Push-forward of a quasi-coherent graded algebra under a proper map

Question

Let $f\colon X \rightarrow Y$ be a proper morphism with $Y$ Noetherian (and even affine, if you wish), and let $\mathscr{A} = \bigoplus_{n \ge 0} \mathscr{A}_n$ be a quasi-coherent graded $\mathscr{O}_X$-algebra of finite type. Is the quasi-coherent graded $\mathscr{O}_Y$-algebra $f_* \mathscr{A}$ of finite type?

An analogue of this for graded modules is EGA III, 3.3.1. Perhaps the version for algebras that I am asking about can somehow be deduced, but I could't see how. I'd be happy to accept any precise reference (as well as any proof or counterexample, of course) as an answer.

Answer 1

No, that is not true, and there are several similar (and standard) examples. For one, let $Y$ be $\text{Spec}(k)$, let $X$ be an elliptic curve over $k$ (there are many other varieties one could use), let $f$ be the unique $k$-morphism, let $\mathcal{L}$ be an invertible sheaf on $X$ that is algebraically equivalent to $0$ but not torsion, and let $\mathcal{M}$ be an invertible sheaf of positive degree. Let $\mathcal{E}$ be $\mathcal{L}\oplus \mathcal{M}$, and let $\mathcal{A}$ be the symmetric algebra, $\text{Sym}^{\bullet}_{\mathcal{O}_X}(\mathcal{E})$. Thus, we have a direct sum decomposition, $$ \mathcal{A}_n = \text{Sym}^n_{\mathcal{O}_X}(\mathcal{L}\oplus \mathcal{M}) = \bigoplus_{l,m\geq 0, \ l+m=n} \left(\mathcal{L}^{\otimes l}\otimes_{\mathcal{O}_X} \mathcal{M}^{\otimes m}\right). $$ As is evident from this decomposition, $\mathcal{A}$ is actually $\mathbb{Z}_{\geq 0} \times \mathbb{Z}_{\geq 0}$-graded; the decomposition is, $$ \mathcal{A} = \bigoplus_{l,m\geq 0} \mathcal{A}_{l,m}, \ \ \mathcal{A}_{l,m} = \mathcal{L}^{\otimes l}\otimes_{\mathcal{O}_X} \mathcal{M}^{\otimes m}. $$ Since cohomology commutes with direct sum of coherent sheaves on a Noetherian scheme, we have a $\mathbb{Z}_{\geq 0}\times \mathbb{Z}_{\geq 0}$-grading of the corresponding algebra of global sections, $$ H^0(X,\mathcal{A}) = \bigoplus_{l,m\geq 0} H^0(X,\mathcal{A}_{l,m}), \ \ H^0(X,\mathcal{A}_{l,m}) = H^0(X,\mathcal{L}^{\otimes l}\otimes_{\mathcal{O}_X} \mathcal{M}^{\otimes m}). $$ Using Riemann-Roch and the hypotheses on $\mathcal{L}$ and $\mathcal{M}$, the graded piece $H^0(X,\mathcal{A}_{l,m})$ is nonzero if and only if either $(l,m) = (0,0)$ or $m>0$. In particular, this gives a subsemigroup of $\mathbb{Z}_{\geq 0} \times \mathbb{Z}_{\geq 0}$ that is not finitely generated as a subsemigroup. Therefore $H^0(X,\mathcal{A})$ cannot be finitely generated as a graded $k$-algebra.

Edit. Probably this example has appeared previously on MathOverflow -- it is a standard example. There are other examples with various other interesting features.

Answer 2

What you're overlooking for the motivating situation (which should have been mentioned in the question as posed so that readers would have more context) is the final part of EGA III$_1$ 3.3.1. (Note that $Y'$ and $g$ there play no role whatsoever, by the way.)

In the setup of that result, set $\mathscr{S}$ to be the symmetric algebra over $\mathscr{O}_Y$ for the coherent sheaf $f_{\ast}(\mathscr{L}^{\otimes d})$ for an integer $d > 0$ (to be determined); this is graded quasi-coherent algebra of finite type on $Y$, generated in degree 1. Take $\mathscr{M}$ there to be $\mathscr{A} := \oplus_{n \ge 0} \mathscr{L}^{\otimes n}$ and set $p=0$. Observe that $\mathscr{M}$ is naturally a graded module over $f^{\ast}(\mathscr{S})$ by defining $\mathscr{M}_n = \oplus_{nd \le j < (n+1)d} \mathscr{L}^{\otimes j}$.

Hence, by (3.3.1.1) in loc. cit. if $\mathscr{M}$ is of finite type as an $f^{\ast}(\mathscr{S})$-module then there is an integer $k_0 > 0$ so that for $k\ge k_0$ we have $$\oplus_{(k+r)d \le j < (k+r+1)d} f_{\ast}(\mathscr{L}^{\otimes j})= \mathscr{S}_r \cdot \oplus_{kd \le j < (k+1)d} f_{\ast}(\mathscr{L}^{\otimes j})$$ for all $r \ge 0$. This means the natural multiplication map ${\rm{Sym}}^r(f_{\ast}(\mathscr{L}^{\otimes d})) \otimes f_{\ast}(\mathscr{L}^{\otimes j}) \rightarrow f_{\ast}(\mathscr{L}^{\otimes(rd+j)})$ is surjective for all $r \ge 0$ and all $j \ge k_0d$. It would then follow that the quasi-coherent $\mathscr{O}_Y$-algebra $f_{\ast}(\mathscr{A}) = \oplus_{n \ge 0} f_{\ast}(\mathscr{L}^{\otimes n})$ is generated by its (coherent) terms in degrees $\le k_0d$ as an $\mathscr{O}_Y$-algebra, so it would be of finite type as an $\mathscr{O}_Y$-algebra.

Thus, one just has to find $d>0$ for which the finite type graded $\mathscr{O}_X$-algebra $\oplus_{n \ge 0} \mathscr{L}^{\otimes n}$ is finite type as a graded module over the symmetric algebra on $f^{\ast}(f_{\ast}(\mathscr{L}^{\otimes d}))$. For this purpose it is certainly enough that $\mathscr{L}^{\otimes d}$ is generated by $f_{\ast}(\mathscr{L}^{\otimes d})$. In other words, it suffices that the natural map $f^{\ast}(f_{\ast}(\mathscr{L}^{\otimes d})) \rightarrow \mathscr{L}^{\otimes d}$ is surjective for some $d > 0$. That in turn is exactly one of the hypotheses in the motivating situation (which is presently not given in the question as posed).