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I guess my general question is exactly what's in the title, but let me explain why I'm asking and how I came to it.

Consider the ideal $I=\langle x,y \rangle \subset k[x,y]$ for a field $k$. Just to be safe, let's assume $k$ is infinite and of zero characteristic. Then $I$ is not flat. The proof I came up with uses Lemma 6.4 in Eisenbud's "Commutative Algebra", or equivalently Lemma 36.10 in The Stacks Project: just notice that the relation $y(x) + (-x)y=0$ is nontrivial.

But then I thought: it seems this sort of logic extends to any ideal in $k[X_1,\ldots,X_n]$. If $I=\langle f_1,\ldots,f_n\rangle$ then the relation $$(f_2\cdots f_n)f_1 + (-f_1f_3\cdots f_n)f_2 + \cdots + (-f_1\cdots f_{n-1})f_n = 0 $$ is nontrivial for $n$ even and if $n$ is odd, just replace the last coefficient with $0$. But once I wrote this, I thought this could be applied to any Noetherian ring $R$. So in a Noetherian ring, every ideal is not flat. But this sounds too strong. Since Prufer Domains exist, I think it is indeed wrong (one of the characterizations of a Prufer domain $R$ is that every ideal of $R$ is flat).

So I think I'm going wrong in one of a few places. First: my "proof" that $\langle x,y\rangle \subset k[x,y]$ is not flat is wrong and indeed this relation is trivial. I don't think this is true since, using the Stacks Project's notation, the $y_j \in I$ so must be divisible by $x$ or $y$. Then we would want $a_{ij} \in k \subset k[x,y]$ so get something like $y=2y + x - x - y$, for example, since otherwise $x_i = \sum_j a_{ij} y_j$ is impossible. But if $a_{ij} \in k$ then there's no way for $\sum_ia_{ij} f_i = 0$ since $f_1=y, f_2=-x$. I realize this is sort of handwavy, and indeed this may be why I'm confused since I haven't formalized it: maybe it can't be formalized.

Second, maybe the proof is correct, but extending this and concluding that every ideal in $k[X_1,\ldots,X_n]$ is not flat doesn't hold. If this is the case, can someone give an example of such an ideal which is flat?

Finally, maybe every ideal in the polynomial ring isn't flat for the reasons above, but maybe there are rings which contain finitely generated ideals such that this logic does not extend. If so, can someone provide an example?

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Lots of Noetherian rings have ideals which are even free! For example, all principal ideals in noetherian domains provide examples of this. Similarly, Dedekind rings provide examples where all ideals are projective, and most of them are not principal. – Mariano Suárez-Alvarez Jun 1 '12 at 19:42
Well, I guess that answers the general question. Thanks to both of you! Any thoughts on where my reasoning breaks down though? – Derek Allums Jun 1 '12 at 20:14
up vote 8 down vote accepted

An ideal $I$ of an noetherian (commutative) ring $A$ is flat if and only if it is locally free of (local) rank $\le 1$ (i.e. $IA_P=0$ or $\simeq A_P$ for every prime ideal $P$).

The if part is clear. Suppose $I$ is flat. Then it is locally free and finitely generated. Let $P$ be a prime ideal. So $IA_P$ is an ideal free of some rank $r$. Your reasonning shows that $r\ge 1$: $e_1, e_2\in IA_P$ are not zero, then $\{ e_1, e_2\}$ is not free because $e_2.e_1+(-e_1).e_2=0$.

To make connection to the examples in the comments, it is easy to see that all ideals of $A$ are flat if and only if $A$ is a finite product of Dedekind domains (including fields).

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Ok, so as a very particular case, this shows that the ideal $\langle X_1,\ldots, X_n \rangle$ in the polynomial ring over a field $k[X_1,\ldots,X_n]$ is not flat, right? But in general, there are examples of (non principal) ideals in the polynomial ring which are flat? – Derek Allums Jun 2 '12 at 16:18
@unit3000-21: yes if $n\ge 2$. For the second question: as the polynomial rings with coefficients in a field are UFD, the only flat (hence locally free of rank 1) ideals are principal ideals. – user18119 Jun 2 '12 at 20:55
Thanks for the answer, QiL. – Derek Allums Jun 2 '12 at 21:23

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