I'm having some trouble with Exercise III 4.2 a) in Hartshorne's Algebraic geometry. It is

Let $f: X \to Y$ be a finite surjective morphism of integral noetherian schemes. Show that there is a coherent sheaf $\mathscr{M}$ on $X$, and a morphism of sheaves $\alpha: \mathcal{O}^r_Y \to f_* \mathscr{M}$ for some $r > 0$, such that $\alpha$ is an isomorphism at the generic point.

I solved the affine case: If $Y = \text{Spec } A$, then $X = \text{Spec }B$ and $f$ comes from a homomorphism of integral domains $A \to B$, where $B$ is finitely generated as an $A$-algebra. This means there is a surjection $A^n \to B$ of $A$-modules. Localizing is exact, so at the generic point this is still a surjection $Q(A)^n \to B_{(0)}$. But $Q(A)$ is a field, so we can choose a basis $Q(A)^r \subset Q(A)^n$, which gives an isomorphism $Q(A)^r \to B_{(0)}$. The corresponding map $A^r \to B$ is the desired morphism.

So the "obvious" generalization would be to take the structure sheaf $\mathscr{M} = \mathcal{O}_X$. But this cannot work, as the following counterexample illustrates:

Consider $f: \mathbb{P}^1 \to \mathbb{P}^1$, defined by $[x:y] \mapsto [x^2:y^2]$. Giving a morphism $\alpha: \mathcal{O}_Y^r \to f_* \mathscr{M}$ is the same as choosing $r$ elements from $\Gamma(X, \mathscr{M})$. But $\Gamma(X, \mathcal{O}_X) = k$, so we can only choose $k$-linearly dependent elements. No morphism $\mathcal{O}_Y^r \to f_*\mathcal{O}_X$ can give an isomorphism at the generic point $\eta$, because $(f_*\mathcal{O}_X)_\eta \cong \mathcal{O}_{Y,\eta}^2$, and the image of $\mathcal{O}_Y^r \to f_*\mathcal{O}_X $ will only be $1$-dimensional at $\eta$.

I'm a bit clueless at this point. Maybe one has to glue the affine case together and yields an invertible sheaf $\mathscr{M}$? But I don't think that replacing $\mathcal{O}_X$ by $\mathcal{O}_X(d)$ really changes anything in the example. How to proceed here?

  • $\begingroup$ could you kindly clarify your argument "But $\Gamma$..."? If you take $P^1_{\mathbb{C}}$ and take $F_1=O$, $F_2=O(1)$, then we have a morphism $F_1^{\oplus 2}\rightarrow F_2$ which is surjective on global sections. $\endgroup$ – user251240 Apr 23 at 1:36
  • $\begingroup$ also, can't you just run the following argument? $f$ is affine, so we can pick an affine open neighbourhood of the generic point of $Y$ (which is a Noetherian integral domain), its inverse image is also an affine open. Stalks can be evaluated on non-empty open sets, so your reasoning for affine case does the job. $\endgroup$ – user251240 Apr 23 at 1:56
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    $\begingroup$ I have reached a somewhat disturbing conclusion. Some googling (math.stackexchange.com/questions/2432718/…) convinced me that a coherent sheaf on $P^1$ is a direct sum of a locally free sheaf and a torsion sheaf (i.e. supported at a finite number of closed points, in our case). Coherent locally free sheaves on $P^1$ are direct sums of locally free sheaves of rank 1 (depending on your preference, that's Birkhoff or Grothendieck or whatever). $\endgroup$ – user251240 Apr 23 at 18:31
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    $\begingroup$ The torsion summand should not matter because it does not contribute to the generic stalk. Your argument applies to locally free sheaves of rank 1, and by splitting, to all coherent locally free sheaves. If somebody could set me straight, that would be great. $\endgroup$ – user251240 Apr 23 at 18:31
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    $\begingroup$ @schematic_boi I did the calculation in the $\mathbb{P}^1 \to \mathbb{P}^1$ example and I think it actually works out. See my answer for this. $\endgroup$ – red_trumpet Apr 23 at 20:05

Copied from my answer on mathoverflow:

Suppose we can solve the problem for affine schemes, and choose an open affine $j: V \hookrightarrow Y$ and let $i: U \hookrightarrow X$ be its preimage. Suppose we have a morphism $$ \alpha_V: \mathcal{O}_V^n \to (f|_U)_*\mathcal{O}_U$$ which is an isomorphism at the generic point $\eta$ of $Y$.

