# A solution to the fun question in the last post

Recall the question:

Show that $x^4-200x^2+36$ and $x^4-240x^2+4$ are irreducible over $\mathbf{Q}$, but reducible in $\mathbf{F}_p$.

First we try the standard methods; the rational root test shows that there aren’t any rational roots. Also, you can’t use Eisenstein here. So we’ll resort to Galois theory :o)

The polynomial $x^4-200x^2+36$ is the polynomial $x^4-2(p+q)x^2+(p-q)^2$ in the case $p=53$, $q=47$, and similarly for $x^4-240x^2+4$. We will prove the required result for $x^4-2(p+q)x^2+(p-q)^2$ where $p$ and $q$ are primes.

To prove irreducibility over $\mathbf{Q}$, we use Galois theory, by generalizing the same argument used to prove that $x^4-10x^2+1$ is irreducible (this is case $p=2,q=3$). I’d only seen the argument for $x^4-10x^2+1$ before, and realized that a generalization was possible. This was a fun exercise.

Let $x=\sqrt{p}+\sqrt{q}$. Then it is easily checked that:

$(\sqrt{p}+\sqrt{q})^4-2(p+q)(\sqrt{p}+\sqrt{q})^2+(p-q)^2 = 0$

This motivates the following lemma. Here by $\sqrt{a}$ we mean the positive square root.

Lemma: There’s an equality $\mathbf{Q}(\sqrt{p},\sqrt{q})=\mathbf{Q}(\sqrt{p}+\sqrt{q})$.

Proof. Clearly $\sqrt{p}+\sqrt{q}\in \mathbf{Q}(\sqrt{p}.\sqrt{q})$, so $\mathbf{Q}(\sqrt{p}+\sqrt{q})\subseteq\mathbf{Q}(\sqrt{p},\sqrt{q})$. We need to prove the converse. Let $\sigma\in\mathrm{Gal}(\mathbf{Q}(\sqrt{p},\sqrt{q})/\mathbf{Q})$. Then $\sigma(\sqrt{p})=\pm\sqrt{p}$, and analogously for $\sigma(\sqrt{q})$. Suppose $\sigma(\sqrt{p}+\sqrt{q})=\sqrt{p}+\sqrt{q}$. Then $\sigma(\sqrt{p})+\sigma(\sqrt{q})=\sqrt{p}+\sqrt{q}$, so we need $\sigma(\sqrt{p})=\sqrt{p}$ and $\sigma(\sqrt{q})=\sqrt{q}$.

Now, $\mathbf{Q}(\sqrt{p},\sqrt{q})/\mathbf{Q}(\sqrt{p}+\sqrt{q})$ is a Galois extension, so the above argument shows that $\#\mathrm{Gal}(\mathbf{Q}(\sqrt{p},\sqrt{q})/\mathbf{Q}(\sqrt{p}+\sqrt{q}))=1$ by the Galois correspondence. Thus $|\mathbf{Q}(\sqrt{p},\sqrt{q}):\mathbf{Q}(\sqrt{p}+\sqrt{q})|=1$, so that $\mathbf{Q}(\sqrt{p},\sqrt{q})=\mathbf{Q}(\sqrt{p}+\sqrt{q})$. QED.

We also prove the following result.

Lemma: $\dim_\mathbf{Q}\mathbf{Q}(\sqrt{p},\sqrt{q})=4$, i.e. $[\mathbf{Q}(\sqrt{p},\sqrt{q}):\mathbf{Q}]=4$.

Proof. First note that $\sqrt{q}\not\in \mathbf{Q}(\sqrt{p})$. Indeed, suppose otherwise. Then $\sqrt{q}=a+b\sqrt{p}$, i.e., $q=a^2+pb^2+2ab\sqrt{p}$, so that $ab=0$, i.e., $a=0$ or $b=0$. If $b=0$, then $\sqrt{q}$ is rational. This is impossible because $x^2-q$ is irreducible over $\mathbf{Q}$ by Eisenstein. If $a=0$, then $\sqrt{q}=b\sqrt{p}$. This is also impossible because $px^2-q$ is also irreducible, again by Eisenstein.

We also know that $\mathbf{Q}(\sqrt{p},\sqrt{q})=\mathbf{Q}(\sqrt{p})(\sqrt{q})$. By our analysis above, $[\mathbf{Q}(\sqrt{p})(\sqrt{q}):\mathbf{Q}(\sqrt{p})]=2$, and since $[\mathbf{Q}(\sqrt{p}):\mathbf{Q}]=2$, it follows that $[\mathbf{Q}(\sqrt{p})(\sqrt{q}):\mathbf{Q}]=4$. QED.

This means that the minimal polynomial of $\sqrt{p}+\sqrt{q}$ must be monic and irreducible of degree $4$, and it must (by definition) divide $x^4-2(p+q)x^2+(p-q)^2$. Therefore $x^4-2(p+q)x^2+(p-q)^2$ is the minimal polynomial of $\sqrt{p}+\sqrt{q}$, i.e., $x^4-2(p+q)x^2+(p-q)^2$ is an irreducible polynomial over $\mathbf{Q}$.

Let’s move on to the second part. Let $a$ be a prime. Then, modulo $a$, either $p$ is a square, $q$ is a square, or $pq$ is a square. This follows from quadratic reciprocity: if both $p$ and $q$ aren’t squares modulo $p$, then $\displaystyle\left(\frac{pq}{a}\right)=\left(\frac{p}{a}\right)\left(\frac{q}{a}\right)=1$ since $\displaystyle\left(\frac{p}{a}\right)=\left(\frac{q}{a}\right)=-1$. We can now work case-by-case, where everything is now modulo $a$.

Suppose $p$ is a square $b^2$ modulo $a$. Then it is direct to check that:

$x^4-2(p+q)x^2+(p-q)^2=(x^2-2bx+q-p)(x^2+2bx+q-p)$

Suppose $q$ is a square $c^2$ modulo $a$. Then it is direct to check that:

$x^4-2(p+q)x^2+(p-q)^2=(x^2-2cx+p-q)(x^2+2cx+p-q)$

Suppose $pq$ is a square $d^2$ modulo $a$. Then it is direct to check that:

$x^4-2(p+q)x^2+(p-q)^2=(x^2-p-q-2d)(x^2-p-q+2d)$

This proves the desired statement. QED.

Some comments: I think there’s a more elementary solution, but this really is (I think) the simplest. Also, the same argument (along with another additional nontrivial step, namely that $\sqrt{p_n}$ is not in $\mathbf{Q}(\sqrt{p_1},\cdots,\sqrt{p_{n-1}})$, see eg. http://math.stackexchange.com/questions/30687/the-square-roots-of-different-primes-are-linearly-independent-over-the-field-of) proves the following more general result: if $p_1,\cdots,p_n$ are primes, then $\prod_{\alpha}(x+\alpha)$ is irreducible over $\mathbf{Q}$, where $\alpha$ runs over all $2^n$ conjugates of $p_1+\cdots+p_n$. This is not irreducible over $\mathbf{F}_q$ for prime $q$.