# Irreducible polynomial over $\mathbb Z_p(x)$ and dimension of splitting field

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.

• 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. – Arturo Magidin Aug 4 '19 at 22:09
• 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 $6$. – Arturo Magidin Aug 4 '19 at 22:18
• @ArturoMagidin This raises the question: what about $k(x)$ with $k$ algebraically closed ? – reuns Aug 4 '19 at 22:44

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.