is residue field ever flat over its local ring? Let R be a local ring with maximal ideal m and residue field k.  Is k ever flat over R?  What conditions are needed on R?
Sorry, it's not a very profound question.  It came up in a derived functor calculation.
 A: Consider the exact sequence $0 \to \mathfrak m \to R \to k \to 0.$ Tensoring with $k$ gives
$0 \to \mathfrak m/\mathfrak m^2 \to k  = k \to 0.$ Thus if $k$ is flat over $R$, then
$\mathfrak m = \mathfrak m^2$.  If furthermore $R$ is Noetherian, this implies that
$\mathfrak m = 0$, and hence that $R = k$.
Conclusion: For Noetherian $R$, the desired flatness hold only if $R = k$.
Added: A colleague points out that flat local maps of local rings are always faithfully flat,
hence injective.  Thus even in the non-Noetherian case, the only way for $k$ to be
flat over $R$ is if $R = k$.
In fact, one can directly extend the above argument to the not-necessarily-Noetherian case, as follows:
Let $I$ be any finitely generated ideal contained in $\mathfrak m$.  Since $k$ is 
flat, $k\otimes_R I \hookrightarrow k\otimes_R \mathfrak m,$ the target of which vanishes, as noted above.
Thus $k\otimes _R I$ vanishes, and so by Nakayama's lemma, $I = 0$.  Since $\mathfrak m$
is the union of such $I$, we see that $\mathfrak m = 0$, and thus $R = k$.
