Recall the question:

Show that and are irreducible over , but reducible in .

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 is the polynomial in the case , , and similarly for . We will prove the required result for where and are primes.

To prove irreducibility over , we use Galois theory, by generalizing the same argument used to prove that is irreducible (this is case ). I’d only seen the argument for before, and realized that a generalization was possible. This was a fun exercise.

Let . Then it is easily checked that:

This motivates the following lemma. Here by we mean the positive square root.

Lemma:There’s an equality .

*Proof. *Clearly , so . We need to prove the converse. Let . Then , and analogously for . Suppose . Then , so we need and .

Now, is a Galois extension, so the above argument shows that by the Galois correspondence. Thus , so that . QED.

We also prove the following result.

Lemma:, i.e. .

*Proof. *First note that . Indeed, suppose otherwise. Then , i.e., , so that , i.e., or . If , then is rational. This is impossible because is irreducible over by Eisenstein. If , then . This is also impossible because is also irreducible, again by Eisenstein.

We also know that . By our analysis above, , and since , it follows that . QED.

This means that the minimal polynomial of must be monic and irreducible of degree , and it must (by definition) divide . Therefore is the minimal polynomial of , i.e., is an irreducible polynomial over .

Let’s move on to the second part. Let be a prime. Then, modulo , either is a square, is a square, or is a square. This follows from quadratic reciprocity: if both and aren’t squares modulo , then since . We can now work case-by-case, where everything is now modulo .

Suppose is a square modulo . Then it is direct to check that:

Suppose is a square modulo . Then it is direct to check that:

Suppose is a square modulo . Then it is direct to check that:

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 is not in , 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 are primes, then is irreducible over , where runs over all conjugates of . This is not irreducible over for prime .