Prove that is irreducible over the quadratic field .

We saw previously that is a Euclidean domain with field of fractions . Thus it is a unique factorization domain, and thus by Gauss’ lemma, is irreducible over if and only if it is irreducible over .

Suppose is reducible in .

Suppose has a linear factor; say . Then we must have that divides in . Recall that the Euclidean norm in this ring is given by ; thus divides 4, so that . Evidently, however, we have , , , , , and . So does not have a linear factor.

Suppose instead that has a quadratic factor, say . Using the long division algorithm for polynomials, we have . Since quotients and remainders in the Euclidean algorithm in are unique, we have and . Moreover, we have as in the linear factor case, and similarly . Note that ; computing norms, we see that divides 64. If , then the second equation gives . Computing norms, we have that 4 divides 25, a contradiction. If , then we have , which is similarly a contradiction. If , then we have . Then is either 1, 5, or 25; since divides 64, and so . Then , a contradiction. If , then we have , a contradiction since no such element exists in . If , then the second equation yields , a contradiction since the only units in are 1 and -1. If , then we likewise have , again a contradiction. Thus does not have a quadratic factor.

Thus is irreducible over , and thus over .

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## Comments

Problem statement should be Q adjoined \sqrt{-2} but it is currently listed with a positive.

Thanks!