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April 12th, 2009, 09:48 AM   #1
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algebraic number theory - proving an ideal is prime

I don't understand the following proof of the fact that in the number ring R=Z[sqrt(-5)], the ideal (2, 1+sqrt(-5)) is prime.

Quote:
 This can be seen by observing that |R/(2)| = 4, hence R/(2, 1+sqrt(-5)) has order dividing 4. The only possibility is 2 because (2, 1+sqrt(-5)) contains (2) properly and cannot be all of R (if it were, then its square would also be R). This implies that in fact (2, 1+sqrt(-5)) is maximal as an additive subgroup, hence a maximal ideal, hence a prime.
I'm kind of lost. Why is |R/(2)| 4? and why does R/(2, 1+sqrt(-5)) has order dividing 4?

Lots of thanks.

April 14th, 2009, 12:15 PM   #2
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Re: algebraic number theory - proving an ideal is prime

Quote:
 Originally Posted by nilap I'm kind of lost. Why is |R/(2)| 4?
In Z[sqrt(-5)], an element looks like a+b(sqrt(-5)). If you mod out each term by two, how many choices do you have for a and b? So then how many total values can a number take?

Quote:
 and why does R/(2, 1+sqrt(-5)) has order dividing 4?
(2) is a sub-ideal of (2, 1+sqrt(-5)). So R/(2) contains R/(2,1+sqrt(-5)). Since they are both quotient rings, R/(2,1+sqrt(-5)) is an additive sub-group of R/(2). So, then we appeal to group theory.

Does that help?

Quote:
 Lots of thanks.
No problem,
Cheers.

April 14th, 2009, 04:46 PM   #3
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Re: algebraic number theory - proving an ideal is prime

Quote:
 Originally Posted by cknapp Does that help?
Yes, it helps a lot. Thanks!

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