Galois Theory: Inseparable and Separable elements of a field K.

**fatpolomanjr** · September 9th 2011, 09:44 PM

Problem: If $u \in F$ is separable over field $K$ , where $F$ is an extension of $K$ , and $v\in F$ , then $K(u,v) = K(u+v)$ . If $u, v \neq 0$ , then $K(u,v) = K(uv)$ .

It is clear to me that in characteristic 0 this proof is trivial. In characteristic $p$ , with $p$ prime, one direction of the inclusion is obvious for each case, and that is $K(u+v), K(uv) \subset K(u,v)$ . To get the other direction I was thinking of using the identity $(u + v)^{p^n} = u^{p^n} + v^{p^n}$ , where $n$ is the power that makes $v^{p^n} \in K$ . Therefore, we must have that $u^{p^n} \in K(u + v)$ by subtracting off $v^{p^n}$ (since it is an element of $K$ ). If $u$ is an element of $K(u+v)$ , then we can subtract it off of the element $u + v \in K(u+v)$ , and the reverse containment will follow. If $u \notin K(u+v)$ , then it is purely inseparable over $K(u+v)$ . Similar reasoning can be used to show $v$ is purely inseparable over $K(u+v)$ (otherwise $v$ would be an element of $K(u+v)$ and we'd be done), hence $K(u,v)$ is a purely inseparable extension of $K(u+v)$ .

My issue is that I have no idea where to go with this; I'm hoping to find some kind of contradiction, but I am probably lacking somewhere in my understanding of what it means for an element to be separable. Any ideas? Is my approach even close to correct? I've been on this problem for 2 days, with no luck.

**fatpolomanjr** · September 10th 2011, 10:19 PM

I think I might have an idea:

Continuing on in the same thinking to get my contradiction / show that $u\in K(u+v)$ , since $u\in K\subset K(u+v)$ is separable, that means its minimal polynomial, say $h(x) \in K(u+v)[x]$ factors into simple roots. Since $u\notin K(u+v)$ , $x^{p^n} - u^{p^n} = (x - u)^{p^n}$ is irreducible in $K(u+v)[x]$ . Hence, $h(x) | (x-u)^{p^n}$ . But that means $h(x) = x - u \in K(u+v)[x]$ . Thus $u\in K(u+v)$ .

The only fuzzy part that I cannot get past in this reasoning is the relationship between the minimal polynomial of an extension field (in this case the extension over $K(u+v)$ containing $u$ ) and any irreducible polynomial in the base field. EDIT: Oh, and if $h(x)$ is separable in $K[x]$ then is it still separable in any extension field with $u$ as a root of some irreducible polynomial (not necessarily intermediate between base field and splitting field of the irreducible polynomial?...I mean...I don't think $K(u+v)$ is an intermediate field of $K$ and the splitting field of $h(x)$ ). I'll try to convince myself this answer is correct...

**fatpolomanjr** · September 23rd 2011, 08:29 PM

I think my reasoning was correct. EDIT: in the second post of this thread where I "solved" it, however, it should say $(x-u)^{p^n} | h(x)$ . *my bad*. Problem. Solved.

To answer my own questions, the irreducible polynomial splits in its splitting field as simple roots. One can consider the chain of extensions $K\subset K(u+v)\subset K(u,v)\subset F(v)$ , where $F$ is the splitting field of $u$ 's irreducible polynomial, $f$ , over $K$ . Therefore since $u$ is a simple root of $f$ in $F(v)[x]$ , and $(x-u)^{p^n}$ has to divide $f$ since $K(u+v)$ is an intermediate field, then clearly $p^n$ has to equal 1.

Side note: I find it funny how helpful writing my problem out to a public forum helped me understand it myself; it's similar to discussing the problem with someone else, or to having to prepare a talk on a subject.

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Galois Theory: Inseparable and Separable elements of a field K.

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