin being the only singular point.

3. And again the origin is the only singular point.

Proof:

Looking over the first affine chart \( w \neq 0 \), we have that the coordinate ring is given by \( k[x, y, z] / (y - x^2, z - x^3) \) so that our Jacobian is of the form

which is rank 2 (i.e. corank 1) everywhere, giving us regularity on the first affine patch. Similarly for \( x \neq 0 \), our coordinate ring is of the form \( k[w, y, z] / (wy - 1, z - y^2) \) so our Jacobian is of the form

which again gives us corank 1.

Proof:

Suppose \( g = \sum_i c_i f_i \) for some \( c_i \in k[x_1, \dots, x_n] \), and the \( a \in V(f_1, \dots, f_r) \) is a regular \(k\)-valued point so that \( f_i(a) = 0 \). By the Leibniz rule we necessarily have

so that \( \frac{\partial g}{\partial x_i}(a) \) is a \( k[x_1, \dots, x_n] \)-linear combination of the \( \frac{\partial f}{\partial x_i}(a) \).

Proof:

Letting \( f = y^2 - x^3 \), we have \( \frac{\partial f}{\partial x}(1, 1) = -3(1)^2 = -3 \) and \( \frac{\partial f}{\partial y}(1, 1) = 2(1) = 2 \) giving us

Proof:

The naive way to approach this would be to use the definition affine locally and show that it agrees on affine open subsets. In this vein, let us restrict our attention to a single \( V_+(f) \subset \P^n_k \) and let \( a \) be a regular \( k \)-valued point. Then on any overlap \( U_i \cap U_j \) containing \( a \), we have

In addition, the relation \( x_{k/j} = x_{k/i} \cdot (x_{j/i})^{-1} \) tells us that \( \frac{\partial}{\partial x_{k/i}} = \frac{\partial}{\partial x_{k/j}} \cdot \frac{ \partial x_{k/j} }{ \partial x_{k/i} } = (x_{j/i})^{-1} \cdot \frac{\partial}{\partial x_{k/j}} \). Then

so that the tangent planes are independent of affine chart.

Proof:

If \( X \cap L \) intersect at \( P \) with order \( d \geq 2 \), then the stalk \( \OO_{X \cap L, P} \) is an order-\( d \) point (c.f. double point) \( \spec k[x] / (x^d) \). By Exercise 12.1.I, we know that the data of a map from a double point into \( X \) is the same as a point along with a tangent vector. This gives a natural inclusion of \( L \cap X \) into \( (\mm_p / \mm^2_p)^\vee \). As \( \mm_p / \mm_p^2 \) is generated by the cokernel of the \( \frac{\partial f}{\partial x_i}(a) \), the terms of \( L \) have coefficients that are simply a linear combination of the \( \frac{\partial f}{\partial x_i}(a) \) (that is to say, \(X\) and \(L\) agree at \( p \) up to at least linear order terms). However since \( T_p X \) is the linear space generated at \( p \) by these \( \frac{\partial f}{\partial x_i}(a) \), we must have that \( L \subset T_p X \).

Proof:

As \( \Z \) is a PID every point is a closed point so that any maximal ideal \( \mm \) is of the form \( \mm = (p) \) for some prime integer \( p \). Then \( \mm / \mm^2 = p \Z / p^2 \Z \) which is of dimension 1 in \( \kappa(p) = \mathbb{F}_p \).

Proof:

Since regularity is an affine local condition, we may reduce to the case that \( X = \spec B \) so that \( X \times \A^1 \cong \spec (B \otimes_\Z \Z[x]) \cong \spec B[x] \). Let \( \phi : B \to B[x] \) denote the natural inclusion in our fibred coproduct. We wish to show that \( (B[x])_\pp \) is a regular local ring for every prime \( \pp \subset B[x]\). Consider the prime ideal \( \mathfrak{n} = B \cap \pp \); since \( X \) is assumed to be regular, \( B_\mathfrak{n} \) is a regular local ring with maximal ideal \(\mathfrak{n} \), so that by construction we have that \( \mathfrak{n} B[x] \subset \pp \). By the previous proposition \( (B[x])_\pp \) is a regular local ring, thus proving the claim. It is immediate that \( \A^n_k \) is thus regular by induction since \( \A^1_k \) is regular by Exercise 12.3.A and \( \A^n_k \times_k \A^m_k \cong \A^{n+m}_k \).

Proof:

1. Applying Bertini's Theorem to \( X_{reg} \) (the regular locus) yields a dense open subset \( U \subset (\P^n)^\vee \) such that for any closed point \( [H] \in U \), \( H \) doesnt contain any component of \( X_{reg} \) and \( H \cap X_{reg} \) is \( k \)-smooth. For any singular point \( x in X \backslash X_{reg} \), \( x \in H \) is a closed condition so that we may take an open subset \( U^\prime \subset U \) such that no \( [H] \in U^\prime \) contains \( U \) and \( U^\prime \) is still dense. Repeating this process finitely many times yields the desired open subset.

