Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 23 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 $R = k$.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.