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I am studying section III.9 on flat morphisms of Hartshorne's Algebraic Geometry and stuck in the proof of the following

Theorem 9.9 (Hartshorne, page 261). Let $T$ be an integral noetherian scheme. Let $X \subseteq \mathbb{P}^n_T$ be a closed subscheme. For each point $t \in T$, we consider the Hilbert polynomial $P_t \in \mathbb{Q}[z]$ of the fibre $X_t$ considered as a closed subscheme of $\mathbb{P}^n_{k(t)}$. Then $X$ is flat over $T$ if and only if the Hilbert polynomial $P_t$ is independent of $t$.

My situation is the following. $\mathscr{F}$ is a coherent sheaf on $X=\mathbb{P}^n_T$ for $T=Spec(A)$ with a local noetherian ring $A$ and I want to show that if $H^0(X, \mathscr{F}(m))$ is a free $A$-module of finite rank for $m\gg0$, then $\mathscr{F}$ is flat over $T$.

For this, Hartshorne defines a graded $A[X_0,\dots,X_n]$-module \begin{align*} M = \bigoplus_{m\geq m_0} H^0(X, \mathscr{F}(m)), \end{align*} where $m_0$ is choosen large enough, so that the $H^0(X, \mathscr{F}(m))$ are all free for $m \geq m_0$. (By the way: Don't we need the finiteness condition on the rank?) Then he claims that $\mathscr{F}=M^{\sim}$ by a Proposition (Prop. 5.15 in II.5), which states that there is a natural isomorphism $(\Gamma_*(\mathscr{F}))^{\sim} \cong \mathscr{F}$. But I don't see why this gives what he claims. He says "Note that $M$ is the same as $\Gamma_*(\mathscr{F})$ in degrees $m \geq m_0$." On this I agree with him, since by definition \begin{align*} \Gamma_*(\mathscr{F}) = \bigoplus_{m \in \mathbb{Z}} \Gamma(X, \mathscr{F}(m)), \end{align*} and $\Gamma(X, \mathscr{F}(m)) \cong H^0(X, \mathscr{F}(m))$. But what has happened to the parts of degree less than $m_0$ in the tilde-construction, so that he gets $M^{\sim}=(\Gamma_*(\mathscr{F}))^{\sim}$, which now would imply $M^{\sim} \cong \mathscr{F}$ by the Proposition mentioned above?

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  • $\begingroup$ In your situation, $F$ is flat if and only if $F(m)$ is flat for some $m$. $\endgroup$ – Mohan Jul 23 '17 at 12:51
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    $\begingroup$ Just don't use the Book of Hartshorne. $\endgroup$ – tiefi Jul 23 '17 at 15:20
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    $\begingroup$ Essentially, this will show that $H^0(F(m))$ for large $m$ is all that matters to the question. $\endgroup$ – Mohan Jul 23 '17 at 15:28
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    $\begingroup$ @tiefi: Well, that sounds great.. but what can we do.. $\endgroup$ – Jupp Jul 23 '17 at 15:28
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    $\begingroup$ @Mohan: Okay. But just because this it is not true that $M^{\sim}= (\Gamma_*(\mathscr{F}))^{\sim}$, is it? $\endgroup$ – Jupp Jul 23 '17 at 15:37
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The sheaf on $\mathbb{P}^n$ associated to a graded module $M$ only depends on what the module looks like in sufficiently large degrees. This is because sections of $\widetilde{M}$ are locally given by fractions $m/f$ where $m\in M_d$ and $f$ is a homogeneous polynomial of degree $d$ (which locally does not vanish). You can always assume that the degree $d$ is arbitrarily large in such a fraction, since you can multiply both $m$ and $f$ by a homogeneous polynomial of large degree (which locally does not vanish).

(To my surprise, Hartshorne appears to never actually explicitly mention this fact in the main text. However, it does appear in Exercise II.5.9.)

