The Stacks project

10.140 Smooth algebras over fields

Warning: The following two lemmas do not hold over nonperfect fields in general.

Lemma 10.140.1. Let $k$ be an algebraically closed field. Let $S$ be a finite type $k$-algebra. Let $\mathfrak m \subset S$ be a maximal ideal. Then

\[ \dim _{\kappa (\mathfrak m)} \Omega _{S/k} \otimes _ S \kappa (\mathfrak m) = \dim _{\kappa (\mathfrak m)} \mathfrak m/\mathfrak m^2. \]

Proof. Consider the exact sequence

\[ \mathfrak m/\mathfrak m^2 \to \Omega _{S/k} \otimes _ S \kappa (\mathfrak m) \to \Omega _{\kappa (\mathfrak m)/k} \to 0 \]

of Lemma 10.131.9. We would like to show that the first map is an isomorphism. Since $k$ is algebraically closed the composition $k \to \kappa (\mathfrak m)$ is an isomorphism by Theorem 10.34.1. So the surjection $S \to \kappa (\mathfrak m)$ splits as a map of $k$-algebras, and Lemma 10.131.10 shows that the sequence above is exact on the left. Since $\Omega _{\kappa (\mathfrak m)/k} = 0$, we win. $\square$

Lemma 10.140.2. Let $k$ be an algebraically closed field. Let $S$ be a finite type $k$-algebra. Let $\mathfrak m \subset S$ be a maximal ideal. The following are equivalent:

  1. The ring $S_{\mathfrak m}$ is a regular local ring.

  2. We have $\dim _{\kappa (\mathfrak m)} \Omega _{S/k} \otimes _ S \kappa (\mathfrak m) \leq \dim (S_{\mathfrak m})$.

  3. We have $\dim _{\kappa (\mathfrak m)} \Omega _{S/k} \otimes _ S \kappa (\mathfrak m) = \dim (S_{\mathfrak m})$.

  4. There exists a $g \in S$, $g \not\in \mathfrak m$ such that $S_ g$ is smooth over $k$. In other words $S/k$ is smooth at $\mathfrak m$.

Proof. Note that (1), (2) and (3) are equivalent by Lemma 10.140.1 and Definition 10.110.7.

Assume that $S$ is smooth at $\mathfrak m$. By Lemma 10.137.10 we see that $S_ g$ is standard smooth over $k$ for a suitable $g \in S$, $g \not\in \mathfrak m$. Hence by Lemma 10.137.7 we see that $\Omega _{S_ g/k}$ is free of rank $\dim (S_ g)$. Hence by Lemma 10.140.1 we see that $\dim (S_{\mathfrak m}) = \dim (\mathfrak m/\mathfrak m^2)$ in other words $S_\mathfrak m$ is regular.

Conversely, suppose that $S_{\mathfrak m}$ is regular. Let $d = \dim (S_{\mathfrak m}) = \dim \mathfrak m/\mathfrak m^2$. Choose a presentation $S = k[x_1, \ldots , x_ n]/I$ such that $x_ i$ maps to an element of $\mathfrak m$ for all $i$. In other words, $\mathfrak m'' = (x_1, \ldots , x_ n)$ is the corresponding maximal ideal of $k[x_1, \ldots , x_ n]$. Note that we have a short exact sequence

\[ I/\mathfrak m''I \to \mathfrak m''/(\mathfrak m'')^2 \to \mathfrak m/(\mathfrak m)^2 \to 0 \]

Pick $c = n - d$ elements $f_1, \ldots , f_ c \in I$ such that their images in $\mathfrak m''/(\mathfrak m'')^2$ span the kernel of the map to $\mathfrak m/\mathfrak m^2$. This is clearly possible. Denote $J = (f_1, \ldots , f_ c)$. So $J \subset I$. Denote $S' = k[x_1, \ldots , x_ n]/J$ so there is a surjection $S' \to S$. Denote $\mathfrak m' = \mathfrak m''S'$ the corresponding maximal ideal of $S'$. Hence we have

\[ \xymatrix{ k[x_1, \ldots , x_ n] \ar[r] & S' \ar[r] & S \\ \mathfrak m'' \ar[u] \ar[r] & \mathfrak m' \ar[r] \ar[u] & \mathfrak m \ar[u] } \]

