Let $F$ be a field, let $\omega$ be a primitive $n$th root of unity in an algebraic closure of $F$. If $a \in F$ is not an $m$th power in $F(\omega)$ for any $m\gt 1$ that divides $n$, how to show that $x^n -a$ is irreducible over $F$?
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Below is a classical result:
Theorem $\ $ Suppose $\,c\in F\,$ a field, and $\,0 < n\in\mathbb Z$.
$\ \ \ x^n - c\ $ is irreducible over $F\! \iff c \not\in F^p\,$ for all primes $\,p\mid n,\,$ and $\ c\not\in -4F^4$ when $\, 4\mid n$
A proof is in many Field Theory textbooks, e.g. Karpilovsky, Topics in Field Theory, Theorem 8.1.6.
Bill Dubuque
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7See also Lang's Algebra. There is a whole section about the polynomial $x^n - a$ in the chapter on Galois theory. – KCd Apr 19 '12 at 00:20
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I will assume "$m \geq 1$", since otherwise $a \in F(\omega)$, but $F(\omega)$ is $(n-1)$th extension and not $n$th extension, so $x^n-a$ must have been reducible.
Let $b^n=a$ (from the algebraic closure of $F$).
$x^n-a$ is irreducible even over $F(\omega)$. Otherwise $$f= \prod_{k=0}^n (x-\omega^k b) = (x^p + \cdots + \omega^o b^p)(x^{n-p} + \cdots + \omega^ó b^{n-p}),$$ so $b^p$ and $b^{n-p}$ are in $F(\omega)$. Consequenty $b^{\gcd(p,n-p)}$ is in $F(\omega)$, but $\gcd(p,n-p)$ divides $n$, so $(b^{\gcd})^\frac{n}{\gcd} = a$, a contradiction.
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I don't understand how this proves it. Wouldn't you want to show if $f$ is reducible then a is an $m$th power. It looks like you showed when $f$ is reducible that $a$ is an $n$th power. Also minor thing, the upper index of your product should be $n-1$. – Wyatt Kuehster Jul 22 '19 at 01:40
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@WyattKuehster note it has proved that $b^{gcd}\in F $ so that proved the claim. – mathemather Nov 19 '19 at 14:58
You should have: $(x^2 + 2x + 2)(x^2 - 2x + 2)$.
This is not a counterexample to the original question, because $-4 = (2i)^2$ and $2i \in Q(i)$ and of course $2$ divides $4$. (Anyway, I've given a proof.)