2. Instead of assuming \( X \) is globally cut out by \( f_1, \dots, f_r \), we may restrict our attention to local generators on sufficiently small affine patches. However, as the Jacobian matrix is stalk-local this does not change the argument of dimensionality of \( Z \).

Proof:

Following the idea of Bertini's Theorem, consider the variety \( Z \subset X \times ((\P^m)^\vee)^n \) consisting of points \( (p, H_1, \dots, H_n) \) with \( p \in H_1 \cap \dots \cap H_n \), and either \( \dim (H_1 \cap \dots \cap H_n \cap X) \neq 0 \) (recall we say the empty set has dimension \( - \infty \)) or the intersection is singular at \( p\). We may in fact obtain this variety by applying Bertini's theorem iteratively to the successive varieties \( H \cap X \) which are shown to be \( k \)-smooth. By Exercise 11.3.C(d) (or more generally Krull's principal ideal theorem), we know that the general intersection of \( n \) hyperplanes will be a finite collection of points — however, since Bertini's theorem tells us that these points are smooth, their stalks must be integral domains by Theorem 12.2.13 so that the points are necessarily reduced.

(The proof is a bit hand-waivey in terms of the nonempty open subset that you should end up with. We should interpret Exercise 11.3.C(d) as telling us that there exists a non-empty open subset \( U_1 \) of \( ((\P^n_k)^\vee)^n \) where all hyperplanes have a non-empty intersection of dimension 0, and then use the fact that being a singular point would be a closed condition in this open subset.)

Proof:

Let \( a_0, \dots, a_n \) be the coordinates of \( (\P^n)^\vee \) and suppose that \( b_0 x_0 + \dots + b_n x_n = 0 \) is our hyperplane \(H \subset \P^n \). Then the augmented Jacobian matrix of \( H \) (as in 12.4.2.1) is

Since we require that this matrix have corank \( \geq \dim H = n - 1\) (i.e. rank at most 1), we must have the \( a_i = b_i \) for all \( i \) up to some constant in \(k\). That is to say that \( H^\vee = [b_0, \dots, b_n] \) is simply a point in \( ( \P^n)^\vee \) as expected.

Proof:

From Exercise 12.3.D we know that a quadirc (i.e conic for \( \P^2 \)) is smooth if and only if it has maximal rank. Thus, up to an algebraic change of coordinates we may suppose that \( C = V_+(c_0 x_0^2 + c_1 x_1^2 + c_2 x_2^2) \) for some nonzero \(c_i\). Then the augmented Jacobian is of the form

Since we again want this to have rank at most 1, we see that \( a_i \) is a nonzero scalar multiple of \( 2c_i x_i \). Since \( \textrm{char}\,k \neq 2 \), this tells us that the coordinates \( a_i \) also cut out a conic isomorphic to \( C \), which is again necessarily smooth since none of our coefficients of the \(a_i\) are 0.

Proof:

An arbitrary conic over \( \P^2 \) can be written of the form

Since choosing \( P_1, \dots, P_5 \in \P^2_k \) in such a way that

is an open condition (i.e. look at the vanishing of minors), 5 generically chosen points will give that the above matrix has a 1 dimensional kernel (in \( k^6 \)) which corresponds to a unique point \( [a_0, \dots, a_5] \in \P^5 \). By the previous problem, we know that the dual of a conic \( C \) — call it \( C^\vee \) — is also a smooth conic. In this case, it can also be determined by 5 generically chosen points. By dualizing again (warning: we have not shown that the dual of a dual conic is the original conic), our 5 generic points naturally correspond to 5 hyperplanes (i.e. lines) so that \( C \) is uniquely determined by 5 lines.

Proof:

For the forward direction, assume any / all of (a) - (c). Then any nonzero ideal \( I \subset A \) is of the form \( \mm^n \) by (c) and since \( \mm \) is principal by (a) we have that \( I = (f)^n = (f^n) \) for some \( f \in A \). Conversely, suppose \(A\) is a principal ideal domain. Then \( \mm \) is necessarily principal.

Proof:

The first valuation ring consists of rational numbers \( a / b \) such that after reducing by common denominators, the denominator is not divisible by a power of 5. In other words, we obtain the ring \( \Z_{(5)} \). The second valuation ring consists of rational functions whose numerator has equal or higher degree than the denominator (after applying polynomial division). The third is the exact opposite of this.

Proof:

Let \( \mm \) denote the ideal above (we notice it is indeed an ideal since \( \nu(xy) = \nu(x) + \nu(y) \geq \nu(y) > 0 \) for any \( x \in \OO_\nu \) and \( y \in \mm \), and a similar idea using the other axiom shows it is a subgroup). Then any element \(a\) of the complement \( \OO_\nu - \mm \) is a nonzero element of \( K^\times \) such that \( \nu(a) = 0 \) — in other words, \( \OO_\nu - \mm = \ker \nu - \{ 0 \} \). By the valuation axioms we know that \( 1 \in \OO_\nu - \mm\) since

Since \( \nu \) is a group homomorphism, we know that \( \nu(a^{-1}) = - \nu(a) \), so if \( a \in \OO_\nu - \mm \) we must also have \( a^{-1} \in \OO_\nu - \mm \). Therefore, everything in the complement is invertible and thus \( \mm \) is the unique maximal ideal.