By our choice of $J$ the exact sequence

\[ J/\mathfrak m''J \to \mathfrak m''/(\mathfrak m'')^2 \to \mathfrak m'/(\mathfrak m')^2 \to 0 \]

shows that $\dim ( \mathfrak m'/(\mathfrak m')^2 ) = d$. Since $S'_{\mathfrak m'}$ surjects onto $S_{\mathfrak m}$ we see that $\dim (S_{\mathfrak m'}) \geq d$. Hence by the discussion preceding Definition 10.60.10 we conclude that $S'_{\mathfrak m'}$ is regular of dimension $d$ as well. Because $S'$ was cut out by $c = n - d$ equations we conclude that there exists a $g' \in S'$, $g' \not\in \mathfrak m'$ such that $S'_{g'}$ is a global complete intersection over $k$, see Lemma 10.135.4. Also the map $S'_{\mathfrak m'} \to S_{\mathfrak m}$ is a surjection of Noetherian local domains of the same dimension and hence an isomorphism. Hence $S' \to S$ is surjective with finitely generated kernel and becomes an isomorphism after localizing at $\mathfrak m'$. Thus we can find $g' \in S'$, $g \not\in \mathfrak m'$ such that $S'_{g'} \to S_{g'}$ is an isomorphism. All in all we conclude that after replacing $S$ by a principal localization we may assume that $S$ is a global complete intersection.

At this point we may write $S = k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ with $\dim S = n - c$. Recall that the naive cotangent complex of this algebra is given by

\[ \bigoplus S \cdot f_ j \to \bigoplus S \cdot \text{d}x_ i \]

see Lemma 10.136.13. By Lemma 10.137.16 in order to show that $S$ is smooth at $\mathfrak m$ we have to show that one of the $c \times c$ minors $g_ I$ of the matrix “$A$” giving the map above does not vanish at $\mathfrak m$. By Lemma 10.140.1 the matrix $A \bmod \mathfrak m$ has rank $c$. Thus we win. $\square$

Lemma 10.140.3. Let $k$ be any field. Let $S$ be a finite type $k$-algebra. Let $X = \mathop{\mathrm{Spec}}(S)$. Let $\mathfrak q \subset S$ be a prime corresponding to $x \in X$. The following are equivalent:

  1. The $k$-algebra $S$ is smooth at $\mathfrak q$ over $k$.

  2. We have $\dim _{\kappa (\mathfrak q)} \Omega _{S/k} \otimes _ S \kappa (\mathfrak q) \leq \dim _ x X$.

  3. We have $\dim _{\kappa (\mathfrak q)} \Omega _{S/k} \otimes _ S \kappa (\mathfrak q) = \dim _ x X$.

Moreover, in this case the local ring $S_{\mathfrak q}$ is regular.

Proof. If $S$ is smooth at $\mathfrak q$ over $k$, then there exists a $g \in S$, $g \not\in \mathfrak q$ such that $S_ g$ is standard smooth over $k$, see Lemma 10.137.10. A standard smooth algebra over $k$ has a module of differentials which is free of rank equal to the dimension, see Lemma 10.137.7 (use that a relative global complete intersection over a field has dimension equal to the number of variables minus the number of equations). Thus we see that (1) implies (3). To finish the proof of the lemma it suffices to show that (2) implies (1) and that it implies that $S_{\mathfrak q}$ is regular.

