# Prove that $x^4-2x^2-11$ is irreducible over $\Bbb Q$

To find the minimal polynomial of $i\sqrt{-1+2\sqrt{3}}$, I need to prove that $x^4-2x^2-11$ is irreducible over $\Bbb Q$. And I am stuck. Could someone please help? Thanks so much!

• Rational root theorem. – DHMO Apr 13 '17 at 7:40
• @DHMO I think it can only gives us the result that it has no root. But it does not necessarily imply it is not reducible. – non-abelian group of order 9 Apr 13 '17 at 7:42
• You're right. I'm sorry. – DHMO Apr 13 '17 at 7:43
• If it has a factor ovet Q then the lowest coefficient of the factor is a product of roots and rational. Find all four roots and check that no products of a subset s rational. You only need check products of one or two roots. – user1998586 Apr 27 '17 at 7:53

As there are no rational roots the only possible factorisation is into two quadratics whose coefficients we may assume to be integral by Gauss. As 11 is prime and as there is no $X^3$ term we must have $$(X^2 +\alpha X +\epsilon)(X^2 -\alpha X -11\epsilon)$$ where $\epsilon=\pm1$. Now look at the $X$ term and get $\alpha=0$, and a contradiction.

The rational root theorem (or the quadratic formula, solving for $x^2$), shows that there are no linear factors over $\Bbb Q$. That means that if the polynomial is reducible, then it reduces to two irreducible quadratic polynomials.

However, if that were true, then your number would be a root of one of them. The quadratic formula then says it can be written in the form $a\pm\sqrt b$ with $a,b$ rational. The number is pure imaginary, so $a=0$, which means that $b=1-2\sqrt3$, which clearly isn't rational. This is a contradiction.

If you have a candidate $f$ for the minimal polynomial of an algebraic element $\alpha$ over $\mathbf Q$ but you don't know if $f$ is really irreducible and want to avoid trying possible factorizations, you might want to compute the degree $$[\mathbf Q(\alpha): \mathbf Q]$$ and check if it agrees with the degree of $f$.

For instance, in your question, let $$\alpha = i\sqrt{-1 + 2\sqrt 3}$$.

Exercise: show that $[\mathbf Q(\alpha): \mathbf Q] = 4$ using the inclusions $$\mathbf Q \subset \mathbf Q(\sqrt{3}) \subset \mathbf Q(\alpha).$$

See also this answer for a similar application of this philosophy.

As noted already, the rational root theorem excludes rational roots, which only leaves a product of rational quadratics as a potential factorization.

By the way the polynomial was constructed, it is known that its roots are $\,\pm i \sqrt{2 \sqrt{3}-1}\,$ and $\,\pm \sqrt{2 \sqrt{3}+1}\,$. Any quadratic factor of the quartic would need to have two of those as roots, and among quadratics with rational (non-complex) coefficients the only possible pairings would be between the two real roots, or between the two complex roots, respectively. However, both cases result in quadratics that have irrational coefficients $\,(x^2 - 2 \sqrt{3} - 1)\cdot(x^2 + 2 \sqrt{3} - 1)\,$, thus no factorization over $\mathbb{Q}$ exists.