flatness condition for local noetherian ring without nilpotent elements - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T07:39:12Z http://mathoverflow.net/feeds/question/88863 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88863/flatness-condition-for-local-noetherian-ring-without-nilpotent-elements flatness condition for local noetherian ring without nilpotent elements Lucy 2012-02-18T23:05:55Z 2012-02-19T09:54:54Z <p>In SGA 1, chapter 4, there is a corollary describing a flatness condition for a module of finite type over a local noetherian integral ring. The corollary is number 4.4, and states:</p> <p>Suppose that <code>$A$</code> is a local noetherian integral ring with maximal ideal <code>$I$</code> and residue field <code>$k = A/I$</code> and field of fractions <code>$K$</code>. Let <code>$M$</code> be a module of finite type over <code>$A$</code>. Then saying that <code>$M$</code> is flat (note: SGA actually cites here a previous proposition with equivalent conditions of flatness which in this case is satisfied) is equivalent to saying that <code>$M\otimes_A K$</code> and <code>$M\otimes_A k$</code> are vector spaces of the same dimension.</p> <p>There is a remark immediately following this corollary saying the reader is left to generalize this to the case where <code>$A$</code> is only assumed to be a ring without nilpotent elements. How does one show this more general case?</p> http://mathoverflow.net/questions/88863/flatness-condition-for-local-noetherian-ring-without-nilpotent-elements/88881#88881 Answer by Florian Eisele for flatness condition for local noetherian ring without nilpotent elements Florian Eisele 2012-02-19T02:25:39Z 2012-02-19T02:25:39Z <p>I assume you are referring to <a href="http://arxiv.org/pdf/math/0206203v2.pdf" rel="nofollow">page 78 here</a>. I think the author is only implying that you may drop the condition of $A$ being an integral domain (in favor of it being a ring without nilpotent elements). You should still assume that $A$ is noetherian and local with maximal ideal $I$ (since in the formulation of the corollary left to the reader, the author still refers to the field $k=A/I$, which would be ambiguous otherwise; also there are conditions on $A$ and $I$ in place throughout the entire section which prohibit you from working with an arbitrary ring). </p> <p>So you can apply the criterion for flatness given in Corollary 4.4 to the $A/{\mathfrak p}$-module $M/{\mathfrak p} M$ for each minimal prime $\mathfrak p$ of $A$ (since $A/\mathfrak p$ will be an integral domain again). What you want to show then is that $M$ is flat as an $A$-module if and only if $M/{\mathfrak p}M$ is flat as an $A/{\mathfrak p}$-module for each minimal prime $\mathfrak p$. By the foregoing Corollary 4.3, this is the same as showing that $M$ is free if and only if $M/\mathfrak p M$ is free for any minimal prime $\mathfrak p$ (however, the "and only if"-part is trivial). </p> <p>Here is how I think the proof goes: Since $A$ is local, we may lift a $k$-basis of $k\otimes M$ to a minimal $A$-generating set of $M$ (by Nakayama's lemma). We get an exact sequence $$A^i \stackrel{\varphi}{\longrightarrow} A^{dim_k k\otimes M} \longrightarrow M \longrightarrow 0$$ for some $i\in\mathbb Z_{\geq 0}$ and some $\varphi\in A^{dim_kk\otimes M \times i}$. If $\varphi$ is zero then $M$ is free. So assume $\varphi$ is not zero and $x$ is some non-zero entry of the matrix. Because $A$ has no nilpotent elements there is some minimal prime $\mathfrak p$ such that $x \notin \mathfrak p$. Now, by tensoring $$(A/\mathfrak p)^i \stackrel{\varphi}{\longrightarrow} (A/\mathfrak p)^{dim_k k\otimes M} \longrightarrow M/\mathfrak p M \longrightarrow 0$$ with the fraction field $K$ of $A/\mathfrak p$, we see that $dim_K K\otimes (M/\mathfrak p M) &lt; dim_k k \otimes M= dim_k k\otimes (M/\mathfrak p M)$, i.e. $M/\mathfrak p M$ is not flat/free/projective by Corollary 4.4 (i.e., we have shown: $M$ not free $\Longrightarrow$ $M/\mathfrak p M$ not free for some minimal prime $\mathfrak p$).</p> http://mathoverflow.net/questions/88863/flatness-condition-for-local-noetherian-ring-without-nilpotent-elements/88920#88920 Answer by a-fortiori for flatness condition for local noetherian ring without nilpotent elements a-fortiori 2012-02-19T09:54:54Z 2012-02-19T09:54:54Z <p>For a generalization to arbitrary noetherian reduced rings, see EGA III, 7.6.9. Applied to a projective resolution $P_*\to M$ for a finitely generated $A$-module $M$, it says in particular that $d(x)\colon x\mapsto\dim_{k(x)}(M\otimes k(x))$ is semi-continuous on $\mathrm{Spec}(A)$, and $M$ is flat iff $d$ is locally constant. The statement in SGA is a special case since semi-continuity together with $\dim_K(M\otimes K)=\dim_k(M\otimes k)$ already implies that $d$ is constant. A more elementary statement can be found in Mumford's treatment of semi-continuity in his Abelian Varieties (Lemma 1 in section II.5).</p>