Is $x$ an algebraic element over the field of rational functions $K(x)^p$?

Question

Question: Is $x \in K(x)$ an algebraic element over the field $K(x)^p$?

Edit: Let $K$ be a field with char($K)=p>0$ and let $K(x)$ be the field of rational functions over $K$.

My attempt: I basically tried to answer this by referring to:

Field $K (x)$ of rational functions over $K$, the element $x$ has no $p$th root.

Suppose to the contrary that $x$ is algebraic over $K(x)^p$, and so $x$ is a root of some $p$-degree polynomial such that; $(\frac{f(x)}{g(x)})^p -x = 0$

$f(x)^p=g(x)^p * x$

Here we see the contradiction since the degrees of $f(x)^p= deg(f(x)*p)$ and $g(x)^p*x = \deg(g(x)*p+1)$.

I am totally lost on this, I've been using the 4th edition of Abstract Algebra by Beachy and there is hardly any mention of the field of rationals. Any hints and maybe suggestions on resources where I can find out more about the field of rationals would be greatly appreciated, thanks!

Answer 1

$x$ is in fact algebraic over $K(x)^p$ (note the comments to the question, we only need that $x^p\in K(x)^p$. I think it might be confusing to you in which ring we're trying to find polynomials that have $x$ as a root. To get around this notational issue, let's call $F:=K(x)^p$.

Now $x$ is algebraic over $F$ if there is some polynomial $g\in F[Y]$ s.t. $g(x)=0$. Let's look at the polynomial $g=Y^p-x^p$. We know that $x^p\in F$, so $g\in F[Y]$. Clearly also $g(x)=x^p-x^p=0$, so $x$ is algebraic over $F$.

Answer 2

I’m assuming throughout that you mean for $K$ to have characteristic $p>0$. Perhaps you’re thrown by the possibility that $K$ is not perfect, in which case $\bigl(K(x)\bigr)^p$ is different from $K(x^p)$. Don’t worry, though: for our purposes, it doesn’t matter.

Let’s consider your field field $\mathscr L=\bigl(K(x)\bigr)^p$, in which there is an element $x^p$. I’ll call this element $t$. We note that there’s a field isomorphism $\varphi:K(x)\to\mathscr L$, by $\varphi(f)=f^p$. And the image of the element $x$ of $K(x)$ is $t\in\mathscr L$; just as $x$ has no $p$-th root in $K(x)$, so $t$ has no $p$-th root in $\mathscr L$. Thus the $\mathscr L$-polynomial $X^p-t$ is irreducible ($\dagger$). It has a root back up in $K(x)$, though, namely $x$. And there you are.

($\dagger$) I’ve used the fact that in a field $k$ of characteristic $p$, $X^p-b$ either has a root in $k$ or is $k$-irreducible.