Assume (2). By Nakayama's Lemma 10.20.1 we see that $\Omega _{S/k, \mathfrak q}$ can be generated by $\leq \dim _ x X$ elements. We may replace $S$ by $S_ g$ for some $g \in S$, $g \not\in \mathfrak q$ such that $\Omega _{S/k}$ is generated by at most $\dim _ x X$ elements. Let $K/k$ be an algebraically closed field extension such that there exists a $k$-algebra map $\psi : \kappa (\mathfrak q) \to K$. Consider $S_ K = K \otimes _ k S$. Let $\mathfrak m \subset S_ K$ be the maximal ideal corresponding to the surjection

\[ \xymatrix{ S_ K = K \otimes _ k S \ar[r] & K \otimes _ k \kappa (\mathfrak q) \ar[r]^-{\text{id}_ K \otimes \psi } & K. } \]

Note that $\mathfrak m \cap S = \mathfrak q$, in other words $\mathfrak m$ lies over $\mathfrak q$. By Lemma 10.116.6 the dimension of $X_ K = \mathop{\mathrm{Spec}}(S_ K)$ at the point corresponding to $\mathfrak m$ is $\dim _ x X$. By Lemma 10.114.6 this is equal to $\dim ((S_ K)_{\mathfrak m})$. By Lemma 10.131.12 the module of differentials of $S_ K$ over $K$ is the base change of $\Omega _{S/k}$, hence also generated by at most $\dim _ x X = \dim ((S_ K)_{\mathfrak m})$ elements. By Lemma 10.140.2 we see that $S_ K$ is smooth at $\mathfrak m$ over $K$. By Lemma 10.137.18 this implies that $S$ is smooth at $\mathfrak q$ over $k$. This proves (1). Moreover, we know by Lemma 10.140.2 that the local ring $(S_ K)_{\mathfrak m}$ is regular. Since $S_{\mathfrak q} \to (S_ K)_{\mathfrak m}$ is flat we conclude from Lemma 10.110.9 that $S_{\mathfrak q}$ is regular. $\square$

The following lemma can be significantly generalized (in several different ways).

Lemma 10.140.4. Let $k$ be a field. Let $R$ be a Noetherian local ring containing $k$. Assume that the residue field $\kappa = R/\mathfrak m$ is a finitely generated separable extension of $k$. Then the map

\[ \text{d} : \mathfrak m/\mathfrak m^2 \longrightarrow \Omega _{R/k} \otimes _ R \kappa (\mathfrak m) \]

is injective.

Proof. We may replace $R$ by $R/\mathfrak m^2$. Hence we may assume that $\mathfrak m^2 = 0$. By assumption we may write $\kappa = k(\overline{x}_1, \ldots , \overline{x}_ r, \overline{y})$ where $\overline{x}_1, \ldots , \overline{x}_ r$ is a transcendence basis of $\kappa $ over $k$ and $\overline{y}$ is separable algebraic over $k(\overline{x}_1, \ldots , \overline{x}_ r)$. Say its minimal equation is $P(\overline{y}) = 0$ with $P(T) = T^ d + \sum _{i < d} a_ iT^ i$, with $a_ i \in k(\overline{x}_1, \ldots , \overline{x}_ r)$ and $P'(\overline{y}) \not= 0$. Choose any lifts $x_ i \in R$ of the elements $\overline{x}_ i \in \kappa $. This gives a commutative diagram

\[ \xymatrix{ R \ar[r] & \kappa \\ & k(\overline{x}_1, \ldots , \overline{x}_ r) \ar[lu]^\varphi \ar[u] } \]

of $k$-algebras. We want to extend the left upwards arrow $\varphi $ to a $k$-algebra map from $\kappa $ to $R$. To do this choose any $y \in R$ lifting $\overline{y}$. To see that it defines a $k$-algebra map defined on $\kappa \cong k(\overline{x}_1, \ldots , \overline{x}_ r)[T]/(P)$ all we have to show is that we may choose $y$ such that $P^\varphi (y) = 0$. If not then we compute for $\delta \in \mathfrak m$ that

\[ P(y + \delta ) = P(y) + P'(y)\delta \]

because $\mathfrak m^2 = 0$. Since $P'(y)\delta = P'(\overline{y})\delta $ we see that we can adjust our choice as desired. This shows that $R \cong \kappa \oplus \mathfrak m$ as $k$-algebras! From a direct computation of $\Omega _{\kappa \oplus \mathfrak m/k}$ the lemma follows. $\square$

