Let $A$ be an integral domain and let $K$ field of fractions. Let $k$ be a subfield of $K$ such that $K$ is simple finite extension of $k$. Is it true that there is an element $ a \in A $ such that $ K = k[a]$?
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I think the primitive element theorem may be handy here.. I would say we don't know anything about the relation between $K$ and $A$ and thus, cannot say if there is an element $a\in A$ s.t. $K=k[a].$ (I would have commented, but i am not privileged..)
emma
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Here's a proof when $ k \cap A $ is an infinite set.
Suppose that $ K = k [ c/d] $ for $ c, d \in A $. Consider the algebras $ k[c + \alpha d] $ for $ \alpha \in k \cap A $. These are all fields, since they are finite dimensional integral domains. Then, since there are only finitely many intermediate extensions between $ K $ and $ k $ (as $ K/k $ is simple), two of them must be equal. But then, both $ c $ and $ d $ will be in that extension, and thus that extension must be $ K $.
This leaves the case when $ k \cap A $ is finite i.e. char $ k $ = p, and $ k \cap A $ is equal to a finite field.
Hodge-Tate
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I don't think you even need that $k[A]$ is Artinian. Any integral domain of finite dimension over a field is itself a field, if I am not mistaken. – Robert Lewis May 09 '18 at 23:51
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What I wrote above is wrong. Not all extensions are simple. – Hodge-Tate May 10 '18 at 00:00
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What is $N_{K/k}(a)$? – Robert Lewis May 10 '18 at 00:12
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I wrote something different. – Hodge-Tate May 11 '18 at 05:09