The Stacks project

30.13 Finite morphisms and affines

In this section we use the results of the preceding sections to show that the image of a Noetherian affine scheme under a finite morphism is affine. We will see later that this result holds more generally (see Limits, Lemma 32.11.1 and Proposition 32.11.2).

Lemma 30.13.1. Let $f : Y \to X$ be a morphism of schemes. Assume $f$ is finite, surjective and $X$ locally Noetherian. Let $Z \subset X$ be an integral closed subscheme with generic point $\xi $. Then there exists a coherent sheaf $\mathcal{F}$ on $Y$ such that the support of $f_*\mathcal{F}$ is equal to $Z$ and $(f_*\mathcal{F})_\xi $ is annihilated by $\mathfrak m_\xi $.

Proof. Note that $Y$ is locally Noetherian by Morphisms, Lemma 29.15.6. Because $f$ is surjective the fibre $Y_\xi $ is not empty. Pick $\xi ' \in Y$ mapping to $\xi $. Let $Z' = \overline{\{ \xi '\} }$. We may think of $Z' \subset Y$ as a reduced closed subscheme, see Schemes, Lemma 26.12.4. Hence the sheaf $\mathcal{F} = (Z' \to Y)_*\mathcal{O}_{Z'}$ is a coherent sheaf on $Y$ (see Lemma 30.9.9). Look at the commutative diagram

\[ \xymatrix{ Z' \ar[r]_{i'} \ar[d]_{f'} & Y \ar[d]^ f \\ Z \ar[r]^ i & X } \]

We see that $f_*\mathcal{F} = i_*f'_*\mathcal{O}_{Z'}$. Hence the stalk of $f_*\mathcal{F}$ at $\xi $ is the stalk of $f'_*\mathcal{O}_{Z'}$ at $\xi $. Note that since $Z'$ is integral with generic point $\xi '$ we have that $\xi '$ is the only point of $Z'$ lying over $\xi $, see Algebra, Lemmas 10.35.3 and 10.35.20. Hence the stalk of $f'_*\mathcal{O}_{Z'}$ at $\xi $ equal $\mathcal{O}_{Z', \xi '} = \kappa (\xi ')$. In particular the stalk of $f_*\mathcal{F}$ at $\xi $ is not zero. This combined with the fact that $f_*\mathcal{F}$ is of the form $i_*f'_*(\text{something})$ implies the lemma. $\square$

Lemma 30.13.2. Let $f : Y \to X$ be a morphism of schemes. Let $\mathcal{F}$ be a quasi-coherent sheaf on $Y$. Let $\mathcal{I}$ be a quasi-coherent sheaf of ideals on $X$. If the morphism $f$ is affine then $\mathcal{I}f_*\mathcal{F} = f_*(f^{-1}\mathcal{I}\mathcal{F})$.

Proof. The notation means the following. Since $f^{-1}$ is an exact functor we see that $f^{-1}\mathcal{I}$ is a sheaf of ideals of $f^{-1}\mathcal{O}_ X$. Via the map $f^\sharp : f^{-1}\mathcal{O}_ X \to \mathcal{O}_ Y$ this acts on $\mathcal{F}$. Then $f^{-1}\mathcal{I}\mathcal{F}$ is the subsheaf generated by sums of local sections of the form $as$ where $a$ is a local section of $f^{-1}\mathcal{I}$ and $s$ is a local section of $\mathcal{F}$. It is a quasi-coherent $\mathcal{O}_ Y$-submodule of $\mathcal{F}$ because it is also the image of a natural map $f^*\mathcal{I} \otimes _{\mathcal{O}_ Y} \mathcal{F} \to \mathcal{F}$.

Having said this the proof is straightforward. Namely, the question is local and hence we may assume $X$ is affine. Since $f$ is affine we see that $Y$ is affine too. Thus we may write $Y = \mathop{\mathrm{Spec}}(B)$, $X = \mathop{\mathrm{Spec}}(A)$, $\mathcal{F} = \widetilde{M}$, and $\mathcal{I} = \widetilde{I}$. The assertion of the lemma in this case boils down to the statement that

\[ I(M_ A) = ((IB)M)_ A \]

where $M_ A$ indicates the $A$-module associated to the $B$-module $M$. $\square$

Lemma 30.13.3. Let $f : Y \to X$ be a morphism of schemes. Assume

  1. $f$ finite,

  2. $f$ surjective,

  3. $Y$ affine, and

  4. $X$ Noetherian.

Then $X$ is affine.

Proof. We will prove that under the assumptions of the lemma for any coherent $\mathcal{O}_ X$-module $\mathcal{F}$ we have $H^1(X, \mathcal{F}) = 0$. This will in particular imply that $H^1(X, \mathcal{I}) = 0$ for every quasi-coherent sheaf of ideals of $\mathcal{O}_ X$. Then it follows that $X$ is affine from either Lemma 30.3.1 or Lemma 30.3.2.

Let $\mathcal{P}$ be the property of coherent sheaves $\mathcal{F}$ on $X$ defined by the rule

\[ \mathcal{P}(\mathcal{F}) \Leftrightarrow H^1(X, \mathcal{F}) = 0. \]

We are going to apply Lemma 30.12.8. Thus we have to verify (1), (2) and (3) of that lemma for $\mathcal{P}$. Property (1) follows from the long exact cohomology sequence associated to a short exact sequence of sheaves. Property (2) follows since $H^1(X, -)$ is an additive functor. To see (3) let $Z \subset X$ be an integral closed subscheme with generic point $\xi $. Let $\mathcal{F}$ be a coherent sheaf on $Y$ such that the support of $f_*\mathcal{F}$ is equal to $Z$ and $(f_*\mathcal{F})_\xi $ is annihilated by $\mathfrak m_\xi $, see Lemma 30.13.1. We claim that taking $\mathcal{G} = f_*\mathcal{F}$ works. We only have to verify part (3)(c) of Lemma 30.12.8. Hence assume that $\mathcal{J} \subset \mathcal{O}_ X$ is a quasi-coherent sheaf of ideals such that $\mathcal{J}_\xi = \mathcal{O}_{X, \xi }$. A finite morphism is affine hence by Lemma 30.13.2 we see that $\mathcal{J}\mathcal{G} = f_*(f^{-1}\mathcal{J}\mathcal{F})$. Also, as pointed out in the proof of Lemma 30.13.2 the sheaf $f^{-1}\mathcal{J}\mathcal{F}$ is a quasi-coherent $\mathcal{O}_ Y$-module. Since $Y$ is affine we see that $H^1(Y, f^{-1}\mathcal{J}\mathcal{F}) = 0$, see Lemma 30.2.2. Since $f$ is finite, hence affine, we see that

\[ H^1(X, \mathcal{J}\mathcal{G}) = H^1(X, f_*(f^{-1}\mathcal{J}\mathcal{F})) = H^1(Y, f^{-1}\mathcal{J}\mathcal{F}) = 0 \]

by Lemma 30.2.4. Hence the quasi-coherent subsheaf $\mathcal{G}' = \mathcal{J}\mathcal{G}$ satisfies $\mathcal{P}$. This verifies property (3)(c) of Lemma 30.12.8 as desired. $\square$


Comments (4)

Comment #953 by correction_bot on

In the second sentence of the proof delete "Since", and in the third sentence delete "will".

Comment #961 by on

Many thanks for all the comments 943 -- 953. These are fixed here.

Comment #4057 by Laurent Moret-Bailly on

At the beginning, it would be better to refer to 05YU than 01ZT (the former being more general).


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 01YN. Beware of the difference between the letter 'O' and the digit '0'.