Lemma 10.140.5. Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $\mathfrak q \subset S$ be a prime. Assume $\kappa (\mathfrak q)$ is separable over $k$. The following are equivalent:

  1. The algebra $S$ is smooth at $\mathfrak q$ over $k$.

  2. The ring $S_{\mathfrak q}$ is regular.

Proof. Denote $R = S_{\mathfrak q}$ and denote its maximal by $\mathfrak m$ and its residue field $\kappa $. By Lemma 10.140.4 and 10.131.9 we see that there is a short exact sequence

\[ 0 \to \mathfrak m/\mathfrak m^2 \to \Omega _{R/k} \otimes _ R \kappa \to \Omega _{\kappa /k} \to 0 \]

Note that $\Omega _{R/k} = \Omega _{S/k, \mathfrak q}$, see Lemma 10.131.8. Moreover, since $\kappa $ is separable over $k$ we have $\dim _{\kappa } \Omega _{\kappa /k} = \text{trdeg}_ k(\kappa )$. Hence we get

\[ \dim _{\kappa } \Omega _{R/k} \otimes _ R \kappa = \dim _\kappa \mathfrak m/\mathfrak m^2 + \text{trdeg}_ k (\kappa ) \geq \dim R + \text{trdeg}_ k (\kappa ) = \dim _{\mathfrak q} S \]

(see Lemma 10.116.3 for the last equality) with equality if and only if $R$ is regular. Thus we win by applying Lemma 10.140.3. $\square$

Lemma 10.140.6. Let $R \to S$ be a $\mathbf{Q}$-algebra map. Let $f \in S$ be such that $\Omega _{S/R} = S \text{d}f \oplus C$ for some $S$-submodule $C$. Then

  1. $f$ is not nilpotent, and

  2. if $S$ is a Noetherian local ring, then $f$ is a nonzerodivisor in $S$.

Proof. For $a \in S$ write $\text{d}(a) = \theta (a)\text{d}f + c(a)$ for some $\theta (a) \in S$ and $c(a) \in C$. Consider the $R$-derivation $S \to S$, $a \mapsto \theta (a)$. Note that $\theta (f) = 1$.

If $f^ n = 0$ with $n > 1$ minimal, then $0 = \theta (f^ n) = n f^{n - 1}$ contradicting the minimality of $n$. We conclude that $f$ is not nilpotent.

Suppose $fa = 0$. If $f$ is a unit then $a = 0$ and we win. Assume $f$ is not a unit. Then $0 = \theta (fa) = f\theta (a) + a$ by the Leibniz rule and hence $a \in (f)$. By induction suppose we have shown $fa = 0 \Rightarrow a \in (f^ n)$. Then writing $a = f^ nb$ we get $0 = \theta (f^{n + 1}b) = (n + 1)f^ nb + f^{n + 1}\theta (b)$. Hence $a = f^ n b = -f^{n + 1}\theta (b)/(n + 1) \in (f^{n + 1})$. Since in the Noetherian local ring $S$ we have $\bigcap (f^ n) = 0$, see Lemma 10.51.4 we win. $\square$

The following is probably quite useless in applications.

