Showing that $x^4-18x^2 + 25$ is irreducible over $\mathbb Q$

Question

This is part of a past qualifying exam problem:

Let $\alpha = \sqrt 2 + \sqrt 7$ and let $F=\mathbb Q(\alpha)$. Find the minimal polynomial of $\alpha$ over $\mathbb Q$ and determine $[F: \mathbb Q]$.

Work so far. I was able to calculate the following polynomial over $\mathbb Q$ that has $\alpha$ as a root: $f(x):= x^4 - 18x^2 + 25$. I'm pretty sure that this is the minimal polynomial, but I also need to verify it is irreducible over $\mathbb Q$. Eisenstein's criterion doesn't apply as far as I can tell, and since it's a quartic, then it's not enough to check that $f$ has no roots in $\mathbb Q$ (using, e.g., rational roots test).

That being said, using rational roots test, I can deduce that if $f$ does factor, it must factor as a pair of quadratics. My question is: besides comparing coefficients and setting up a system of equations to rule out a factorization into quadratics, is there any easier way to show that $f$ is irreducible over $\mathbb Q$?

Answer 1

To solve this problem, we can first decompose the polynomial $f(x)$ into a product of 4 first-degree polynomials (Gauss's theorem implies that such a decomposition is unique).

$$ x^4 -18x^2+25=x^4+10x^2-28x^2+25=(x^2+5)^2 - (2x\sqrt{7})^2 = \\\\ = (x^2+5-2x\sqrt{7})(x^2+5+2x\sqrt{7})=(x^2-2x\sqrt7+7-2)(x^2+2x\sqrt{7}+7-2)=\\\\= [(x-\sqrt7)^2 - (\sqrt2)^2]\cdot [(x+\sqrt7)^2 - (\sqrt2)^2] = \\\\=\boxed{\bf (x-\sqrt7-\sqrt2)(x-\sqrt7+\sqrt2)(x+\sqrt7-\sqrt2)(x+\sqrt7+\sqrt2)}.$$

Now, let's assume that there exist two polyminals $g(x)$ and $h(x)$, such that $g(x)\cdot h(x)=f(x)$. These polyminals must, like all 2-degre polynominals, consist of 2 polyminals of degre 1 each - and these polyminals must be divisors of $f(x)$ because divisibility is a transitive relation. There are only 4 divisors of $f(x)$ like that - they are shown above. So, we have to consider only $3 $ possibilites, and none of them matches our search: for all 3 cases, both $g(x)$ and $h(x)$ are not in $\mathbb{Q}$.

So, we can conclude, that the polynominal $f(x) := x^4 - 18x^2 + 25$ is irreductible over $\mathbb{Q}$.

Answer 2

In order to prove that $f(x)=x^4-18x^2+25$ is irreducible over $\mathbb Q$ we can prove that $f(n)$ is prime for some values of $n$. The reason is that if $f(x)=g(x)h(x)$ then $f(n)=g(n)h(n)$ can be prime only when one of the factors $g(n),f(n)$ is equal to $1$ and the other is prime.

Since $f(x)=f(-x)$ each time that $f(n)$ is prime so is $f(-n)$.

We have $$f(2)=-31\\f(4)=-7\\f(6)=673\\f(8)=2969$$ so we have show eight primes which is a proof that $f(x)$, of four degree, is irreducible over $\mathbb Q$.

Note.-According to the Bunyakovsky's conjecture, $f(x)$ is prime for infinite values of $f(n)$

Answer 3

Using elementary methods only...

You can quickly check $f(\pm1)$, $f(\pm5)$, $f(\pm25)$ are all nonzero so by the rational root theorem, $f$ has no linear factor.

So if $f$ factors, it factors as two quadratics: $$x^4-18x^2+25=\left(x^2+ax+b\right)\left(x^2+cx+d\right)$$ With no cubic term, $c=-a$: $$x^4-18x^2+25=\left(x^2+ax+b\right)\left(x^2-ax+d\right)$$ With no linear term, $ad=ab$. Either:

• $a=0$ $$x^4-18x^2+25=\left(x^2+b\right)\left(x^2+d\right)$$ $$x^4-18x^2+25=x^4+(b+d)x^2+bd$$ But there are no integers that multiply to $25$ and sum to $-18$.

Or

• $a\neq0$, so $d=b$: $$x^4-18x^2+25=\left(x^2+ax+b\right)\left(x^2-ax+b\right)$$ Now $b$ must be $\pm5$ considering the linear term: $$x^4-18x^2+25=\left(x^2+ax\pm5\right)\left(x^2-ax\pm5\right)$$ Finally, balancing the quadratic terms, $$-18=\pm10-a^2$$ But neither $8$ nor $28$ is a perfect square.

So this factorization is not possible.