Proof:

To see that it is a group homomorphism, let \( x, y \) be two elements of \( K(A)^\times \) so that \( x = ut^m \) and \( y = vt^n \) for some units \( u, v \in K(A)^\times \). Then

moreover, the homomorphism is clearly surjective. Now suppose without loss of generality that \( m \geq n \) — then we may write \( x + y \) as \( (ut^{m - n} + v) \cdot t^n \) so that \( \nu(x + y) \geq n = \min \{ \nu(x), \nu(y) \} \)

Proof:

Let \( I \) be an ideal of \( (A, \mm) \). We first show that there must necessarily exist a nonzero element \( z \in I \) with minimal valuation — suppose not, so that for every \( k \geq 0\) there exists some \( x_k \neq 0 \) with \( \nu(x_k) = k \). Since \( A - \mm \) consists of invertible elements of valuation 0, this implies that \( I \) contains a unit and is thus all of \(A\). Therefore, we may suppose \( I \) has minimal valuation \( m > 0 \) — thus it immediately follows that \( I \subset I_m \). Now notice that if two elements \( x, y \in K^\times \) both have valuation equal to \( m \), \( xy^{-1} \) has valuation \(0\) and is therefore a unit so that \( (x) = (y) \). Therefore, \( \mm \) is generated by any element with valuation 1 and \( I = I_m = \mm^m \). Notice that this immediately implies \( \mm^1 \) is the only maximal ideal, since otherwise we could take \( x \in \mm \) (of valuation exactly 1), \( x^{k-1} \in \mm^{k-1} \) so that \( x^k \in \mm^k \). Thus, we immediately get that \( (A, \mm) \) is dimension 1. We can also easly see that the ascending chain condition holds, as it is graded by valuation — moreover, as \( \mm \) is generated by any element of valuation 1 it is necessarily principal so that by Theorem 12.5.1 \( (A, \mm) \) is regular.

Proof:

By the above theorem, being a DVR is equivalent to being a Noetherian regular local ring of dimension 1. Suppose \( t \in \mm - \mm^2 \) is a uniformizer, and let \( \nu : K^\times \to \Z \) be a discrete valuation. By Exercise 12.5.C we know that \( t \) must have valuation precisely 1 (as otherwise by surjectivity of \( \nu \) we could find some \( y \in \mm \) of valuation 1 not generated by \(t\)). As before, every element \( x \in A \) can be uniquely written \( x = ut^n \) for some unit \( u \) and \( n \geq 0\). But then by virtue of being a homomorphism, we have \( \nu(x) = \nu(u) + n\cdot \nu(t) = n \) so that our discrete valuation is the same as that described in the text. Notice that if we chose a different uniformizer \( s \) for \( \mm \), we would have \( (s) = \mm = (t) \) so that \( s = u^\prime t \) for some unit \( u^\prime \).

Proof:

Since \( X \) is quasi-compact by Noetherianness, we may reduce to the case that \( X = \spec A \). Suppose \( Y \) is a regular codimension 1 closed subvariety, which corresponds to a minimial prime ideal \( \pp_i \). Then our local ring \( \OO_{X, p} \) is necessarily of the form \( A_{\pp_i} / \pp_i A_{\pp_i} \), and \( \nu_Y(f_i) > 0 \) if and only if \( f_i \in \pp_i \). However, since there are only finitely many minimal primes \( \pp_i \subset A \) we have that \( f_i \) can have only finitely many zeros and poles.

Proof:

Let \( \spec A \) be some affine neighborhood and suppose \( f\vert_\spec A = f_1 / f_2\) for \( f_1, f_2 \in A \). Since \( f_2 \) has no poles, for every codimension \( 1 \) closed subset corresponding to some minimal prime \( \pp_i \subset A \), we must have \( f_2 \notin \pp_i \) so that \( f \in A_{\pp_i} \) for each minimal prime \( \pp_i \). Then by Algebraic Hartogs lemma, we have \( f \in A \). Since our affine neighborhood was arbitrary, \( f \) must be a regular function.

Proof:

We first show that \( A \to k[x, y] \) is a normalization. Notice that we have a ring map \( k[x, y] \to k[u, v, w, y] \) given by \( u = x^3, v = x^2, w = xy \), where the image is precisely \(A\). Now we get relations \( u^2 - v^3, uy - vw, vy^2 - w^2 \), so that away from the origin \( (0, 0) \) (i.e. localizing at \( (x, y) \)) there is an obvious isomorphism \( u/v \mapsto x \), \( y \mapsto y\).

It is somewhat easy to see that \( A \) is not integrally closed, as it contains \( x^2 \) and \( x^3 \) but not \(x\) — to be precise, we could simply take \( t^2 - x^2 \) and notice that this obviously has a root in the field of fractions ( again since \( x = x^3 / x^2 \) ) but does not have a root in \( A \).

To see that it is regular in codimension 1, we notice that away from the origin (as explained above) we get an isomorphism onto \( \A^2_k\) which is indeed a regular surface. Since the origin is codimension 2, the result follows.