Lemma 10.140.7. Let $k$ be a field of characteristic $0$. Let $S$ be a finite type $k$-algebra. Let $\mathfrak q \subset S$ be a prime. The following are equivalent:

  1. The algebra $S$ is smooth at $\mathfrak q$ over $k$.

  2. The $S_{\mathfrak q}$-module $\Omega _{S/k, \mathfrak q}$ is (finite) free.

  3. The ring $S_{\mathfrak q}$ is regular.

Proof. In characteristic zero any field extension is separable and hence the equivalence of (1) and (3) follows from Lemma 10.140.5. Also (1) implies (2) by definition of smooth algebras. Assume that $\Omega _{S/k, \mathfrak q}$ is free over $S_{\mathfrak q}$. We are going to use the notation and observations made in the proof of Lemma 10.140.5. So $R = S_{\mathfrak q}$ with maximal ideal $\mathfrak m$ and residue field $\kappa $. Our goal is to prove $R$ is regular.

If $\mathfrak m/\mathfrak m^2 = 0$, then $\mathfrak m = 0$ and $R \cong \kappa $. Hence $R$ is regular and we win.

If $\mathfrak m/ \mathfrak m^2 \not= 0$, then choose any $f \in \mathfrak m$ whose image in $\mathfrak m/ \mathfrak m^2$ is not zero. By Lemma 10.140.4 we see that $\text{d}f$ has nonzero image in $\Omega _{R/k}/\mathfrak m\Omega _{R/k}$. By assumption $\Omega _{R/k} = \Omega _{S/k, \mathfrak q}$ is finite free and hence by Nakayama's Lemma 10.20.1 we see that $\text{d}f$ generates a direct summand. We apply Lemma 10.140.6 to deduce that $f$ is a nonzerodivisor in $R$. Furthermore, by Lemma 10.131.9 we get an exact sequence

\[ (f)/(f^2) \to \Omega _{R/k} \otimes _ R R/fR \to \Omega _{(R/fR)/k} \to 0 \]

This implies that $\Omega _{(R/fR)/k}$ is finite free as well. Hence by induction we see that $R/fR$ is a regular local ring. Since $f \in \mathfrak m$ was a nonzerodivisor we conclude that $R$ is regular, see Lemma 10.106.7. $\square$

Example 10.140.8. Lemma 10.140.7 does not hold in characteristic $p > 0$. The standard examples are the ring maps

\[ \mathbf{F}_ p \longrightarrow \mathbf{F}_ p[x]/(x^ p) \]

whose module of differentials is free but is clearly not smooth, and the ring map ($p > 2$)

\[ \mathbf{F}_ p(t) \to \mathbf{F}_ p(t)[x, y]/(x^ p + y^2 + \alpha ) \]

which is not smooth at the prime $\mathfrak q = (y, x^ p + \alpha )$ but is regular.

Using the material above we can characterize smoothness at the generic point in terms of field extensions.

Lemma 10.140.9. Let $R \to S$ be an injective finite type ring map with $R$ and $S$ domains. Then $R \to S$ is smooth at $\mathfrak q = (0)$ if and only if the induced extension $L/K$ of fraction fields is separable.

Proof. Assume $R \to S$ is smooth at $(0)$. We may replace $S$ by $S_ g$ for some nonzero $g \in S$ and assume that $R \to S$ is smooth. Then $K \to S \otimes _ R K$ is smooth (Lemma 10.137.4). Moreover, for any field extension $K'/K$ the ring map $K' \to S \otimes _ R K'$ is smooth as well. Hence $S \otimes _ R K'$ is a regular ring by Lemma 10.140.3, in particular reduced. It follows that $S \otimes _ R K$ is a geometrically reduced over $K$. Hence $L$ is geometrically reduced over $K$, see Lemma 10.43.3. Hence $L/K$ is separable by Lemma 10.44.1.

Conversely, assume that $L/K$ is separable. We may assume $R \to S$ is of finite presentation, see Lemma 10.30.1. It suffices to prove that $K \to S \otimes _ R K$ is smooth at $(0)$, see Lemma 10.137.18. This follows from Lemma 10.140.5, the fact that a field is a regular ring, and the assumption that $L/K$ is separable. $\square$

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