The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

28.32 Smooth morphisms

Let $f : X \to Y$ be a continuous map of topological spaces. Consider the following condition: For every $x \in X$ there exist open neighbourhoods $x \in U \subset X$ and $f(x) \in V \subset Y$, and an integer $d$ such that $f(U) \subset V$ and such that we obtain a commutative diagram

\[ \xymatrix{ X \ar[d] & U \ar[l] \ar[d] \ar[r]_-\pi & V \times \mathbf{R}^ d \ar[ld] \\ Y & V \ar[l] } \]

where $\pi $ is a homeomorphism onto an open subset. Smooth morphisms of schemes are the analogue of these maps in the category of schemes. See Lemma 28.32.11 and Lemma 28.34.20.

Contrary to expectations (perhaps) the notion of a smooth ring map is not defined solely in terms of the module of differentials. Namely, recall that $R \to A$ is a smooth ring map if $A$ is of finite presentation over $R$ and if the naive cotangent complex of $A$ over $R$ is quasi-isomorphic to a projective module placed in degree $0$, see Algebra, Definition 10.135.1.

Definition 28.32.1. Let $f : X \to S$ be a morphism of schemes.

  1. We say that $f$ is smooth at $x \in X$ if there exists an affine open neighbourhood $\mathop{\mathrm{Spec}}(A) = U \subset X$ of $x$ and affine open $\mathop{\mathrm{Spec}}(R) = V \subset S$ with $f(U) \subset V$ such that the induced ring map $R \to A$ is smooth.

  2. We say that $f$ is smooth if it is smooth at every point of $X$.

  3. A morphism of affine schemes $f : X \to S$ is called standard smooth if there exists a standard smooth ring map $R \to R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ (see Algebra, Definition 10.135.6) such that $X \to S$ is isomorphic to

    \[ \mathop{\mathrm{Spec}}(R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)) \to \mathop{\mathrm{Spec}}(R). \]

A pleasing feature of this definition is that the set of points where a morphism is smooth is automatically open.

Note that there is no separation or quasi-compactness hypotheses in the definition. Hence the question of being smooth is local in nature on the source. Here is the precise result.

Lemma 28.32.2. Let $f : X \to S$ be a morphism of schemes. The following are equivalent

  1. The morphism $f$ is smooth.

  2. For every affine opens $U \subset X$, $V \subset S$ with $f(U) \subset V$ the ring map $\mathcal{O}_ S(V) \to \mathcal{O}_ X(U)$ is smooth.

  3. There exists an open covering $S = \bigcup _{j \in J} V_ j$ and open coverings $f^{-1}(V_ j) = \bigcup _{i \in I_ j} U_ i$ such that each of the morphisms $U_ i \to V_ j$, $j\in J, i\in I_ j$ is smooth.

  4. There exists an affine open covering $S = \bigcup _{j \in J} V_ j$ and affine open coverings $f^{-1}(V_ j) = \bigcup _{i \in I_ j} U_ i$ such that the ring map $\mathcal{O}_ S(V_ j) \to \mathcal{O}_ X(U_ i)$ is smooth, for all $j\in J, i\in I_ j$.

Moreover, if $f$ is smooth then for any open subschemes $U \subset X$, $V \subset S$ with $f(U) \subset V$ the restriction $f|_ U : U \to V$ is smooth.

Proof. This follows from Lemma 28.13.3 if we show that the property “$R \to A$ is smooth” is local. We check conditions (a), (b) and (c) of Definition 28.13.1. By Algebra, Lemma 10.135.4 being smooth is stable under base change and hence we conclude (a) holds. By Algebra, Lemma 10.135.14 being smooth is stable under composition and for any ring $R$ the ring map $R \to R_ f$ is (standard) smooth. We conclude (b) holds. Finally, property (c) is true according to Algebra, Lemma 10.135.13. $\square$

The following lemma characterizes a smooth morphism as a flat, finitely presented morphism with smooth fibres. Note that schemes smooth over a field are discussed in more detail in Varieties, Section 32.25.

