Prime & Maximal Ideals of $\mathbb Z_7$ & $\mathbb Z_3 \times \mathbb Z_5$

[DanielThrice]

I'm working on prime an maximal ideals. My partner and I are studying for our final exam and got conflicting answers.

The question was to find all of the prime and maximal ideals of $$\mathbb Z_7$$. My answer was that because a finite integral domain is a field, the prime and maximal ideals coincide, but that there are no prime and maximal ideals for $$\mathbb Z_7$$.

As for $$\mathbb Z_3 \times \mathbb Z_5$$, what are the prime and maximal ideals, and more importantly, how in the world do we know that we have found them all?

 

[micromass]

I'm working on prime an maximal ideals. My partner and I are studying for our final exam and got conflicting answers.

The question was to find all of the prime and maximal ideals of $$\mathbb Z_7$$. My answer was that because a finite integral domain is a field, the prime and maximal ideals coincide, but that there are no prime and maximal ideals for $$\mathbb Z_7$$.

This is not correct. A (nontrivial) ring ALWAYS has maximal ideals. In this case, {0} is the unique maximal ideal, and it is also prime.

As for $$\mathbb Z_3 \times \mathbb Z_5$$, what are the prime and maximal ideals, and more importantly, how in the world do we know that we have found them all?

There may be others who answer in this thread with sexier solutions, but I would solve this by some kind of brute force:

Every element in $$\mathbb{Z}_3\times \mathbb{Z}_5$$ generates a principal ideal:
- (0,0) generates the zero ideal
- (1,0) generates the same ideal as (2,0)
- (0,1) generates the same ideal as (0,2), (0,3) and (0,4)
- all the other elements are invertible and generate the entire ring.

So we can look at three principal ideals: <(0,0)>, <(1,0)> and <(0,1)>. A quick inspection shows that <(0,0)> is not prime, but that <(1,0)> and <(0,1)> are. Furthermore, we are lucky because these two last ideals are maximal.

This is of course easily generalized: for two field F and K, the ideal $$\{0\}\times K$$ and $$F\times \{0\}$$ are the only prime/maximal ideals of $$F\times K$$.