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I have a flat, quasicompact, and separated map $p : X \to \mathbb{A}^1$ and I know that $R^i p_* \mathcal{O}_X$ vanishes everywhere except possibly $0 \in \mathbb{A}^1$.

Q1: Does "cohomology and base change" automatically hold here? I.e., is the natural map $(R^i p_* \mathcal{O}_X)_s \otimes_{\mathcal{O}_{\mathbb{A}^1, s}} k(s) \to H^i(X_s, \mathcal{O}_{X, s})$ an isomorphism for all $s \in \mathbb{A}^1$? This is clear for all nonzero $s$ and I believe it should follow for $0 \in \mathbb{A}^1$ automatically.

Q2: Does it then follow that $R^i p_* \mathcal{O}_X = 0$? Theorem 1.1 in Conrad's notes would then imply that $R^i p_* \mathcal{O}_X$ was locally free, hence constant rank 0, hence 0.

Q3: If $X \overset{v}{\to} Y \to \mathbb{A}^1$ are proper maps and both $X$ and $Y$ are flat over $\mathbb{A}^1$ ($v$ is proper birational, n.n. flat), then does cohomology of $\mathcal{O}_X$ along $v$ commute with base change in $\mathbb{A}^1$? The versions of Cohomology and Base Change I've seen have one map which must be both proper and flat, as opposed to a composite being flat and the first map proper. This is my original situation, and I got to the above by localizing in $Y$.

I've been trying to do this descending induction, where $R^N p_* \mathcal{O}_X$ is zero for $N \gg 0$ for dimensional reasons, so $R^{N-1} p_* \mathcal{O}_X$ commutes with base change, but I also need $R^{N-2} p_* \mathcal{O}_X$ to commute with base change to conclude $R^{N-1}p_* \mathcal{O}_X$ was locally free, hence zero, and continue.

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  • $\begingroup$ It seems to me that you are assuming your hypothesis also for $i=0$, which is impossible — $p_*\mathscr{O}_X$ contains $\mathscr{O}_{\mathbb{A}^1}$. $\endgroup$
    – abx
    Sep 10 '20 at 5:18
  • $\begingroup$ What happens for the projection from $\mathbb{A}^3\setminus\{0\}$ to $\mathbb{A}^1$? $\endgroup$ Sep 10 '20 at 6:59
  • $\begingroup$ @abx I understand, but I'd like to at least get it for $i \geq 2$ if possible -- I'll take what I can get. $\endgroup$
    – Leo Herr
    Sep 10 '20 at 17:04
  • $\begingroup$ @PiotrAchinger Thank you for bringing up this example. I thought I checked it but I made a mistake. I don't think that arises in my situation because I got there by (Zariski) localizing a genuinely proper map $X \to Y$ in $Y$ and then I have an affine $Y$ over $\mathbb{A}^1$ (as in Q3). I don't believe $\mathbb{A}^3 \setminus 0$ can factor as a projective map to an affine over $\mathbb{A}^1$ because the map to an affine would have to factor through $\mathbb{A}^3$ and then the inclusion $\mathbb{A}^3 \setminus 0 \subseteq \mathbb{A}^3$ would have to be closed. I'll rewrite the question better. $\endgroup$
    – Leo Herr
    Sep 10 '20 at 17:09

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