Lemma 28.32.3. Let $f : X \to S$ be a morphism of schemes. If $f$ is flat, locally of finite presentation, and all fibres $X_ s$ are smooth, then $f$ is smooth.

Proof. Follows from Algebra, Lemma 10.135.16. $\square$

Lemma 28.32.4. The composition of two morphisms which are smooth is smooth.

Proof. In the proof of Lemma 28.32.2 we saw that being smooth is a local property of ring maps. Hence the first statement of the lemma follows from Lemma 28.13.5 combined with the fact that being smooth is a property of ring maps that is stable under composition, see Algebra, Lemma 10.135.14. $\square$

Lemma 28.32.5. The base change of a morphism which is smooth is smooth.

Proof. In the proof of Lemma 28.32.2 we saw that being smooth is a local property of ring maps. Hence the lemma follows from Lemma 28.13.5 combined with the fact that being smooth is a property of ring maps that is stable under base change, see Algebra, Lemma 10.135.4. $\square$

Proof. This is true because an open immersion is a local isomorphism. $\square$

Proof. See Algebra, Lemma 10.135.10. $\square$

Proof. True because a smooth ring map is of finite presentation by definition. $\square$

The following lemma says locally any smooth morphism is standard smooth. Hence we can use standard smooth morphisms as a local model for a smooth morphism.

slogan

Lemma 28.32.11. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$ be a point. Let $V \subset S$ be an affine open neighbourhood of $f(x)$. The following are equivalent

  1. The morphism $f$ is smooth at $x$.

  2. There exist an affine open $U \subset X$, with $x \in U$ and $f(U) \subset V$ such that the induced morphism $f|_ U : U \to V$ is standard smooth.

Proof. Follows from the definitions and Algebra, Lemmas 10.135.7 and 10.135.10. $\square$

Lemma 28.32.12. Let $f : X \to S$ be a morphism of schemes. Assume $f$ is smooth. Then the module of differentials $\Omega _{X/S}$ of $X$ over $S$ is finite locally free and

\[ \text{rank}_ x(\Omega _{X/S}) = \dim _ x(X_{f(x)}) \]

for every $x \in X$.

Proof. The statement is local on $X$ and $S$. By Lemma 28.32.11 above we may assume that $f$ is a standard smooth morphism of affines. In this case the result follows from Algebra, Lemma 10.135.7 (and the definition of a relative global complete intersection, see Algebra, Definition 10.134.5). $\square$

Lemma 28.32.12 says that the following definition makes sense.

Definition 28.32.13. Let $d \geq 0$ be an integer. We say a morphism of schemes $f : X \to S$ is smooth of relative dimension $d$ if $f$ is smooth and $\Omega _{X/S}$ is finite locally free of constant rank $d$.

In other words, $f$ is smooth and the nonempty fibres are equidimensional of dimension $d$. By Lemma 28.32.14 below this is also the same as requiring: (a) $f$ is locally of finite presentation, (b) $f$ is flat, (c) all nonempty fibres equidimensional of dimension $d$, and (d) $\Omega _{X/S}$ finite locally free of rank $d$. It is not enough to simply assume that $f$ is flat, of finite presentation, and $\Omega _{X/S}$ is finite locally free of rank $d$. A counter example is given by $\mathop{\mathrm{Spec}}(\mathbf{F}_ p[t]) \to \mathop{\mathrm{Spec}}(\mathbf{F}_ p[t^ p])$.

Here is a differential criterion of smoothness at a point. There are many variants of this result all of which may be useful at some point. We will just add them here as needed.

Lemma 28.32.14. Let $f : X \to S$ be a morphism of schemes. Let $x \in X$. Set $s = f(x)$. Assume $f$ is locally of finite presentation. The following are equivalent:

  1. The morphism $f$ is smooth at $x$.

  2. The local ring map $\mathcal{O}_{S, s} \to \mathcal{O}_{X, x}$ is flat and $X_ s \to \mathop{\mathrm{Spec}}(\kappa (s))$ is smooth at $x$.

  3. The local ring map $\mathcal{O}_{S, s} \to \mathcal{O}_{X, x}$ is flat and the $\mathcal{O}_{X, x}$-module $\Omega _{X/S, x}$ can be generated by at most $\dim _ x(X_{f(x)})$ elements.

