How can I show that the polynomial $t^n-x$ is irreducible in $F(x)[t]$?

Question

In the setup of the problem, it is given that $F$ is a field, $F(x)$ is the field of rational functions with coefficients in the field $F$, and $n$ is a positive integer. I also understand that $F(x)[t]$ is the ring of polynomials in the variable $t$ with coefficients given by the elements in $F(x)$.

I began to approach this problem by setting up a proof by contradiction:

Suppose that $t^n-x$ is reducible in $F(x)[t]$. Then, $t^n-x=a(t)*b(t)$ for some irreducible, non-units $a(t)$, $b(t)$ in $F(x)[t]$. We know that the degrees of $a(t)$, $b(t)$ are strictly less than $n$.

But after this, I am stumped.

Any help is much appreciated!

Edit: I also want to work this out a bit on my own, so instead of a full answer or proof maybe I'd prefers a few hints or maybe observations that I'm missing?

Answer 1

Here are two observations which may useful. First, let $\overline{F}$ be the algebraic closure of $F$. If the polynomial $h(t) = t^n - x$ factors in $F(x)[t]$, then it also factors in $\overline{F}(x)[t]$. So without loss of generality you can assume that $F$ is algebraically closed.

Second, it might be useful to imagine the roots of $h(t)$ as already living in some field containing $F(x)$. That is start with an indeterminate $y$, let $L = F(y)$, and define $x = y^n$, letting $K = F(y^n) = F(x)$. The object $y^n$ works just as well as any indeterminate, so $K = F(x)$ is the field you started with in your question, and is a subfield of $L$.

The polynomial $h(t)$ lies in the polynomial ring $K[t]$. This polynomial has a root in $L$, namely $y$. Since $F$ is algebraically closed, $F$ contains all the $n$th roots of unity, say $\zeta, \zeta', \zeta''$ etc. There are at most $n$ distinct $n$th roots of unity (less than $n$ if and only if the characteristic of $F$ divides $n$) Then all the roots of $h(t)$ are in $L$: they are $y, \zeta y, \zeta' y, \zeta''y$ etc.

Answer 2

Hint: The Eisenstein criterion can be generalized to arbitrary integral domains and (in the case of UFDs) their fraction field as well.

Answer 3

Suppose $F$ has characteristic $0$ or at least than its characteristic is prime to $n$. Suppose for the moment that $F$ contains $n$ $n^{th}$ roots of unity, and let $\zeta$ be a primitive such root. Consider $F(x^n)$, a subfield of $F(x)$. The automorphism of $F(x)$ taking $x$ to $\zeta x$ has order $n$ and fixed field $F(x^n)$. (This is an endomorphism because you get to map the indeterminate to anything. Then you have to check (easy) that it's onto.) This shows that $t^n-x^n$ is irreducible over $F(x^n)$. But since $F(x)$ is isomorphic to $F(x^n)$ by the map taking $x$ to $x^n$, we also have $t^n-x$ irreducible over $F(x)$.

Since the degree of $k(a,b)$ over $k(a)$ is never larger than the degree of $k(b)$ over $k$, you can readily see that this argument holds even if $F$ does not contain the desired roots of unity, as long as they can be adjoined.

For the case $(n, char F)>1$, you can handle that, too, but it will be a different argument since it's not a separable extension. I leave that as an exercise.

Answer 4

A short way of seeing this:

By Gauss' Lemma, $t^n - x$ is irreducible in $F(x)[t]$ if and only if it is irreducible in $F[x][t]$. But $F[x][t] = F[t][x]$ and $t^n - x$ is a polynomial of degree $1$ in $F[t][x]$ so is irreducible in $F[t][x]$ and therefore in $F[x][t]$ so as well in $F(x)[t]$.

Answer 5

I think I may have come up with a different solution, and probably one my professor was looking for. Let me know what you guys think though.

In class, we discussed primitive polynomials, and proved that in UFDs, a primitive polynomial is irreducible. In F(x)[t], we see that the coefficients in the monomials of $t^n-x$ are 1 and x. Hence, in F(x)[t], the content of $t^n-x$ is $\gcd(1,x)=1$ and $t^n-x$ is primitive and therefore irreducible in F(x)[t].

Edit: nah this is wrong