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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: $\mathrm{char}(F) = p$ and $\mathrm{char}(F) \ne p$.]

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

user26857
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user79709
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  • Have you done it in the case where the characteristic of $F$ is $p$? – Tobias Kildetoft May 27 '13 at 15:04
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    yes with freshman's dream – user79709 May 27 '13 at 15:06
  • possible duplicate of $x^p -a$ irreducible in a field of char 0 – vonbrand May 27 '13 at 15:32
  • 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. – Tobias Kildetoft May 27 '13 at 15:50

2 Answers2

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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 $F$.

Camilo Arosemena-Serrato
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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 $a \in F$.

user26857
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Varadharajan R
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