  4. The local ring map $\mathcal{O}_{S, s} \to \mathcal{O}_{X, x}$ is flat and the $\kappa (x)$-vector space

    \[ \Omega _{X_ s/s, x} \otimes _{\mathcal{O}_{X_ s, x}} \kappa (x) = \Omega _{X/S, x} \otimes _{\mathcal{O}_{X, x}} \kappa (x) \]

    can be generated by at most $\dim _ x(X_{f(x)})$ elements.

  5. There exist affine opens $U \subset X$, and $V \subset S$ such that $x \in U$, $f(U) \subset V$ and the induced morphism $f|_ U : U \to V$ is standard smooth.

  6. There exist affine opens $\mathop{\mathrm{Spec}}(A) = U \subset X$ and $\mathop{\mathrm{Spec}}(R) = V \subset S$ with $x \in U$ corresponding to $\mathfrak q \subset A$, and $f(U) \subset V$ such that there exists a presentation

    \[ A = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c) \]

    with

    \[ g = \det \left( \begin{matrix} \partial f_1/\partial x_1 & \partial f_2/\partial x_1 & \ldots & \partial f_ c/\partial x_1 \\ \partial f_1/\partial x_2 & \partial f_2/\partial x_2 & \ldots & \partial f_ c/\partial x_2 \\ \ldots & \ldots & \ldots & \ldots \\ \partial f_1/\partial x_ c & \partial f_2/\partial x_ c & \ldots & \partial f_ c/\partial x_ c \end{matrix} \right) \]

    mapping to an element of $A$ not in $\mathfrak q$.

Proof. Note that if $f$ is smooth at $x$, then we see from Lemma 28.32.11 that (5) holds, and (6) is a slightly weakened version of (5). Moreover, $f$ smooth implies that the ring map $\mathcal{O}_{S, s} \to \mathcal{O}_{X, x}$ is flat (see Lemma 28.32.9) and that $\Omega _{X/S}$ is finite locally free of rank equal to $\dim _ x(X_ s)$ (see Lemma 28.32.12). Thus (1) implies (3) and (4). By Lemma 28.32.5 we also see that (1) implies (2).

By Lemma 28.31.10 the module of differentials $\Omega _{X_ s/s}$ of the fibre $X_ s$ over $\kappa (s)$ is the pullback of the module of differentials $\Omega _{X/S}$ of $X$ over $S$. Hence the displayed equality in part (4) of the lemma. By Lemma 28.31.12 these modules are of finite type. Hence the minimal number of generators of the modules $\Omega _{X/S, x}$ and $\Omega _{X_ s/s, x}$ is the same and equal to the dimension of this $\kappa (x)$-vector space by Nakayama's Lemma (Algebra, Lemma 10.19.1). This in particular shows that (3) and (4) are equivalent.

Algebra, Lemma 10.135.16 shows that (2) implies (1). Algebra, Lemma 10.138.3 shows that (3) and (4) imply (2). Finally, (6) implies (5) see for example Algebra, Example 10.135.8 and (5) implies (1) by Algebra, Lemma 10.135.7. $\square$

Lemma 28.32.15. Let

\[ \xymatrix{ X' \ar[r]_{g'} \ar[d]_{f'} & X \ar[d]^ f \\ S' \ar[r]^ g & S } \]

be a cartesian diagram of schemes. Let $W \subset X$, resp. $W' \subset X'$ be the open subscheme of points where $f$, resp. $f'$ is smooth. Then $W' = (g')^{-1}(W)$ if

  1. $f$ is flat and locally of finite presentation, or

  2. $f$ is locally of finite presentation and $g$ is flat.

Proof. Assume first that $f$ locally of finite type. Consider the set

\[ T = \{ x \in X \mid X_{f(x)}\text{ is smooth over }\kappa (f(x))\text{ at }x\} \]

and the corresponding set $T' \subset X'$ for $f'$. Then we claim $T' = (g')^{-1}(T)$. Namely, let $s' \in S'$ be a point, and let $s = g(s')$. Then we have

\[ X'_{s'} = \mathop{\mathrm{Spec}}(\kappa (s')) \times _{\mathop{\mathrm{Spec}}(\kappa (s))} X_ s \]

In other words the fibres of the base change are the base changes of the fibres. Hence the claim is equivalent to Algebra, Lemma 10.135.18.

