Sub-quotients and Gaussian Integer rings

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This has to do with number theory along with group and set theory, but the main focus of the proof is number theory, so forgive me if I'm in the wrong place. I've been struggling to understand a piece of a proof put forth in my book. I know what the Gaussian integers are exactly, and what a subquotient group and isomorphism is (although probably not perfectly, right?), but I don't know what pZ is exactly.

p is a prime.

"We know that p is reducible [ie p=(a+bi)(a-bi), a,b∈Z] iff (p)=pZ is not prime.
Consider the isomorphisms
Z≅Z[X]/(X2+1)
Z/(p)≅Z[X]/(X2+1,p)
Z/(p)≅(Z[X]/(p))/(X2+1)
Z/(p)≅Fp[X]/(X2+1)


If p≠2, we have
(p) reducible⇔X2+1 factors in Fp[X]
⇔-1∈(Fp*)2, the group of squares in Fp*
⇔p congruent to 1 mod(4) (Euler's criterion)​
"


I just barely know where to start with what he's getting at. What is (p), exactly? What is Z[X] and Fp[X]?
 

Answers and Replies

  • #2
morphism
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(p) is the ideal generated by p in Z.

Z[X] and F_p[X] are the rings of polynomials in the indeterminate X with coefficients in the rings Z and F_p, resp.

What your book is doing is showing that just because p is a prime in Z (so the ideal generated by p in Z is a prime ideal), it doesn't follow that the ideal generated by p is prime in Z. The book is (essentially) trying to determine when this is the case, which is precisely when the quotient Z/(p) is a domain.

P.S. You probably don't mean to say "subquotient" here; "quotient" will do just fine.
 
  • #3
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(p) is the ideal generated by p in Z.

Z[X] and F_p[X] are the rings of polynomials in the indeterminate X with coefficients in the rings Z and F_p, resp.

What your book is doing is showing that just because p is a prime in Z (so the ideal generated by p in Z is a prime ideal), it doesn't follow that the ideal generated by p is prime in Z. The book is (essentially) trying to determine when this is the case, which is precisely when the quotient Z/(p) is a domain.

P.S. You probably don't mean to say "subquotient" here; "quotient" will do just fine.

Thanks, you really cleared up some concepts, then. Also, yes, I must have meant quotient. Not sure where the sub came from.
 

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