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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.

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up vote 24 down vote accepted

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 $R = k$.

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You have fine colleagues, Matt, and vice-versa. – Georges Elencwajg Jun 3 '11 at 7:25
Dear Georges, Thank you for your kind remark. Best wishes, Matt – Emerton Jun 3 '11 at 11:37

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