Thread: Prime Ideals, Maximal Ideals

Is every prime ideal of also a maximal ideal?
If so, how can this be justified?
If not, can anyone think of example?

Obviously, every maximal ideal is also prime. Also, if we were in an finite integral domain, then every prime ideal would be maximal... but has zero divisors, so we are not in an integral domain, and it is also infinite.

Any help much appreciated!
Thanks.

referring to the integers... sorry, I am not sure how to type-face it as bold.

Every ideal of a finite product of rings is a product of ideals in each ring. Such an ideal is prime if and only if it is a product of a prime ideal in one factor and the entire ring in each other factor. Since prime ideals of Z are maximal, the answer is yes.

Note that is \mathbb{Z}.

Thanks, for both the mathematical and formatting aid!

Tinyboss, you wouldn't happen to know where I could find a proof/statment of those claims anywhere, would you?

Thanks again.

You're quite welcome for the formatting tip. Cheers!

try this

http://www.maths.nuigalway.ie/MA416/section3-4.pdf

Hi... I am a little unsure of this result...

I think I have found a prime ideal which is not maximal.

is a prime ideal of

but

is (properly) contained in , also an ideal of , so is not maximal...

so I think there is an issue here?

Originally Posted by matt.qmar

Hi... I am a little unsure of this result...

I think I have found a prime ideal which is not maximal.

is a prime ideal of

but

is (properly) contained in , also an ideal of , so is not maximal...

so I think there is an issue here?

No, you are correct.

It is true that is an ideal if and only if , each an ideal of . Also, is prime implies each is prime in or equals ; however, the converse is not necessarily true (e.g., ).

I don't believe anything can necessarily be said about maximal ideals.

In this case we have (where the first two run over all primes). The latter two are not gonna be maximal, as you have noticed. The others are all maximal, though.