Thus case (1) follows because in case (1) $T$ is the (open) set of points where $f$ is smooth by Lemma 28.32.14.

In case (2) let $x' \in W'$. Then $g'$ is flat at $x'$ (Lemma 28.24.6) and $g \circ f$ is flat at $x'$ (Lemma 28.24.4). It follows that $f$ is flat at $x = g'(x')$ by Lemma 28.24.12. On the other hand, since $x' \in T'$ (Lemma 28.32.5) we see that $x \in T$. Hence $f$ is smooth at $x$ by Lemma 28.32.14. $\square$

Here is a lemma that actually uses the vanishing of $H^{-1}$ of the naive cotangent complex for a smooth ring map.

Lemma 28.32.16. Let $f : X \to Y$, $g : Y \to S$ be morphisms of schemes. Assume $f$ is smooth. Then

\[ 0 \to f^*\Omega _{Y/S} \to \Omega _{X/S} \to \Omega _{X/Y} \to 0 \]

(see Lemma 28.31.9) is short exact.

Proof. The algebraic version of this lemma is the following: Given ring maps $A \to B \to C$ with $B \to C$ smooth, then the sequence

\[ 0 \to C \otimes _ B \Omega _{B/A} \to \Omega _{C/A} \to \Omega _{C/B} \to 0 \]

of Algebra, Lemma 10.130.7 is exact. This is Algebra, Lemma 10.137.1. $\square$

Lemma 28.32.17. Let $i : Z \to X$ be an immersion of schemes over $S$. Assume that $Z$ is smooth over $S$. Then the canonical exact sequence

\[ 0 \to \mathcal{C}_{Z/X} \to i^*\Omega _{X/S} \to \Omega _{Z/S} \to 0 \]

of Lemma 28.31.15 is short exact.

Proof. The algebraic version of this lemma is the following: Given ring maps $A \to B \to C$ with $A \to C$ smooth and $B \to C$ surjective with kernel $J$, then the sequence

\[ 0 \to J/J^2 \to C \otimes _ B \Omega _{B/A} \to \Omega _{C/A} \to 0 \]

of Algebra, Lemma 10.130.9 is exact. This is Algebra, Lemma 10.137.2. $\square$

Lemma 28.32.18. Let

\[ \xymatrix{ Z \ar[r]_ i \ar[rd]_ j & X \ar[d] \\ & Y } \]

be a commutative diagram of schemes where $i$ and $j$ are immersions and $X \to Y$ is smooth. Then the canonical exact sequence

\[ 0 \to \mathcal{C}_{Z/Y} \to \mathcal{C}_{Z/X} \to i^*\Omega _{X/Y} \to 0 \]

of Lemma 28.31.18 is exact.

Proof. The algebraic version of this lemma is the following: Given ring maps $A \to B \to C$ with $A \to C$ surjective and $A \to B$ smooth, then the sequence

\[ 0 \to I/I^2 \to J/J^2 \to C \otimes _ B \Omega _{B/A} \to 0 \]

of Algebra, Lemma 10.132.7 is exact. This is Algebra, Lemma 10.137.3. $\square$

Lemma 28.32.19. Let

\[ \xymatrix{ X \ar[rr]_ f \ar[rd]_ p & & Y \ar[dl]^ q \\ & S } \]

be a commutative diagram of morphisms of schemes. Assume that

  1. $f$ is surjective, and smooth,

  2. $p$ is smooth, and

  3. $q$ is locally of finite presentation1.

Then $q$ is smooth.

