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?