Couple more quadratic field questions

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April 27th 2010, 02:39 PM
machack

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))?
April 27th 2010, 03:09 PM
Bruno J.

You seem to be confusing the notions of prime and irreducible. What you have shown is that $3$ is irreducible in $\mathbb{Q}(\sqrt{7})$ . 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, $\mathbb{Q}(\sqrt{7})$ is a unique factorization domain and hence 3 is indeed prime.
April 27th 2010, 03:15 PM
machack

Quote:

Originally Posted by Bruno J.

You seem to be confusing the notions of prime and irreducible. What you have shown is that $3$ is irreducible in $\mathbb{Q}(\sqrt{7})$ . 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, $\mathbb{Q}(\sqrt{7})$ 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)
April 28th 2010, 11:43 AM
Bruno J.

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

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