# Proving that $x^3-2$ is irreducible over $\mathbb{Q}$

I need to prove that $x^3-2$ is irreducible over $\mathbb{Q}$. It is the same as proving that $\mathbb{Q}[x]\cdot(x^3-2)$ is a maximal ideal of $\mathbb{Q}[x]$. This is the same as saying that every polynomial in $\mathbb{Q}[x]$, when multiplied by a member of the set $\mathbb{Q}[x]\cdot(x^3-2)$, is in $\mathbb{Q}[x]$, AND no an ideal which contains this one is either $\mathbb{Q}[x]$ or the own ideal.

What is a technique to prove it? I've seen that if you can prove that the unit polynomial is in there, then this ideal is maximal, but I couldn't do it. Any ideas?

Please, using the maximal ideal theorem I cited

• I don't think that maximal ideal theorem will ever help you show a polynomial is irreducible. It's more useful the other way - find out the polynomial is irreducible, then you know the factor ring is a field. I mean you could show $\mathbb{Q}[2^{1/3}]$ is a field by hand to prove the polynomial is irreducible, but that's a lot of work. Instead, use the rational root theirem. Since the degree of the polynomial is only 3, it is enough to show it has no rational roots. – Noah Olander Jan 10 '16 at 1:47
• Note that if this splits over $\Bbb Q$ then it must have a root, for it must split as either three linear factors or a quadratic and a linear factor. – Pedro Tamaroff Jan 10 '16 at 1:55
• @PedroTamaroff I think you mean "as either three linear factors or a quadratic and a linear factor". – user236182 Jan 10 '16 at 2:06
• @user236182 Yes, that was one "or" too many. – Pedro Tamaroff Jan 10 '16 at 2:07
• "I've seen that if you can prove that the unit polynomial is in there, then this ideal is maximal". If the unit polynomial is there, then the ideal is not maximal by definition, since it is the whole ring! – Maffred Jan 10 '16 at 2:22

No need to use ideals, nor Eisenstein. If $x^3 - 2$ were reducible, it could be written as the product of two monic non-zero polynomials of degree less than $3$, so one of them must have degree $1$.

But $x - r$ divides $x^3-2$ if and only if $r$ is a root of $x^3-2$ over $\Bbb{Q}$. By the rational root theorem $r$ can only be $\pm1$ or $\pm2$, but clearly none of those is a cubic root of $2$.

Note: Since $\Bbb{Q}[x]$ is a PID, this is exactly the same thing as proving that $x^3-2$ cannot lie in a proper ideal containing $(x^3-2)$. My point is that you don't need the notion of ideal here: it doesn't simplify the proof in any meaningful way, it just adds a bit to the stuff you need to write.

Hint: Use the Eisenstein criteria for the prime 2.

If you don't know eisenstein you can do this: If $(x^3-2)$ is not maximal it is contained in an ideal generated by a polynomial $P$ since $Q[X]$ is principal. The degree of $P$ is 1 implies that $X^3-2$ has a root in $Q$ this is not true, but if the degree of $P$ is 2, by dividing $X^3-2$ by $P$ you obtain a polynomial of degree 1 of $Q[X]$ which divides $X^3-2$ impossible again since $X^3-2$ does not have a root in $Q$

• yes, but since this comes before Eisenstein criteria in my book, I must use the theorem I cited – Guerlando OCs Jan 10 '16 at 1:40

Since that has degree 3, in order to be "reducible", it would have to be equal to the product of three linear factors or a linear factor and a quadratic factor. In either case, there would be a linear factor! And to have a linear factor, there would have to be a rational root. By the "rational root" theorem, any rational root, of the form p/q, the denominator, q, would have to divide the leading coefficient, 1, and the numerator, p, would have to divide the constant term, 2. So the only possible rational roots are 1, 2, -1, and -2. None or those satisfy the equation so there is no rational root so no linear factor so the polynomial is irreducible.

By eisenstein's criterion, $2$ divides $2$ but does not divide 1 (the coefficient of $x^3$. $2^2$ does not divide 2. Therefore it is irreducible.

• thank you, but the idea is to use maximal ideals since it comes before in my book – Guerlando OCs Jan 10 '16 at 1:44