The problem is that $i_* \mathcal{O}$ is not in general coherent on $X$. But $\alpha_V$ chooses $n$ global sections $s_1,\dotsc,s_n \in \Gamma(X, i_* \mathcal{O}_U) = \Gamma(U, \mathcal{O}_U)$, which can be used to define a morphism $\alpha_X: \mathcal{O}_X^n \to i_* \mathcal{O}_U$. Let $\mathscr{G}$ be the image of this morphism. Then $\mathscr{G}$ is coherent, because for every open affine $\text{Spec }A = W \subset X$, $\mathscr{G}(W) \subset (i_*\mathcal{O}_U)(W)$ is the $A$-submodule generated by $s_1|_W,\dotsc,s_n|_W$.

This allows us to define the morphism $$ \alpha_Y: \mathcal{O}_Y^n \to f_* \mathscr{G} \subset (f i)_* \mathcal{O}_U$$ by takting the same global sections $s_1,\dots,s_n \in \Gamma(Y, f_* \mathscr{G})$. At the generic point this yields $$ \mathcal{O}^n_{Y, \eta} \xrightarrow{\alpha_{Y, \eta}} (f_*\mathscr{G})_\eta \hookrightarrow ((fi)_* \mathcal{O}_U)_\eta, $$ and the composition is $\alpha_{U, \eta}$ which is an isomorphism. Hence $\alpha_{Y, \eta}$ is an isomorphism as well.

This is not a full answer, I just think that in the example $f: \mathbb{P}^1 \to \mathbb{P}^1, [x_0:x_1] \mapsto [x_0^2:x_1^2]$ the morphism $\alpha: \mathcal{O}_{\mathbb{P}^1}^2 \to f_*(\mathcal{O}_{\mathbb{P}^1}(1))$, which chooses the two generators of $\Gamma(\mathbb{P}^1, \mathcal{O}(1))$ does in fact work:

To differentiate the two instances of $\mathbb{P}^1$, let $x_0, x_1$ be the coordinates on the domain $X$ of $f$, and let $y_0, y_1$ be the coordinates on the image $Y = \mathbb{P}^1$. Then $\Gamma(X, \mathcal{O}(1)) = \langle x_0, x_1 \rangle_k$, and $\alpha: \mathcal{O}_{\mathbb{P}^1}^2 \to f_* \mathcal{O}_{\mathbb{P}^1}(1)$ is defined on global sections by sending the generator $e_i$ to $x_i$.

Choosing $U = D_+(y_1) \subset Y$, a principle open affine, we have $V = f^{-1}(U) = D_+(x_1)$. We want to study $\alpha$ restricted to $U$. Thus $U = \text{Spec } k\left[\frac{y_0}{y_1}\right]$, and $V = \text{Spec } k\left[\frac{x_0}{x_1}\right]$. The homomorphism of rings determining $f: V \to U$ is given by $$ \frac{y_0}{y_1} \mapsto \left(\frac{x_0}{x_1}\right)^2.$$ The restriction $\langle x_0, x_1 \rangle_k = \Gamma(X, \mathcal{O}(1)) \to \Gamma(V, \mathcal{O}(1)|_V) = k\left[\frac{x_0}{x_1}\right]$ is given by mapping $x_0 \mapsto \frac{x_0}{x_1}$, and $x_1 \mapsto 1$.

If we consider $k\left[\frac{x_0}{x_1}\right]$ as a $k\left[\frac{y_0}{y_1}\right]$-module we see that it has two generators, namely $1$ and $\frac{x_0}{x_1}$. This is because with the $k\left[\frac{y_0}{y_1}\right]$-action we only hit the even degrees (or odd degrees, depending where you start).

Both generators are hit by $\alpha|_U$, which is therefor surjective on $U$-sections (I think it even is an isomorphism).

  • $\begingroup$ Some additional info: if you take a degree $n$ self-map for $P^1$, then $f_*O(n-1)=O(n-1)^{\oplus r}$ for some $r$ I am too lazy to compute. Shown here: mathoverflow.net/a/329812/138661 $\endgroup$ – user251240 Apr 24 at 4:41

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