1. Is any family of hypersurfaces of fixed degree d (in a projective space of dimension n) over a non-reduced base flat? This is true over reduced bases (assume everything Noetherian), since the Hilbert polynomial is constant. Is there any criterion of flatness of the form: If the Hibert polynomial is constant and (extra conditions that do not assume the base is reduced) then the family is flat?

  2. Is there any result that guarantees that push forwards and higher push-forwards (via a proper mophism of Noetherian schemes) of a flat over the base sheaf are locally free if the base is non-reduced? This is about the Cohomology and Base Change statements that appear in Mumford (Abelian Varieties) and Hartshorne. In both places the base is assumed to be reduced. However, it seems that this is used often even when the base is non-reduced (such as when constructing the Hilbert scheme - although I must be misunderstanding this proof).

Thanks a lot!

  • $\begingroup$ I wouldn't have thought that the family "p=0" over the ring Z/p^2Z would be flat, so you might need extra conditions in 1. $\endgroup$ – Kevin Buzzard Feb 26 '10 at 7:40

(1) A family of hypersurfaces of degree $d$ over an affine scheme $Spec\ R$ means the following: it is a closed subscheme of $\mathbb P^n_R$ given by a homogeneous polynomial $f(x_0,\dots,x_n)$ of degree $d$ satisfying the following condition:

Every fiber is a hypersurface of degree $d$. This means that for every prime ideal $m\subset R$, the reduction $\bar f\in k[x_0,\dots,x_n]$ is a polynomial of degree $d$, where $k=R/m$.

Now, let $m$ be a maximal ideal, so that $k$ is a field, and look at the graded ring $k[x_0,\dots,x_n]/(f)$. For each $a\ge d$, the degree-$a$ part is a vector space of dimension $\binom{a+n}{n}-\binom{a-d+n}{n}$. Thus, some $\binom{a-d+n}{n}$ monomials can be written as linear combinations, with coefficients in $k$, of the remaining monomials. This is done by solving a system of linear equations obtained by setting $x^m f=0$, where $x^m$ are monomials of degree $a-d$.

Now, consider the same system of linear equations with coefficients in $R$. For each of the $\binom{a-d+n}{n}$ monomials as above you get a principal minor $M$ of your matrix, and the reduction of $\det M$ in $k$ is not zero. Thus, over an open set $Spec\ R[1/\det M]$, this determinant is invertible, and the monomial can be eliminated.

Thus, over an open neighborhood $Spec\ A$ of the point $[m]\in Spec\ R$, the degree-$a$ part of the ring $A[x_0,\dots,x_n]/(f)$ is a free $A$-module. Recalling how $Proj$ is covered by $Spec$'s and that a free module is flat, this implies that $Proj\ S[x_0,\dots,x_n]/(f)$ is flat over $Spec\ S$. (We will assume $R$ and so $A$ to be Noetherian here for simplicity.)

This is the main trick for proving flatness over a non-reduced base: you prove freeness instead. For a finitely generated module over a Noetherian ring, flatness and freeness are equivalent. So for a projective morphism $f:X\to Y$ a coherent sheaf $F$ on $X$ is flat over $Y$ iff the sheaves $f_* F(a)$ on $X$ are locally free for $a\gg0$.

In (2), you are mistaken about Hartshorne: Theorem III.12.11 (Cohomology and Base Change) has no assumption for the base to be reduced. So if $H^i(X_y,F_y)=0$ for $i>0$ then $f_*F$ is locally free, for any (Noetherian) base $Y$ and coherent sheaf $F$ on $X$, flat over $Y$.

For higher direct images, $H^i(X_y,F_y)=0$ for $i\ge i_0$ implies that $R^{i_0}f_*F=0$. But you can have $H^i(X_y,F_y)=0$ for $i> i_0$ and $H^{i_0}(X_y,F_y)$ non-constant, and still have $R^{i_0}f_*F=0$ (compare the Poincare line bundle on $A\times A^t$, as in Fourier-Mukai).


The answer of VA is quite simple to understand. In fact the result is of local nature.

Proposition: Let $R$ be a noetherian (commutative unitary) ring, let $B$ be a flat noetherian $R$-algebra, and let $f\in B$ be an element such that for any maximal ideal $m$ of $R$, the image of $f$ in $B/mB=B\otimes_R R/m$ is a regular element. Then $B/fB$ is flat over $R$.

This can be found for example in Matsumura, page 151, (20.F) (taking $M=B$). It is also in Milne's "Etale cohomology", first chapter.

To apply to your concrete situation, $B$ is a polynomial ring over $R$, and $f$ is a polynomial which is non-zero modulo $m$ (cf. explanation of VA), so it is regular modulo $m$ because $B/mB$ is an integral domain. Therefore $B/fB$ is flat over $R$.

  • $\begingroup$ I think this is only true for $R$ and $B$ local rings. To reduce to the local case, we need that $f$ is regular in $B/(n \cap R)B$ for all maximal ideal $n$ of $B$, cf Matsumura, Commutative Ring Theory, Theorem 22.6. $\endgroup$ – Damien Robert Jul 30 '20 at 9:41

In response to #2: you mention that cohomology and base change in Mumford and Hartshorne requires reducedness of the target. I don't have either with me right now, but I agree with VA: it's possible that you saw versions in these books that had this hypothesis (the proof is easier in this case --- I've seen it called Grauert's Theorem), but that there is a later version (in the same sources) that works for a nonreduced base. Here is one possible statement (whose hypotheses can be relaxed as usual, e.g. Noetherian hypotheses often can be traded in for finite presentation ones). Suppose $\pi: X \rightarrow Y$ is a proper morphism to a locally Noetherian scheme, $\mathcal{F}$ is coherent and flat over $Y$, and for each point $y \in Y$, the natural map $\phi^p_y$ from the fiber of the $p$th pushforward to the $p$th cohomology of the fiber is surjective. Then (i) for any $\rho: Z \rightarrow Y$, restriction to $Z$ commutes with the $p$th pushforward. In particular, $\phi^p_y$ is an isomorphism. (ii) Furthermore, $\phi^{p-1}_y$ is surjective for all $y$ if and only if the $R^p \pi \mathcal{F}$ is locally free. This in turn implies that $h^p$ is constant.

Better discussion and proof is given here (see for example 25.8-9 in the October 21, 2011 version).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.