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I am trying to understand the following statement.

Let $A$ be a noetherian commutative ring and $\mathfrak a\subset A$ is an ideal. Suppose that the ring $A/\mathfrak a$ is flat over $A$, then $V(\mathfrak a)$ is open in $\operatorname{Spec} (A)$. How to prove this (just using the standard definition of flatness)?

agleaner
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  • I don't think this is supposed to be easy. See e.g. here. – Zhen Lin Apr 01 '14 at 14:25
  • Thank you for this comment. This statement is given in a footnote on page 42 of Milne's "Primer in commutative algebra" http://www.jmilne.org/math/xnotes/ca.html , version version 3.00, May 2, 2013. Since Milne states this without any explanation I still believe that this should not be too hard... – agleaner Apr 01 '14 at 16:17
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    By the proof in http://math.stackexchange.com/questions/487214/is-r-nr-a-faithfully-flat-r-module $a=a^2$. If $A$ is noetherian, this implies that $V(a)$ is open. But in general this is false. – Cantlog Apr 01 '14 at 22:03
  • Cantlog, thanks a lot for your comment! Indeed $A$ is noetherian in Milne, so I changed the question and the link you gave answers this question completely. – agleaner Apr 01 '14 at 22:18

2 Answers2

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Since $A/\mathfrak{a}$ is $A$-flat, the ideal $\mathfrak{a}$ is generated by a single idempotent element $e$ (so $e^2=e$). This is a consequence of Nakayama's lemma and uses that $\mathfrak{a}$ is finitely generated (a consequence of the assumption that $A$ is Noetherian). (I'm pretty sure this is proved somewhere on the site.) You can now check that $\mathrm{Spec}(A)=D(e)\cup D(e-1)$ is a partition. Indeed, from $e^2=e$ we get $e(e-1)=0$, so for all prime ideals $\mathfrak{p}$, either $e\in\mathfrak{p}$ or $e-1\in\mathfrak{p}$, and we cannot have both (since $1\notin\mathfrak{p}$). So $D(e-1)=\mathrm{Spec}(A)\setminus D(e)=V(\mathfrak{a})$ is open.

Keenan Kidwell
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In general, if $A$ is any ring (not necessarily Noetherian), then $A/I$ is flat over $A$ iff $\text{Supp}(I) \cap \text{Supp}(A/I) = \emptyset$ iff $\text{Spec}(A) = \text{Supp}(I) \sqcup \text{Supp}(A/I)$. This follows from the fact that a flat local map is faithfully flat, hence injective, so if $p \in \text{Supp}(A/I) = V(I)$, then $A_p \to A_p/I_p$ is injective, so $I_p = 0$. Thus, $V(I)$ is open iff $\text{Supp}(I)$ is a closed set, which happens e.g. if $I$ is finitely generated.

zcn
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  • Thank you for this answer! Could you please tell me the definition of $\text{Supp}$? In particular, what is $\text{Supp}(I)$? – agleaner Apr 01 '14 at 23:21
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    For any $A$-module $M$, $\text{Supp}(M) := {p \in \text{Spec}(A) \mid M_p \ne 0}$. This is called the support of $M$, and is equal to $V(\text{ann}(M))$ if $M$ is finitely generated – zcn Apr 01 '14 at 23:27
  • I would like to ask you one more question. If $A$ is any ring, and $M$ is any module over $A$, is it correct that $\text{Supp}(M)$ is a closed subset of $\text{Spec}(A)$? If not, would could you please give me an example? – agleaner Apr 02 '14 at 13:20
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    This answer gives an example of a module whose support is not closed. To be honest though, I don't currently have an example of an ideal with non-closed support – zcn Apr 02 '14 at 22:43