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SiddharthM
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i'm working through the following text and I think I found an error please let me know if I'm totally wrong.
Janusz, Gerald J. Algebraic Number Fields
and I'm starting with the 3rd exercise on page 3. It is as following:
let R be an integral domain and p a prime ideal of R. Show there is an isomorphism between the fields R/p and R_p/(pR_p).
Note that R_p is the localization at the multiplicative set S=R-p (this is multiplicative b/c p is a prime ideal). Now I believe that pR_p is a maximal ideal: this follows because if it were strictly contained in a maximal ideal of R_p then this would be a prime ideal of R_p different from pR_p but this corresponds to a prime ideal of R distinct from p. This corresponding prime ideal in R is in our multiplicative set at which we localized so we have invertible elements in a maximal ideal, a contradiction (maximality implies proper containment).
I used the fact that there is a bijection between prime ideals of R_p and prime ideals of R that don't intersect S. Also that all maximal ideals are prime.
So I believe that R_p/pR_p is a field (because it's a quotient of a ring by a maximal ideal). But there is nothing to suggest that p is a maximal ideal (in a general ring with a general ideal a quotient R/p is a field if and only if p is maximal) unless we are in a PID (because then prime ideals are maximal). So if this result were true it would mean that in ANY integral domain a prime ideal is maximal i.e. that the quotient of an integral domain by one of it's prime ideals is a field, there is an easy counterexample.
Z[x] is an integral domain and <x> is prime (because Z[x]/<x>=Z is an integral domain) but Z[x]/<x>=Z is most definitely NOT a field.
What do you think?
Cheers
Janusz, Gerald J. Algebraic Number Fields
and I'm starting with the 3rd exercise on page 3. It is as following:
let R be an integral domain and p a prime ideal of R. Show there is an isomorphism between the fields R/p and R_p/(pR_p).
Note that R_p is the localization at the multiplicative set S=R-p (this is multiplicative b/c p is a prime ideal). Now I believe that pR_p is a maximal ideal: this follows because if it were strictly contained in a maximal ideal of R_p then this would be a prime ideal of R_p different from pR_p but this corresponds to a prime ideal of R distinct from p. This corresponding prime ideal in R is in our multiplicative set at which we localized so we have invertible elements in a maximal ideal, a contradiction (maximality implies proper containment).
I used the fact that there is a bijection between prime ideals of R_p and prime ideals of R that don't intersect S. Also that all maximal ideals are prime.
So I believe that R_p/pR_p is a field (because it's a quotient of a ring by a maximal ideal). But there is nothing to suggest that p is a maximal ideal (in a general ring with a general ideal a quotient R/p is a field if and only if p is maximal) unless we are in a PID (because then prime ideals are maximal). So if this result were true it would mean that in ANY integral domain a prime ideal is maximal i.e. that the quotient of an integral domain by one of it's prime ideals is a field, there is an easy counterexample.
Z[x] is an integral domain and <x> is prime (because Z[x]/<x>=Z is an integral domain) but Z[x]/<x>=Z is most definitely NOT a field.
What do you think?
Cheers