# Prove $f=x^p-a$ either irreducible or has a root. (arbitrary characteristic) (without using the field norm) [duplicate]

Let $K$ be an arbitrary field, $p$ a prime and $a\in K$. Show $f=x^p-a$ is either irreducible in $K[x]$ or has a root in $K$.

My strategy was to split this up into a case for each characteristic.

The characteristic $p$ case is easy. Assume $f$ is reducible $=gh$ both smaller degree $g$ irreducible, with $\alpha$ a root of $g$. Then in $K(\alpha)[x]$, $f=(x-\alpha)^p$. So $g=(x-\alpha)^k$, and all the other factors of $f$ are of the same form. Hence they are all equal to the $minpoly(\alpha)$ and the degree of $minpoly(\alpha)|p$ so it is $1$ or $p$ and we are done.

The characteristic $0$ case and characteristic $q\ne p$ I get stuck on. If $K$ contains a primitive $p$th root of unity $\zeta$, then over $K(\alpha)$, $f$ splits into linear factors. So the degree of the field extension for any root of $f$ is the same, so the degree of such a root is either $1$ or $p$. If $K$ does not have a primitive $p$th root of unity, then simply adjoin one. We know $[K(\zeta):K]|p-1$ and by the previous step either $f$ is irreducible over $K(\zeta)$ or splits into linear factors. If it is irreducible we are done, if not we have it splits into linear factors of the form $x-\zeta^ia^{1/p}$, but I am unsure what to do from here. The characteristic $q$ case seems similar to the characteristic $0$ case but again I am stuck trying to finish the proof.

Some googling has lead me to believe that this problem is generally approached using the field norm and if an accessible online reference exists I would love to see it, but I feel that in the context I have seen it should be entirely possible that this is done without needing the norm. As I have not seen the field norm before, I would like to know if there is a possible way to prove this without using it and hopefully that follows in the same vein my attempt at a proof did.

• That does not give a method without the use of the field norm, and it really does not seem this problem should need one. It seems to me that this should be possible without the field norm, and if it is not, I am asking at least if there is an accessible online exposition of the field norm. A page number in Lang really does me little good. – PVAL-inactive Aug 6 '13 at 9:30
• If you can read something online, why can't you read something in a book? – Gerry Myerson Aug 6 '13 at 9:58
• The answer to the earlier question gives a lot more than a page number in Lang --- it gives the whole proof. It seems to me you can 1) edit the question to make it ultra-clear that you want a proof without norms, and/or 2) ask a new question just asking for an accessible online exposition of field norms. – Gerry Myerson Aug 6 '13 at 10:41
• I added a new answer to the other question. It does not use the field norm. – Makoto Kato Aug 6 '13 at 11:16

Starting where you have left off, if $f=gh$, then $g$ must be a product of terms of the form $x-\zeta^i\beta$, where my $\beta$ is your $\alpha^{1/p}$ (which I think is supposed to be $a^{1/p}$). Given $i$ and $j$ where $x-\zeta^i\beta$ is a factor of $g$ and $x-\zeta^j\beta$ isn't, there's an automorphism taking $\zeta^i$ to $\zeta^j$. The automorphism fixes $g$, since the coefficients of $g$ are in $K$, but it doesn't fix the factors of $g$, contradiction, since we have unique factorization of polynomials over a field.
• Does this same idea work for $K$ characteristic $q\ne p$? Since there $x^p-1$ is separable, so I can adjoin a $p$th root of unity? – PVAL-inactive Aug 6 '13 at 10:29
• @GerryMyerson Can you be more clear about how you know $\zeta^i$ and $\zeta^j$ are Galois conjugates over $K$? Definitely true if $K = \mathbb{Q}$, but why over any field? – Eric Auld Feb 8 '17 at 23:20
• @Eric, they aren't necessarily conjugate over $K$. For one thing, they could both be in $K$. But they are conjugate over the rationals, so there is a ${\bf Q}$-automorphism of $K[\zeta]$ taking the one to the other. – Gerry Myerson Feb 9 '17 at 1:56
• @GerryMyerson But you claim the automorphism fixes $K$, don't you? Also, I thought we were working in arbitrary characteristic $\neq p$. But I'd be happy just to nail down characteristic $0$. – Eric Auld Feb 9 '17 at 2:28