Factor principal ideals in a field with class number 2

[R.P.F.]

Homework Statement

Let K be a number field. Let a be an irreducible element of K. Prove that if <a>(the ideal generated by a) is not a prime ideal, then it is a product of two prime ideals.

Homework Equations

The Attempt at a Solution

I'm actually quite clueless... I have been thinking maybe it is possible for me to obtain some information about the extension K:Q from the condition that the class number is 2, like the degree of the extension, etc..?
Any help is appreciated!

 

[morphism]

The class number of K and the degree of the extension K/Q have nothing to do with each other. (There are quadratic extensions of Q with arbitrarily large class number.)

In your problem I assume by an "irreducible element of K" you really mean an irreducible element in the ring of integers of K. Anyway, just start with what you're given: (a) is not prime, so it factors into a product of prime ideals. Some of these will be principal, some not. Try to use the fact that the class group is Z/2Z to simplify your factorization.

 

[R.P.F.]

Thanks for your reply!

In your problem I assume by an "irreducible element of K" you really mean an irreducible element in the ring of integers of K.

Yes. This is what I meant.

Some of these will be principal, some not.

Could they really be principal? Would that imply that a has a proper factor, which is a contradiction?

 

[morphism]

Could they really be principal? Would that imply that a has a proper factor, which is a contradiction?

Good! Can you use this observation to finish off the solution?

 

[R.P.F.]

Figured it out! Finally can take a break after 8 hours in the library.

Thanks so much.

 

[morphism]

Well done!