Irreductibility of $x^p-t$ for fields $F$ with characteristic $p$ and such that $t$ has not a $p$-root in $F$

Question

I am having problems to understand the the proof of following Lemma 9.14.2 in: https://stacks.math.columbia.edu/tag/09HF

Lemma 9.14.2. Let $p$ be a prime number. Let $F$ be a field of characteristic p. Let $t\in F$ be an element which does not have a $p$-th root in $F$. Then the polynomial $x^p−t$ is irreducible over $F$.

Proof. To see this, suppose that we have a factorization $x^p−t=fg$. Taking derivatives we get $f′g+fg′=0$. Note that neither $f′=0$ nor $g′=0$ as the degrees of $f$ and $g$ are smaller than $p$. Moreover, $\deg(f′)<\deg(f)$ and $\deg(g′)<\deg(g)$.

We conclude that $f$ and $g$ have a factor in common. (WHY?)

Thus if $x^p−t$ is reducible, then it is of the form $x^p−t=cf^n$ for some irreducible $f$, $c\in F^*$, and $n>1$. (WHY?)

Since $p$ is a prime number this implies $n=p$ and $f$ linear, which would imply $x^p−t$ has a root in $F$. Contradiction. $\square$

I suppose that it is immediate but I don't see the reason to conclude that $f$ and $g$ have a factor in common and that $x^p-t=cf^n$

Answer 1

Suppose $x^p-t=fg$. Then we have $f'g+fg'=0$, hence $f'g=-fg'$, and none of the factors is equal to $0$.

Now let $q$ be the greatest common divisor of $f'$ and $f$, and let $p$ be the greatest common divisor of $g$ and $g'$. Then we can write $f'=qF'$, $f=qF$, with $\gcd(F,F')=1$; and similarly $g=pG$ and $g'=pG'$ with $\gcd(G,G')=1$. Substituting in, we have $$qF'pG = f'g = fg' = qFpG'.$$ Cancelling $q$ and $p$, we have $F'G=FG'$. Thus, $F\mid F'G$; since $\gcd(F,F')=1$, then $F\mid G$. But $F$ divides $f$, and $G$ divides $g$. Moreover, because $\deg(f')\lt \deg(f)$, it follows that we $\deg(q)\leq \deg(f')\lt\deg(f)$, so $\deg(F)\geq 1$. Thus, $F$ is a nontrivial common factor of $f$ and $g$.

Why does this mean that a factorization must be a power of a single irreducible? Suppose $f$ is an irreducible factor of $x^p-t$. Then we can write $x^p-t = fg$ for some $g$; but then $f$ and $g$ have a common factor as above, and that factor must be $f$ since $f$ is irreducible. Continuing this way we can write $x^p-t = f^rg'$ for some $g'$ that does not have any factors of $f$. But then if $g'$ is not a unit, then again we have that $f^r$ and $g'$ must have a common factor by the previous observation... which is impossible since the only common factors would be powers of $f$. Thus, $g$ is a unit and in fact we have $x^p-t = cf^r$ for some $r$.

Answer 2

Here is a proof of a more general result that does not rely on derivatives, but instead unique factorization in polynomials over a field.

Theorem. Let $F$ be a field of prime characteristic $p$ and $t \in F$. If $t$ is not a $p$th power in $F$, then $x^{p^m}-t$ is irreducible in $F[x]$ for all $m \geq 1$.

You might want to work through the proof first in the special case $m = 1$, which is the case you are asking about.

Proof. We'll show the contrapositive: if $x^{p^m} - t$ is reducible in $F[x]$ for some $m \geq 1$ then $t$ is a $p$th power in $F$.

Since $x^{p^m} - t$ is monic, being reducible in $F[x]$ means $x^{p^m} - t = f(x)g(x)$ for some nonconstant monic $f(x)$ and $g(x)$ in $F[x]$. Let $r$ be a root of $x^{p^m} - t$ in some extension field $E$ of $F$, so $r^{p^m} = t$. Then in $E[x]$, $$ x^{p^m} - t = x^{p^m} - r^{p^m} = (x-r)^{p^m} $$ since $E[x]$ is a ring of characteristic $p$. From the equation $$ f(x)g(x) = (x-r)^{p^m} $$ and unique factorization in $E[x]$, we must have $f(x) = (x-r)^d$ where $0 < d < p^m$. Factor the biggest power of $p$ from $d$: $d = p^nt$ where $n \geq 0$ and $p \nmid t$, so $\boxed{n < m}$ and $t \in \mathbf F_p^\times$ in characteristic $p$. In $E[x]$, $$ f(x) = (x-r)^d = (x-r)^{p^nt} = (x^{p^n} - r^{p^n})^t = x^{p^nt} - tr^{p^n}x^{p^n(t-1)} + \cdots. $$ Since $f(x) \in F[x]$, looking at its coefficients tells us $tr^{p^n} \in F$. Since $t \in \mathbf F_p^\times \subset F^\times$, $r^{p^n} \in F$. Thus $$ t = r^{p^m} = (r^{p^n})^{p^{m-n}} \subset F^{p^{m-n}} \subset F^p $$ because $m-n > 0$.

Remark. There is a related result in all characteristics: if $p$ is prime, $K$ is a field (of arbitrary characteristic), and $x^p - a$ is irreducible in $K[x]$, then $x^{p^m}-a$ is irreducible in $K[x]$ for all $m \geq 1$ when $p \not= 2$, but there are counterexamples when $p=2$: $x^2+4$ is irreducible in $\mathbf Q[x]$ but $x^4+4$ is reducible since $$ x^4 + 4 = (x^2 + 2x + 2)(x^2 -2x + 2). $$