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Suppose $L/K$ is a finite extension of fields. Is is always true that $$\min_{\substack{\{\alpha_1, \ldots, \alpha_n\} \\ L = K(\alpha_1, \ldots, \alpha_n)}} \left(\prod_{i=1}^n [K(\alpha_i): K] \right) = [L:K]?$$ By the primitive element theorem, this is certainly true if $L/K$ is separable. But what if $L/K$ is an arbitrary finite extension?

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OP told me via personal communication that his motivation for this question was to find a proof of$$|\text{Aut}(L/K)| \le [L: K]$$for all finite extensions $L/K$ without invoking linear independence of automorphisms. Indeed, there is such a proof—an easy one just using induction on the number of generators one needs, replacing, for each $i$, both $K$'s in $[K(\alpha_i): K]$ by $K(\alpha_1, \ldots, \alpha_{i - 1})$. Then the product is clearly equal to $[L:K]$, and the rest of the proof is plain sailing.

As pointed out in the MathOverflow version of the question here, a paper of Sweedler here seems to suggest the condition$$\min_{\substack{\{\alpha_1, \ldots, \alpha_n\} \\ L = K(\alpha_1, \ldots, \alpha_n)}} \left(\prod_{i=1}^n [K(\alpha_i): K] \right) = [L:K]$$does not hold in general for purely inseparable extensions, and as such, OP asks for an explicit counterexample.

Unfortunately, I am not a specialist in inseparable extensions. I would try hunting for the fabled counterexample in the following way. Start from a field $K$ of characteristic $p > 0$ for which $[K: K^p]$ is "large enough" (here $K^p = \{x^p : x \in K\}$)—infinity would probably do, $1$ or $p$ won't, $p^2$ seems risky. Take $E$ purely inseparable of degree $p$ over $K$. Construct $L$ purely inseparable of degree $p^2$ over $E$ such that $L^p \subset E$ (so that $L$ has no primitive element over $E$) and such that $L^p \cap K = E^p$. I am not sure (yet) how to make such an $L$, but I am pretty sure it can be done, also for some quite explicit fields $K \subset E$; if necessary, I would invoke Jacobson-Bourbaki theory on intermediate extensions of $E \subset L$—a theory I've never heard of until I started searching around on Google a few hours ago. Anyway, once one has such an $L$, then each $\alpha$ in $L$ that is not in $K$ has degree $p^2$ over $K$, and at least two such $\alpha$'s are needed, so that $L$ would give the desired example.

Maybe you can close the gap in this argument.

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