58
$\begingroup$

Hungerford's book of algebra has exercise $6$ chapter $3$ section $6$ [Probably impossible with the tools at hand.]:

Let $p \in \mathbb{Z}$ be a prime; let $F$ be a field and let $c \in F$. Then $x^p - c$ is irreducible in $F[x]$ if and only if $x^p - c$ has no root in $F$. [Hint: consider two cases: char$(F) = p$ and char$(F)$ different of $p$.]

I have attempted this a lot. Anyone has an answer?

$\endgroup$
4
  • $\begingroup$ Have you done it in the case where the characteristic of $F$ is $p$? $\endgroup$ May 27, 2013 at 15:04
  • 1
    $\begingroup$ yes with freshman's dream $\endgroup$
    – user79709
    May 27, 2013 at 15:06
  • $\begingroup$ possible duplicate of $x^p -a$ irreducible in a field of char 0 $\endgroup$
    – vonbrand
    May 27, 2013 at 15:32
  • $\begingroup$ Ok, so since we now assume that the characteristic of $F$ is not $p$, there exists an extension of $F$ which has a primitive $p$'th root of unity $\xi$. If $b$ is any $p$'th root of $c$ in some extension of $F$, then $\xi b$ is a primitive $p$'th root of $c$, and so there exists an extension of $F$ of degree $p$ in which the polynomial $x^p - c$ has a root. But this means that $c$ is a root of some field lying between $F$ and this larger field, which means that this intermediate field has degree dividing $p$, so it must be $F$ since we are assuming that $x^p - c$ is reducible. $\endgroup$ May 27, 2013 at 15:50

2 Answers 2

88
$\begingroup$

Perhaps the simplest tool I can think of is the following:

Let $F$ be a field and $f(x)$ an irreducible polynomial over $F$, then there is a field $K\geq F$ where $f(x)$ has a root; as $f(x)$ is irreducible in $F[x]$, a principal ideal domain, then $\langle f(x)\rangle$ is a maximal ideal of $F[x]$, hence $K=F[x]/\langle f(x)\rangle$ is a field, $\bar{x}$ is a root of $f(x)$, and it is easy to see how to embed $F$ into $K$. Now given a polynomial $f(x)\in F[x]$ it is clear how to construct a field $K\geq F$ such that $f(x)$ factors into linear polynomials in $K[x]$.

Now your question can be answered as follows:

Let $K\geq F$ be a field where $x^p-c$ factors into linear polynomials, say $x^p-c=(x-z_1)\cdots(x-z_p)$. Suppose $x^p-c$ is not irreducible in $F[x]$, then there are polynomials $f(x),g(x)\in F[x]$ of degree $\geq 1$ such that $x^p-c=f(x)g(x)$, then we may assume $f(x)=(x-z_1)\cdots(x-z_n)$, where $\deg f(x)=n<p$.

Put $z=z_1\cdots z_n$, then $z$ is $\pm$ the constant term of $f(x)$, so $z\in F$, and $z^p=(z_1\cdots z_n)^p=z_1^p\cdots z_n^p=c^n$. As $p$ is prime there are integers $a,b$ such that $1=ap+bn,$ then $$(c^az^b)^p=c^{ap}z^{bp}=c^{ap}c^{bn}=c,$$ but $c^az^b\in F$, so $x^p-c$ has a root in $F$.

$\endgroup$
4
  • 6
    $\begingroup$ Why is $\;z^p=c^n\;$ ? I just can't see it...and clearly, even! $\endgroup$
    – Timbuc
    Apr 19, 2014 at 20:11
  • 6
    $\begingroup$ @Timbuc because $z=z_1\cdots z_n$, however $z_i^p=c$ for $i=1,\ldots,n$, so that $z^p=(z_1\cdots z_n)^p=z_1^p\cdots z_n^p=c^n$. $\endgroup$ Apr 20, 2014 at 20:37
  • $\begingroup$ Any idea why there was a hint to consider the characteristic $p$ separately? $\endgroup$
    – Alexey
    Jan 2, 2019 at 20:04
  • $\begingroup$ See also here on $\large \ c^n = z^p\, \Rightarrow\ c = (c^a z^b)^p\,$ for $\,a,b\in\Bbb Z\ \ $ $\endgroup$ Nov 21, 2019 at 1:41
1
$\begingroup$

Let $\mathrm{char}(F)=p$ and $P(X)=X^p-c$ be reducible in $F[X]$ and $L=K(a)$, where $a$ is a root of $P(X)$. Note that in $L[X]$ can write $P(X)=(X-a)^p$. Let $f(X)$ be an irreducible factor of degree $1\leq n<p$ in $F[X]$. Now, $\gcd(P(X),f(X))=f(X)$ and gcd is independent of the field extension, so $f(X)=(X-a)^n$. Hence, $a^n, na^{n-1} \in F$. Since $n$ is coprime to $p$, it follows $a^{n-1}\in F$ and $a \in F$.

$\endgroup$
3
  • $\begingroup$ this is a very nice proof but how did you get $na^{n-1}\in F$? $\endgroup$
    – nomadd
    Feb 12 at 2:46
  • $\begingroup$ @nomadd $na^{n-1}$ is the coefficient of $X$ in $f(X)=(X-a)^n$ which belongs to $F[X]$. $\endgroup$
    – user26857
    May 15 at 9:21
  • 1
    $\begingroup$ The answerer could have considered the coefficient of $X^{n-1}$ in $f(X)=(X-a)^n$ and conclude that $na\in F$. Then use that $\gcd(n,p)=1$. $\endgroup$
    – user26857
    May 15 at 9:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.