Prove that a polynomial is irreducible or the field contains a $p$th root

Solution 1:

One side:

If $a=b^{p}$ for $b\in k$ then: $$x^{p^{n}}-a=x^{p^{n}}-b^{p}=\left(x^{p^{n-1}}-b\right)^{p}$$ showing that the polynomial is not irreducible.

Let $k^{\mathbf{a}}$ denote an algebraically closed fied that contains $k$ and is algebraic over $k$.

Let it be that $\alpha\in k^{\mathbf{a}}$ is a root of $x^{p^{n}}-a$.

Then $\alpha^{p^{n}}-a=0$ so that $x^{p^{n}}-a=x^{p^{n}}-\alpha^{p^{n}}=\left(x-\alpha\right)^{p^{n}}$.

This reveals that polynomial $x^{p^{n}}-a$ has only one root in $k^{\mathbf{a}}$.

The other side:

For convenience define $f\left(x\right):=x^{p^{n}}-a$ and let $g\left(x\right)\in k\left[x\right]$ denote the minimal polynomial of $\alpha$.

Then $f\left(x\right)=g\left(x\right)^{m}h\left(x\right)$ for some positive integer $m$ and some $h\left(x\right)\in k\left[x\right]$.

Taking $m$ maximal $g\left(x\right)$ will not divide $h\left(x\right)$ so that $\alpha$ is not a root of $h\left(x\right)$. But $\alpha$ is the only candidate for being a root of $h\left(x\right)$ so we conclude that $h\left(x\right)$ has no roots at all. This allows the conclusion that: $$f\left(x\right)=g\left(x\right)^{m}$$ If $d$ denotes the degree of $g$ then $md=p^{n}$ showing that $m$ and $d$ are both powers of $p$. So nonnegative integers $u,v$ exist with $u+v=n$, $d=p^{u}$ and $m=p^{v}$. Then: $$x^{p^{u}}-\alpha^{p^{u}}=\left(x-\alpha\right)^{p^{u}}=g\left(x\right)\in k\left[X\right]$$ Proved is now that $\alpha^{p^{u}}\in k$.

If $f$ is not irreducible then $p^{v}=m>1$ so that $v\geq1$.

Defining $b:=\alpha^{p^{n-1}}=\left(\alpha^{p^{u}}\right)^{p^{v-1}}\in k$ we find:

$$a=\alpha^{p^{n}}=b^{p}$$

Solution 2:

Rename the field $k$ as $K$. Rename the indeterminate $x$ as $X$. Thus we must prove that either the polynomial $X^{p^n}-a$ is irreducible over $K$ or $a$ is a $b$-th power.

In an extension field of $K$, if $b$ is a root of $X^{p^n}-a$, then $(X-b)^{p^n}=X^{p^n}-b^{p^n} =X^{p^n}-a$. Therefore, any proper monic factor of $X^{p^n}-a$ is $(X-b)^k$ where $0<k<p^n$. If $(X-b)^k$ has coefficients in $K$, then $b^k\in K$. But as $b^{p^n}=a\in K$, then $b^g\in K$, where $g=\gcd(k,p^n)$. Then $g=p^m$ where $0\le m\le n-1$, and this implies $b^{p^{k-1}}\in K$, and that's a $p$-th root of $a$.