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If $K$ is a field, then as is well known every finite separable extension $L$ of $K$ is of the form $L=K(\alpha)$ for some $\alpha \in L$.

A similar statement can be made about an extension of discrete valuation rings with separable residue field extension.

These statements very much resemble the statement "every projective module over a principal ideal domain $A$ is free". This last statement can be interpreted as the vanishing of a certain cohomology group. Now my question is: can the primitive element theorem be interpreted as the vanishing of a cohomology group?

Thank you!

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What vanishing statement is the same as "projective=free over a PID"? – Mariano Suárez-Alvarez Mar 21 '13 at 2:54
Mariano: Presumably the statement is that $H^1(Spec(R),GL_n)=0$ for all $n$. – Steven Landsburg Mar 21 '13 at 2:57
This is the sort of thing that I have in mind: For a Dedekind domain $A$, every projective $A$-module is free if and only if $\text{Pic}(A) = H^1(X, \mathcal O_X^*) = 0$ (where $X=\text{Spec }A$). – Bruno Joyal Mar 21 '13 at 2:58
up vote 11 down vote accepted

The vanishing of the cohomology group $H^1(Spec(R),GL_n)$ doesn't actually say that all projectives of rank $n$ are free; it says only that all projectives of rank $n$ are isomorphic. Combining this with the observation that at least one such projective is free, we get that they're all free.

But in the case of field extensions, it is not true that all finite separable extensions are isomorphic, even though they're all generated by primitive elements. Therefore, I think the analogy you're seeing is largely illusory.

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That's a good point! – Bruno Joyal Mar 21 '13 at 3:07
Ok this rules out an interpretation that a cohomology group vanishes. But perhaps the primitive element theorem gives a surjective map to some cohomology group? – Martin Brandenburg Mar 22 '13 at 1:29
@Martin, I also have not completely given up on the idea. If you think of anything, please share! Regards, – Bruno Joyal Mar 22 '13 at 16:50

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