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Let $m$ be an integer, squarefree, $m\neq 1$. Prove that $x^3-m$ is irreducible in $\mathbb{Q}[X]$.

My thoughts: since $m$ is squarefree, i have the prime factorization $m=p_1\cdots p_k$. Let $p$ be any of the primes dividing $m$. Then $p$ divides $m$, $p$ does not divide the leading coefficient, $p^2$ does not divide $m$. Hence $x^3-m$ is irreducible over $\mathbb{Q}$ by Eisenstein.


1) Do you think it's correct?

2) Is there some different way to prove irreducibility of this polynomial.

Thanks to all.

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Your solution is perfectly fine. I'm not sure about the second question. – Clayton Sep 24 '13 at 12:22
I might note that you need to exclude $m=1$ here. – Alexander Sep 24 '13 at 12:25
up vote 3 down vote accepted

$x^3-m$ is reducible iff it has a factor of degree 1 iff it has a root iff $m$ is a cube. In particular, $x^3-m$ is irreducible when $m$ is squarefree.

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Since your polynomial is of degree $3$ you can say that it's irreducible if it doesn't have any roots. Notice that this is not true for degrees higher than $3$. For example $x^4+1$ has no real roots, but it is reducible and you have $x^4+1 = (x^2 - \sqrt{2}x+1)(x^2+\sqrt{2}x+1)$.

The reason that this is true for degree $2$ and $3$ polynomials is that if $f$ is irreducible then you'll have a first degree factor, and every first degree polynomial $ax+b$ has a root in a field, namely $-b/a$.

So you can reason that if $x^3-m$ is reducible, then it will have a root. so, if $x=a$ is a root of the equation, we will have $m=a^3$, but that means the exponent of prime factors of $m$ must be divisible by $3$ and this contradicts $m$ being squarefree.

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