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Consider field $K = \mathbb Z_p(x)$. Does there exist an irreducible polynomial over $K$ whose splitting field has dimension strictly greater than degree of polynomial.

This is not true in case of finite fields and is true over $\mathbb Q$. But I have not seen anything related to infinite field of positive characteristic.

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    $\begingroup$ I take it $\mathbb{Z}_p$ represents $\mathbb{F}_p$, the field with $p$-elements. I would suggest a notation that does not lend itself to confusion with the $p$-adic integers in this context. $\endgroup$ – Arturo Magidin Aug 4 at 22:09
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    $\begingroup$ Pick a prime $p$ such that $x^2+x+1$ is irreducible over $\mathbb{F}_p$,, and consider $t^3 - x$. If I’m not mistaken, the splitting field will be given by $x^{1/3}$ and $\omega$, where $\omega$ is a root of $t^2+t+1$ (that is, a primitive cubic root of unity), which should be an extension of degree $6$. $\endgroup$ – Arturo Magidin Aug 4 at 22:18
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    $\begingroup$ @ArturoMagidin This raises the question: what about $k(x)$ with $k$ algebraically closed ? $\endgroup$ – reuns Aug 4 at 22:44
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To answer the question of @reuns in their comment, what about $k(x)$ when $k$ is algebraically closed, even of characteristic $p>0$.

If you take a geometric viewpoint, you realize that you’re talking about covers of the projective line, and it’s “well known” that most covers are not Galois. So it should be easy to find a non-Galois cover, necessarily of degree $>2$.

You can forget about geometry now. Let $k$ be your algebraically closed constant field, and call $K=k(x)$. Just look at the polynomial $f(T)=T^3-T-x\in K[T]$, clearly irreducible because it’s irreducible in $k[x,T]$.

Now let $L=K(\tau)$, where $\tau$ is a root of $f$. I claim that the extension $L\supset K$ is not Galois, and I’ll show it by showing that the other roots of $f$ are not in $L$.

Indeed, Euclidean division gives you $f(T)=(T-\tau)(T^2+\tau T +\tau^2-1)$. And what about the quadratic factor? You see that $k(\tau,t)=k(\tau)$, because $t=\tau^3-\tau$. The discriminant of the quadratic factor is $\tau^2-4(\tau^2-1)=4-3\tau^2\in k[\tau]$. But if the characteristic is neither $3$ nor $2$, that discriminant is not a square in $k[\tau]$. Thus the roots of the quadratic factor are not in $\text{frac}\bigl(k[\tau]\bigr)=L$, and the extension is not Galois.

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  • $\begingroup$ very interesting answer $\endgroup$ – M. A. SARKAR Aug 5 at 4:39

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