A problem in algebraic number theory

  • Thread starter eof
  • Start date
  • #1
eof
58
0

Main Question or Discussion Point

I'm trying to do the homework for a course I found online. A problem on the first homework goes as follows:

Suppose A is an integral domain which is integrally closed in its fraction field K. Suppose q in A is not a square, so that K(sqrt(q)) is a quadratic extension of K. Describe the conditions on r,s in K which are necessary and sufficient for r+s*sqrt(q) to be integral over A in L.

I have absolutely no clue how to approach this as A is not even assumed to be a UFD. The proof for A=Z uses the fact that Z is a UFD, so the minimal polynomial over the fraction field equals the minimal polynomial over A for every integral element (Gauss lemma). Does anyone have any ideas on how to approach this?

Thanks.
 

Answers and Replies

  • #2
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
17
I confess I don't see the problem -- I suppose I've been doing this stuff for too long (or maybe not long enough?). Can you narrow it down? What is the argument you would like to use and where do you think it might not work?
 
  • #3
eof
58
0
I confess I don't see the problem -- I suppose I've been doing this stuff for too long (or maybe not long enough?). Can you narrow it down? What is the argument you would like to use and where do you think it might not work?
Ok, so I know that an element of that form satisfies the equation:

x^2-2rx+r^2-s^2q

For a UFD, this would also have to be the polynomial giving the smallest integral relation for r+s*sqrt(q) over A. Thus, we are reduced to when these coefficients belong to A, which gives us conditions on r and s.

The only reason I know this works for a UFD is because given a monic irreducible polynomial f(X) over A[X] having r+s*sqrt(d) as root, then the minimal polynomial over the fraction field divides this in K[X], but Gauss' lemma tells us that f(X) is irreducible, so f(X) equals the minimal polynomial. This reduces the problem above to checking that the coefficients of the polynomial I have written down are actually in A. I don't see how this approach can be used for a general integral domain.

If there is some approach using some theorems I don't know about please tell me, I'd like to do some reading about those (this is my first exposure to this subject).
 
  • #4
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
17
Hrm. Is this an equivalent statement of what's giving you trouble?

You're worried that the following two statements might be true:
  • x^2-2rx+r^2-s^2q is not an element of A[X]
  • r+s*sqrt(q) is a root of some monic higher degree polynomial in A[X]

I'm pretty sure "integrally closed" tells us this is impossible, but I don't recall the precise details. (and FYI, I'm about to leave)
 
  • #5
eof
58
0
Yup, that's exactly the problem I have.
 
  • #6
eof
58
0
I actually figured out why this is impossible... thanks.
 

Related Threads for: A problem in algebraic number theory

Replies
3
Views
1K
Top