1. ## Couple more quadratic field questions

Why is 3 not a prime in Q(sqrt7) despite that there is no integer in Q(sqrt7) of norm 3?

Like, if 3 can be expessed as a*b where a and b are non-unit integers in Q(sqrt7), then |N(a)|>=2, |N(b)|>=2 (they are not units), and |N(a)||N(b)| = N(3) = 9. Thus it must be that |N(a)|=|N(b)|=3. Contradictions, contradictions.. please help.

Also, what exactly are the prime factors of 3 (in Q(sqrt7))?

2. You seem to be confusing the notions of prime and irreducible. What you have shown is that $\displaystyle 3$ is irreducible in $\displaystyle \mathbb{Q}(\sqrt{7})$. A prime element is always irreducible, but an irreducible element is not necessarily prime. If the ring is a unique factorization domain, then the two notions coincide. In this case, $\displaystyle \mathbb{Q}(\sqrt{7})$ is a unique factorization domain and hence 3 is indeed prime.

3. Originally Posted by Bruno J.
You seem to be confusing the notions of prime and irreducible. What you have shown is that $\displaystyle 3$ is irreducible in $\displaystyle \mathbb{Q}(\sqrt{7})$. A prime element is always irreducible, but an irreducible element is not necessarily prime. If the ring is a unique factorization domain, then the two notions coincide. In this case, $\displaystyle \mathbb{Q}(\sqrt{7})$ is a unique factorization domain and hence 3 is indeed prime.
Then how do you tell if a number is prime? Also, the book i'm using says 3 is NOT a prime in Q(sqrt7)

4. I'm sorry, I made a mistake. Your book is correct; $\displaystyle 3$ is not prime, because it divides $\displaystyle (1+\sqrt{7})(1-\sqrt{7})$ but divides neither of the factors.

In fact we $\displaystyle do$ have an element of norm $\displaystyle -3$. Consider $\displaystyle 2+\sqrt{7}$; its norm is $\displaystyle 2^2-7=-3$. Think this through!