Irreducibility of a polynomial if it has no root (Capelli) [duplicate]

Let $F$ be a field of arbitrary characteristic, $a\in F$, and $p$ a prime number. Show that $$f(X)=X^p-a$$ is irreducible in $F[X]$ if it has no root in $F$.

This answer to a related question mentions the result is due to Capelli.

I can prove the result if $F$ has characteristic $p$ as follows. Suppose $f$ is reducible:
$f(X)=g(X)h(X)$ with $g(X)$ an irreducible factor of degree $m$, $1\le m<p$.
Then if $\alpha$ is a root of $g$ in some extension field $K$ of $F$, we have
$$f(X)=X^p-\alpha^p=(X-\alpha)^p$$
so its divisor $g(X)$ must be of the form $(X-\alpha)^m$. Since the coefficient of $X^{m-1}$ in $g$ is in $F$, we have $m\alpha\in F$. So $\alpha\in F$ because $m$ is invertible modulo $p$.

How would you show the result in other characteristics?

Answer

A proof of this result can be found on page 297 of Lang’s Algebra and goes as follows.

Let $F$ be a field of characteristic $q \neq p$. If $f(x)$ has no root in $F$, then it must be the case that $a$ is not a $p$ – th power in $F$. Suppose that $f(x)$ is reducible. By passing the larger extension $K = F(\alpha)$ , we see that $\alpha$ must have degree $d$ where $d < p$. Then $\alpha^p = a$ and by applying $N_{K/F}(-)$ gives that $N_{K/F}(\alpha)^p = a^d$ by multiplicativity of the field norm. Since $(d,p) = 1$ this means that $a$ is a power of $p$ in $F$, a contradiction.

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Source : Link , Question Author : PatrickR , Answer Author : Community

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