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I have problems to prove the following statement.

Let $K$ be a field and let $K(u,v)$ be an algebraic extension of $K$. If $u$ is separable over $K$ then $K(u,v)$ is a simple extension.

(My incomplete) proof: We can suppose that $char K=p$, because in characteristic $0$ all finite extensions are simple. Let's divide the problem in $3$ cases (the third is problematic):

1) $v$ is separable. In this case $K(u,v)$ is a separable extension of $K$ and so it is simple.

2) $v$ is purely inseparable. We prove that $K(u,v)=K(u+v)$; by definition exists $m\in\mathbb N$ such that $v^{p^m}\in K$, so $(u+v)^{p^m}=u^{p^m}+v^{p^m}\in K(u^{p^m})$. So $K(u^{p^m})\subseteq K(u+v)$, but $u$ is separable on $K$ and follows that $K(u^{p^m})=K(u)$. Moreover $v=(u+v)-u\in K(u+v)$ and we can conclude that $K(u+v)=K(u,v)$.

3) $v$ is neither separable nor purely inseparable. ???

I would appreciate any hint about the point 3), but also a more organic proof without any subdivision.

Thanks in advance.

Dubious
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2 Answers2

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Look at the usual proof in the separable case. The only place they use separability is to show the gcd of the minimal polynomials has distinct roots. This only really requires one of the polynomials to be separable.

user26857
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Brian Rushton
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    However it is the exercise 7 at page 59, from the Kaplansky's Book "Fields and Rings". This is the hint: prove that there are only finitely many intermediate fields. It can be assumed that $K(u)$ is the maximal separable subfield. Analyze an intermediate field $L$ by showing that it lies between its maximal separable subfield $L_0$ and $L_0(v)$. – Dubious Jul 01 '13 at 21:07
  • @Dubious Do you mean that $L$ lies between $K$ and $L_0$ or $L_0$ and $L_0(v)$? – Stephen Landsittel Jul 26 '21 at 22:33
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This question is exercise 14. of Chapter V.7 of Hungerford's Algebra. Here is another solution.

There exists $n$ such that $v^{p^n}$ is separable over $K$. Then $K(u^{p^n},v^{p^n})$ is separable over $K$, and so there exists only finitely many intermediate field between $K(u^{p^n},v^{p^n})$ and $K$.

We may assume $K$ is an infinite field, and so $K^{p^n}=\{a^{p^n}|a\in K\}$ is also infinite. So there is a $a\in K^*$, such that $K(u^{p^n},v^{p^n})=K(u^{p^n}+a^{p^n}v^{p^n})$(By the discussion of last paragraph, there are distinct elements $a,b\in K$, $K(u^{p^n}+a^{p^n}v^{p^n})=K(u^{p^n}+b^{p^n}v^{p^n})$).

Since $K(u+av)\supset K(u^{p^n}+a^{p^n}v^{p^n})= K(u^{p^n},v^{p^n})\supset K(u^{p^n})=K(u)$, it contains $u+av-u=av$, and hence $F=K(u+av)$.

Yunsong WEI
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