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Suppose that A is a local ring (commutative with unit), finite over a field k. Let L be the residue field A / m where m is the unique maximal ideal of A.

Does the dimension of L (as a k-vector space) divide the dimension of A (as a k-vector space)? What about if k is perfect?

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If the characteristic of $k$ is zero then isn't $A$ already an $L$-algebra, so you're OK in this case. If the characteristic is $p$ but $k$ is perfect then $A$ will be a $W(L)$-algebra ($W$ the Witt vectors) in which $p=0$ so again $A$ is an $L$-algebra and you're still OK. In the imperfect residue field case I'm much less sure of myself. – Kevin Buzzard Aug 17 '10 at 16:41
$A$ is a $k$-algebra and $L$ is finite over $k$ so you can lift $L$ to $A$ (i.e. split the map $A\to L$) using Hensel's Lemma on polynomials with coefficients in $k$ and roots in $L$. I'm pretty sure that's how the argument should go. – Kevin Buzzard Aug 17 '10 at 16:58
More simply: The quotient $m^j/m^{j+1}$ is an $L$ vector spaces, so its $k$-dimension is divisible by the $k$-dimension of $L$. Sum over $j$. – Tom Goodwillie Aug 17 '10 at 17:18
Kevin, no Hensel if $L/k$ not sep'ble. Anon, Cohen's thm (sec. 29, Matsumura CRT) ensures complete local noeth. ring admits some alg. structure over residue fld (lifting residual id. map); may not respect given alg. structure over subfld of residue fld. For example, if $k$ imperfect of char. $p > 0$ and $a \in k$ not $p$-power, consider completion $A$ of $k[x]_{(x^p-a)}$: dvr with residue field $k' = k(a^{1/p})$ with no $k'$-alg. structure over its $k$-alg. structure (since $A \otimes_k k'$ is completion of $k'[x]_ {(x-a^{1/p})}$, hence dvr, so reduced, but $k' \otimes_k k'$ non-reduced!)... – BCnrd Aug 17 '10 at 18:23
...and artinian qt of $A$ can fail to have $k'$-alg. structure over their $k$-alg. structure. Let $t$ be uniformizer of $A$ (e.g., $x^p-a$) and $A' := A \otimes_k k'$, and let $A_n$ and $A'_n$ denote qts by $t^n$, so $A'_n = A_n \otimes_k k'$. If $A_n$ has $k'$-alg. structure over its $k$-alg. structure (lifting residual id. map) then $A_n = k'[t]/(t^n)$, so $A'_n = (k' \otimes_k k')[t]/(t^n) = k'[x,t]/(x^p,t^n)$. This artinian qt of dvr $A'$ has length $np$, so as abstract ring its $k'[u]/(u^{np})$, so $p$-power map doesn't kill max ideal if $n > 1$. Contradicts $k'[x,t]/(x^p,t^n)$ if $n=p$. – BCnrd Aug 17 '10 at 18:32

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