1. The problem statement, all variables and given/known data Show that [itex] p(x) = x^2 - \sqrt2 [/itex] is irreducible in [itex] \mathbb Z[\sqrt 2] [/itex]. 3. The attempt at a solution I think I have this, but I just want to make sure my reasoning is correct. I'm sure there are other ways. Firstly, it is sufficient to show that p(x) is irreducible in the field of fractions, which is [itex] \mathbb Q[\sqrt 2] [/itex]. Next, since it has degree 2, it is sufficient to show that p(x) has no root in [itex] \mathbb Q[\sqrt 2] [/itex]. Finally, we apply the Field of Fractions Root Test (my name for the rational roots test extended to general UFD's) to say that the only possible roots of p(x) are [itex] \pm \sqrt2 [/itex]. Since neither is a root, we're done. I think the only thing I'm not really sure about here is the "rational roots test," and perhaps the fact that [itex] \mathbb Q[\sqrt 2][/itex] is a UFD. As a matter of fact, a good question might be, if R is a UFD, is [itex] R[\alpha] [/itex] a UFD? I think this would be true, though I'm not certain. In particular, if we move to field theory and S is a field extension of R with [itex] \alpha [/itex] transcendental over R, then [itex] R[\alpha] \cong R[x] [/itex] and since R is a UFD, then R[x] is a UFD, so [itex] R[\alpha] [/itex] is a UFD. (In fact we can so more right? Since then [itex] R[\alpha] [/itex] is actually a Euclidean domain). But what if alpha is algebraic? It seems to me that this would actually be weaker and should still hold. What then do we do when we move out of fields and back into UFDs? Can a similar argument be applied? Edit: Certainly in this case, [itex] \mathbb Q[/itex] is a field so we needn't consider UFD's exclusively, but it's interesting to me as a general question.