$\textbf{Question: }$ Suppose $X \xrightarrow{\phi} Y$ is a proper faithfully flat map of noetherian schemes and let $X \xrightarrow{f} Y' \xrightarrow{g} Y$ be the Stein factorization. Is either of $g$ or $f$ necessarily flat? I’d really like to see examples if these are false. In general, for a composite as above with $f$ faithfully flat, we have that $g$ is faithfully flat or flat iff $\phi$ is.

More generally, it seems like if $\phi = g \circ f$ is a factorization of a proper flat map with $g$ finite then $g$ is also flat. But I can't prove this.

This came up when I was thinking about how I would prove the connectedness principle in algebraic geometry (Let $X\rightarrow Y$ be a proper faithfully flat map with $Y$ the spectrum of a DVR. Then if the generic fiber is connected, so is the special fiber), and you can prove this with ZMT. The above result would give you a different proof. If $f$ in the Stein factorization above is faithfully flat then so is $g$ (by general properties) but because the generic fiber is connected $g$ is also an isomorphism on an open set, hence an isomorphism.

Googling around shows that it also came up in the comments here Global sections of flat scheme also flat? but without a resolution.

  • $\begingroup$ This might or might not be helpful: If $Y$ is reduced, then the flatness of $f_*\mathcal{O}_X$ is equivalent to its formation being compatible with arbitrary base-change. $\endgroup$ – Keerthi Madapusi Pera Jan 8 '13 at 4:12
  • 1
    $\begingroup$ @Keerthi: What you say isn't true; e.g., if $Y$ is a dvr then $f_{\ast}O_X$ is always flat but its formation need not commute with any base change. $\endgroup$ – user29720 Jan 8 '13 at 9:04
  • $\begingroup$ @LMN: Your statement of the connectedness principle is incorrect: let $Y = {\rm{Spec}}(R)$ for a dvr $R$ with fraction field $K$ and let $X = {\rm{Spec}}(R')$ for the integral closure $R'$ of $R$ in a quadratic Galois extension $K'/K$ in which the place of $R$ on $K$ is totally split. The connectedness principle is essentially a deep result of Zariski which historically is related to the origin of Zariski's "Main Theorem". It seems unlikely that there is a proof of the connectedness principle based on a flatness principle; I expect your question has a negative answer, but have no proof yet. $\endgroup$ – user29720 Jan 8 '13 at 9:11
  • 1
    $\begingroup$ @LMN: Tong's answer in the MO thread you linked to is a counterexample for you (take his $X_n$ which is proper faithfully flat over $R/\pi^nR$). $\endgroup$ – Qing Liu Jan 8 '13 at 15:03
  • $\begingroup$ @kreck: Thanks for the correction. I should also say $\dim_{k(Y_0)} H^0(\mathcal{O}_{X_0}) = 1$ where $Y_0$ is the generic point of $Y$ and $X_0$ the fiber of the generic point. The proof (below) also works for the version of the connectedness thm. in Hartshorne (Ex III.11.4). There, assuming $k = \bar{k}$ you automatically have the dimension condition above, (because an open set of points have $k$ as the residue field) which is how I forgot it. $\endgroup$ – LMN Jan 8 '13 at 19:15

It seems to me that your "general guess" is not true, because of the following example.

A nodal cubic $Y$ has an étale (hence flat) double cover $\phi \colon X \to Y$, where $X$ is the union of two copies of the normalization $Y^{\nu}$ of $Y$, intersecting at two points.

Then $\phi$ factors as $\phi=\nu \circ f$, where $f \colon X \to Y^{\nu}$ is a non-normal double cover branched at $\nu^{-1}(p)$, where $p \in Y$ is the node, and $\nu \colon Y^{\nu} \to Y$ is the normalization map.

In particular, $\nu$ is finite but not flat.

Of course, this does not give any counterexample about your question on Stein factorization, since the Stein factorization of $\phi \colon X \to Y$ is $\phi$ itself.

| cite | improve this answer | |

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.