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    $\begingroup$ Thanks for your answer! Well, the fact that Hartshorne does not explicitly mention an important fact is not very suprising to me! ;-) I will think about this and then try it again. $\endgroup$ – Jupp Jul 23 '17 at 16:16
  • $\begingroup$ Do you mind giving my try a read through? Thanks in advance for your time! $\endgroup$ – Jupp Jul 26 '17 at 19:17
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(was too long to post it as a comment)

I think I now understand what is the meaning of what you said (or wanted to tell me), so I will give it a try.

We have a graded morphism of graded $A[X_0,\dots,X_n]$-modules \begin{align*} \varphi \colon M \longrightarrow \Gamma_*(\mathscr{F}), \end{align*} with the property that in degrees greater than or equal to $m_0$ it gives an isomorphism, say $\varphi_m$ for $m\geq m_0$, simply because \begin{align*} \varphi_m \colon H^0(X, \mathscr{F}(m)) \overset{\sim}{\longrightarrow} \Gamma(X, \mathscr{F}(m)). \end{align*} What we want to show is that $M^{\sim} \cong \mathscr{F}$. Since $\mathscr{F} \cong (\Gamma_*(\mathscr{F}))^{\sim}$ it sufficies to show that $M^{\sim} \cong (\Gamma_*(\mathscr{F}))^{\sim}$.

The morphism $\varphi$ induces a morphism \begin{align*} \varphi^{\sim} \colon M^{\sim} \longrightarrow (\Gamma_*(\mathscr{F}))^{\sim} \end{align*} of sheaves on $X = \mathbb{P}^n_{T} \cong \mathbb{P}^n_A = Proj(A[X_0, \dots , X_n])$. We check that $\varphi^{\sim}$ is an isomorphism on the principle open subsets $D_+(X_i)$, which give an open cover of $X$. Under the natural isomorphisms \begin{align*} M^{\sim}(D_+(X_i)) \cong M_{(X_i)} \;\;\; \text{and} \;\;\; (\Gamma_*(\mathscr{F}))^{\sim}(D_+(X_i)) \cong (\Gamma(\mathscr{F}))_{(X_i)}, \end{align*} we see that $\varphi^{\sim}_{D_+(X_i)}$ corresponds to the map \begin{align*} M_{(X_i)} &\longrightarrow (\Gamma_*(\mathscr{F}))_{(X_i)}, \\ \frac{\alpha_d}{X_i^d} &\longmapsto \frac{\varphi_d(\alpha_d)}{X_i^d}. \end{align*} This map is now surjective, since given any $\frac{\beta_d}{X_i^d} \in (\Gamma_*(\mathscr{F}))_{(X_i)}$, let $e \geq 0$, such that $d+e \geq m_0$, and consider the homogenous element $X_i^e \beta_d \in \Gamma_*(\mathscr{F})$ of degree $d+e$. Then there exists a unique element $\alpha_{d+e} \in M$ of degree $d+e$, such that $\varphi_{d+e}(\alpha_{d+e})=X_i^e \beta_d$, hence $\frac{\alpha_{d+e}}{X_i^{d+e}}$ gets mapped (under the map corresponding to $\varphi^{\sim}_{D_+(X_i)}$) to $\frac{X_i^e\beta_d}{X_i^{d+e}}=\frac{\beta_d}{X_i^d}$. This shows surjectivity. It is also straightforward to check that the map is injective (using the fact that for any homogeneous $\alpha_g \in M$ of degree $g$ one has $X_i^h\varphi_g(\alpha_g)=\varphi_{g+h}(X_i^h\alpha_g)$). By this we can now conclude that $\varphi^{\sim}$ is an isomorphism, hence the claim follows.

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  • $\begingroup$ Looks good to me! $\endgroup$ – Eric Wofsey Jul 26 '17 at 19:34
  • $\begingroup$ Great, thank you very much! $\endgroup$ – Jupp Jul 26 '17 at 19:35

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