# Is the primitive element theorem a cohomological statement?

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

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.