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Let $\phi : A \to B$ be a morphism of rings and let $f : X \to S$ be the corresponding morphism of schemes. let $\mathfrak{q}$ be a prime ideal of $A$ corresponding to a point $s$ of $S$.

Let $T = Spec(\mathscr{O}_{S,s}/\mathfrak{m}_s^n)$ be a thickening of $s$ (we have a natural morphism $T \to S$). I would like to understand $f^{-1}(T) = X \times_S T$ in terms of commutative algebra.

In this mathoverflow answer (https://mathoverflow.net/questions/86048/verifying-claims-in-the-proof-of-the-rigidity-lemma-mumford-git) they say that $f^{-1}(t)$ corresponds to the ideal $Q = \phi(\mathfrak{q})B$ of $B$ and I think they say that $f^{-1}(T)$ corresponds to $Q^n$ but i'm not to sure how to see this (and I feel they probably use some hypothesis on f).

Community
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DonDon
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1 Answers1

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$f^{-1}(T)$ is the spectrum of $$B \otimes_A A_{\mathfrak{q}}/\mathfrak{q}^n A_\mathfrak{a} = B \otimes_A (A/\mathfrak{q}^n)_{\mathfrak{q}} \cong (B/\mathfrak{q}^n B)_{\mathfrak{q}}$$ In general, this localization is necessary. However, when $s$ is closed i.e. $\mathfrak{q}$ is maximal, we don't need it: Then $A/\mathfrak{q}^n$ is local with maximal ideal $\mathfrak{q}/\mathfrak{q}^n$, so that every element outside of it is invertible. This means that we localize at units, i.e. nothing changes. Therefore we get $B/\mathfrak{q}^n B$ in that case.

Martin Brandenburg
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  • Thanks a lot ! what's weird is that in the question I link they don't seem to use the fact that $s$ is closed. They "only" assume that $p$ is flat, and geometrically integral and that $S$ is connected. (Sorry I can't vote up your answer, I don't have the reputation) – DonDon Feb 01 '14 at 22:50
  • He also assumes that $f$ is a closed map. – DonDon Feb 02 '14 at 09:55