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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?

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PatrickR
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3 Answers3

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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.

  • I'm curious though because I don't see where the hypothesis that the characteristic is not equal to $p$ is used. –  Dec 28 '12 at 01:33
  • What I think: If the field is of characteristic $p$ then for any $a\in F$ there is $b\in F$ s.t $b^p=a$ hence $F(b)=b^p-a=a-a=0$ hence the polynomial is reducible – Belgi Dec 28 '12 at 01:39
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    @Belgi The Frobenius automorphism is surjective IIRC iff the field is finite. –  Dec 28 '12 at 01:40
  • hmm yes, but we are not given that $F$ is not finite – Belgi Dec 28 '12 at 01:41
  • @Belgi Nor are we given that $F$ is finite. The point now is that $F$ is an arbitrary field of arbitrary characteristic. –  Dec 28 '12 at 01:42
  • But shouldn't the argument fail in a finite field ? – Belgi Dec 28 '12 at 01:43
  • @Belgi Which argument are you referring to? –  Dec 28 '12 at 01:44
  • You asked about where does the assumption that $q\neq p$. If $F$ is finite and $p=q$ I showed the polynomial is reducible – Belgi Dec 28 '12 at 01:45
  • @Belgi Your argument doesn't contradict the claim, if you *can write* $a = b^p$ for some $b \in F$ then this means that the polynomial a priori already has a root which is against the hypothesis that *to start out with* $f$ has no root. –  Dec 28 '12 at 01:53
  • If $p=q$ and the field is finite the polynomials always have a root. I am not saying the answer is wrong, I am commenting to yourr first comment – Belgi Dec 28 '12 at 01:55
  • @BenjaLim The Frobenius automorphism is surjective iff the field is perfect. For instance the Frobenius automorphism is surjective on $\overline{\mathbb F_p}$. – JSchlather Dec 28 '12 at 02:05
  • This question is an exercise in Hungerford's chapter on Rings, where nothing much is supposed to be known about fields. I was wondering if there was a simpler proof. But it will do. I'll have to do some reading about norms. – PatrickR Dec 28 '12 at 02:38
  • @PatrickR The problem is already a problem about polynomial rings over *fields* and already your first line begins with "let $F$ be a field..." so the problem *is very much* about fields. –  Dec 28 '12 at 03:29
  • @BenjaLim You absolutely right. Sorry I am a little slow, but one clarification for the end: $a=\alpha^p$ and $a^d = b^p$ for some $b\in F$ ($b$ = the norm). So how do you conclude that $\alpha\in F$? – PatrickR Dec 28 '12 at 04:07
  • @BenjaLim I've got it now. I think at the end you meant $a$ (and not $\alpha$) is a $p$-th power in $F$: $a^d$ is a $p$-th power and $a^p$ is a $p$-th power, and $d$ and $p$ are relatively prime, hence $a$ is a $p$-th power in $F$. – PatrickR Dec 28 '12 at 04:47
  • @BenjaLim Thanks a lot for the answer. I have accepted it. Can you make the minor correction at the end? – PatrickR Dec 28 '12 at 05:12
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Suppose the characteristic of $F$ is not $p$. Let $\Omega$ be the algebraic closure of $F$. By the assumption on the characteristic of $F$, $\Omega$ has a primitive $p$-th root of unity $\zeta$. Let $\alpha$ be a root of $x^p - a$ in $\Omega$. Since $\alpha$ is not contained in $F$, $\alpha \neq 0$. Hence $\alpha, \alpha\zeta, \cdots, \alpha\zeta^{p-1}$ are distinct roots of $x^p - a$. Suppose $x^p - a = g(x)h(x)$, where $g(x)$ and $h(x)$ are monic polynomials in $F[x]$ and $1 \le$ deg $g(x) \lt p$. Let $k =$ deg $g(x)$. Let $b$ be the constant term of $g(x)$. Then $b = (-1)^k \alpha^k \zeta^m$, where $m$ is an integer. Hence $b^p = (-1)^{kp} a^k$ If $(-1)^{kp} = 1$, then let $c = b$. Suppose $(-1)^{kp} = -1$. If $p$ is odd, then let $c = -b$. If $p = 2$, then let $c = b$. In either case, $c^p = a^k$.

Let $\Gamma$ be the multiplicative group of $F$. Let $\Gamma^p = \{x^p |\ x \in \Gamma\}$. $\Gamma^p$ is a subgroup of $\Gamma$. Let $\pi$ be the canonical homomorophism $\Gamma \rightarrow \Gamma/\Gamma^p$. Let $\beta = \pi(a)$. Since $x^p - a$ does not have a root in $F$, $\beta \neq 1$. On the other hand, $\beta^p = \pi(a^p) = 1$. Hence the order of $\beta$ is $p$. However, since $c^p = a^k$, $\beta^k = 1$. This is a contradiction. Hence $x^p - a$ is irreducible in $F[x]$.

Makoto Kato
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1) I can give you the details of a paper by Chebotarev translated from russian to german in which appears the Capelli Lemma, yet

2) There's also a paper in italian by Capelli, from 1904, which details I can give you, yet

3) The book "Algebra I", by Rédei, has the same lemma with extensions. Alas, the book was written in hungarian, though it seems to be there's a translation to german with, perhaps, the help of Halmös.

Googling around a little there are several references to that lemma but, as far as I could see, none of the first ones, at least, brings the version you want.

DonAntonio
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