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Let $K$ be a field of positive characterstic. I need to show that there exists $a$ and $b$ of the same degree over $K$ but $K(a)$ and $K(b)$ are not isomorphic.

I thought of an example where they are not $K$-isomorphic. Take $K=F_p(X)$ the function field in one variable over $F_p$, $a$ to be a root of a separable irreducible polynomial and $b$ a root of a non-separable one of the same degree.

How can I show that examples exist for which $K(a)$ and $K(b)$ are not isomorphic?

Spook
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  • Your approach will not work, because over the finite field $F_p$ (I assume you mean the field with $p$ elements) all irreducible polynomials are separable. However if you replace $F_p$ with a non-perfect field, then this works. It then remains to treat the case of a separably closed field $K$, that is a field that possesses only purely inseparable extensions ... – Hagen Knaf May 15 '13 at 11:52
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    Can you choose $K$ yourself, or is only the characteristic given? In addition to the problem with finite fields pointed out by Hagen you will also run into problems, if $K$ happens to be algebraically closed. Please specify what parameters are fixed, and what are we free to choose. – Jyrki Lahtonen May 15 '13 at 15:16
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    For a plethora of non-isomorphic extensions of degree $p$ of $F_p((X))$ I refer you to this answer by Keith Conrad. – Jyrki Lahtonen May 15 '13 at 15:22

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This is not true as stated. Indeed, if $K$ is a finite field, then for any positive integer $n$ there is a unique extension of $K$ of degree $n$, up to isomorphism (namely, the splitting field of $x^m-x$ where $m=|K|^n$). Or if $K$ is algebraically closed, then it has no nontrivial finite extensions at all.

For a simple example where you get to choose the field $K$, take $K=\mathbb{F}_p(X)$, let $K(a)=\mathbb{F}_p(\sqrt{X})$ and let $K(b)=\mathbb{F}_{p^2}(X)$ (so $a=\sqrt{X}$ and $b$ is some element of $\mathbb{F}_{p^2}\setminus\mathbb{F}_p$). These both have degree $2$ over $K$, but they are not isomorphic since $\mathbb{F}_{p^2}$ does not embed in $K(a)$ (which is isomorphic to $K$ by sending $\sqrt{X}$ to $X$).

Eric Wofsey
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