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I'd like to fine an example of field $K$ and elements $\alpha, \beta$ such that $\mathrm{char}(K) = p> 0 $, $[K(\alpha):K] = [K(\beta):K]$ but $K(\alpha) \not \cong K(\beta)$.

This obviously can't work if $K$ is a finite field. So I need to find a non-finite $K$. The only ones that pop into my head are $\mathbb F_p(t)$, $\mathbb F_p(t^p)$ and $\overline{\mathbb F}_p$ for $t$ an indererminate, but I'm struggling to find an example.

Any hints would be greatly appreciated.


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$\mathbb{F}_p(t)$ will work. Try quadratic extensions. – Qiaochu Yuan May 8 '12 at 20:52
Skillfully pick two irreducible quadratic polynomials in $F_3(t)[X]$ such that the corresponding quotients are not isomorphic. – Mariano Suárez-Alvarez May 8 '12 at 20:57
@QiaochuYuan Would something simple like $K = \mathbb F_p(t)$, $\alpha = \sqrt{2}, \beta = i$ work? – Jonathan May 8 '12 at 20:58
@Jonathan: if $p$ is such that both of those are quadratic extensions, then both of those extensions are just $\mathbb{F}_{p^2}(t)$. You need to use $t$. – Qiaochu Yuan May 8 '12 at 21:16
@Jonathan: no, those are isomorphic (since $\alpha = t \beta$). Use $t$ in one of the extensions but not the other! – Qiaochu Yuan May 8 '12 at 23:41

Let us try to save this question from the Well of Oblivion of Unanswered Questions, following the solution already detailed in the comments of Qiaochu and the OP Jonathan:

Take $$K:=\mathbb F_3(t)\,,\,\alpha:=\sqrt{t}\,,\,\beta=\sqrt{-1}$$

As $\,\alpha\,,\,\beta\notin K\,\,\,\text{but}\,\,\alpha^2\,,\,\beta^2\in K\,$ , the extensions $\,K(\alpha)\,,\,K(\beta)\,$ are both quadratic ones , so $\,[K:K(\alpha)]=[K:K(\beta)]=2\,$, yet these two fields cannot be isomorphic as, say $\,\beta\notin K(\alpha)\,$ (and also the other way around)

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