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.

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