Proof. By Lemma 28.24.12 we see that $q$ is flat. Pick a point $y \in Y$. Pick a point $x \in X$ mapping to $y$. Suppose $f$ has relative dimension $a$ at $x$ and $p$ has relative dimension $b$ at $x$. By Lemma 28.32.12 this means that $\Omega _{X/S, x}$ is free of rank $b$ and $\Omega _{X/Y, x}$ is free of rank $a$. By the short exact sequence of Lemma 28.32.16 this means that $(f^*\Omega _{Y/S})_ x$ is free of rank $b - a$. By Nakayama's Lemma this implies that $\Omega _{Y/S, y}$ can be generated by $b - a$ elements. Also, by Lemma 28.27.2 we see that $\dim _ y(Y_ s) = b - a$. Hence we conclude that $Y \to S$ is smooth at $y$ by Lemma 28.32.14 part (2). $\square$

In the situation of the following lemma the image of $\sigma $ is locally on $X$ cut out by a regular sequence, see Divisors, Lemma 30.22.7.

Lemma 28.32.20. Let $f : X \to S$ be a morphism of schemes. Let $\sigma : S \to X$ be a section of $f$. Let $s \in S$ be a point such that $f$ is smooth at $x = \sigma (s)$. Then there exist affine open neighbourhoods $\mathop{\mathrm{Spec}}(A) = U \subset S$ of $s$ and $\mathop{\mathrm{Spec}}(B) = V \subset X$ of $x$ such that

  1. $f(V) \subset U$ and $\sigma (U) \subset V$,

  2. with $I = \mathop{\mathrm{Ker}}(\sigma ^\# : B \to A)$ the module $I/I^2$ is a free $A$-module, and

  3. $B^\wedge \cong A[[x_1, \ldots , x_ d]]$ as $A$-algebras where $B^\wedge $ denotes the completion of $B$ with respect to $I$.

Proof. Pick an affine open $U \subset S$ containing $s$ Pick an affine open $V \subset f^{-1}(U)$ containing $x$. Pick an affine open $U' \subset \sigma ^{-1}(V)$ containing $s$. Note that $V' = f^{-1}(U') \cap V$ is affine as it is equal to the fibre product $V' = U' \times _ U V$. Then $U'$ and $V'$ satisfy (1). Write $U' = \mathop{\mathrm{Spec}}(A')$ and $V' = \mathop{\mathrm{Spec}}(B')$. By Algebra, Lemma 10.137.4 the module $I'/(I')^2$ is finite locally free as a $A'$-module. Hence after replacing $U'$ by a smaller affine open $U'' \subset U'$ and $V'$ by $V'' = V' \cap f^{-1}(U'')$ we obtain the situation where $I''/(I'')^2$ is free, i.e., (2) holds. In this case (3) holds also by Algebra, Lemma 10.137.4. $\square$

The dimension of a scheme $X$ at a point $x$ (Properties, Definition 27.10.1) is just the dimension of $X$ at $x$ as a topological space, see Topology, Definition 5.10.1. This is not the dimension of the local ring $\mathcal{O}_{X,x}$, in general.

Lemma 28.32.21. Let $f : X \to Y$ be a smooth morphism of locally Noetherian schemes. For every point $x$ in $X$ with image $y$ in $Y$,

\[ \dim _ x(X) = \dim _ y(Y) + \dim _ x(X_ y), \]

where $X_ y$ denotes the fiber over $y$.

Proof. After replacing $X$ by an open neighborhood of $x$, there is a natural number $d$ such that all fibers of $X \to Y$ have dimension $d$ at every point, see Lemma 28.32.12. Then $f$ is flat (Lemma 28.32.9), locally of finite type (Lemma 28.32.8), and of relative dimension $d$. Hence the result follows from Lemma 28.28.6. $\square$

[1] In fact this is implied by (1) and (2), see Descent, Lemma 34.11.3. Moreover, it suffices to assume $f$ is surjective, flat and locally of finite presentation, see Descent, Lemma 34.11.5.

Comments (2)

Comment #3287 by Dennis Keeler on

For the first commutative diagram, I'm seeing the error message: parse error at or near "Y & \ar[l] V ". I've tried Mac Firefox and Safari.

Comment #3374 by on

Thanks! I've tried to fix this in this commit but as you perhaps know, the XyJax stuff is a bit shaky.


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 01V4. Beware of the difference between the letter 'O' and the digit '0'.