Prime Ideals of direct sum of Z and Z

[joeblow]

I am trying to find nonzero prime ideals of \mathbb{Z} \oplus \mathbb {Z}, specifically those which are not also maximal.

If I try to do direct sums of prime ideals, the resulting set is not a prime ideal. (e.g., 2 \mathbb{Z} \oplus 3 \mathbb{Z} is not prime since (3,3) \cdot (2,2) = (6,6)\in 2 \mathbb{Z} \oplus 3 \mathbb{Z}, but (2,2),(3,3) \notin 2 \mathbb{Z} \oplus 3\mathbb{Z}.)

In fact, I don't think a prime ideal could be constructed in this way since I can always take a product of the form (1,x)(y,1) and obtain (y,x) and since 1 isn't a multiple of any integer other than 1, neither of the factors would have come from the ideal.

Can somebody please help me find the prime ideals? Thanks.

 

[mathwonk]

modding out a prime ideal is supposed to give a domain. does that help?

notice the zero ideal is not prime, why not?

 

[joeblow]

0 is not prime because (a,0)(0,b)=0, but (a,0),(0,b)\notin 0.

Okay, so would \langle (2,3) \rangle be a prime ideal and \langle (4,6) \rangle be a prime ideal that is not maximal?

If you mod out by both, I believe you get an ID, but \langle (4,6) \rangle \subset \langle (2,3) \rangle .

 

[micromass]

We have

(2,5)*(5,3)\in <(2,3)>

but neither (2,5), not (5,3) is in the ideal.

What about (\alpha\mathbb{Z})\times \mathbb{Z}

 

[joeblow]

Okay, then p \mathbb{Z} \oplus \mathbb{Z} is a prime ideal if p is prime, but it would also be maximal... or would it? If I mod out by it, I think I get a field, so it would have to be maximal.

If we have I= rs \mathbb{Z} \oplus \mathbb{Z}, then I can always write (r,1)\cdot (s,1)=(rs,1)\in I, but (r,1),(s,1)\notin I. Thus, composites don't work.

 

[joeblow]

Okay, I have proven that for any rings R and S, R \oplus S can only have a prime ideal of the form I \oplus S where I is a prime ideal of R or R \oplus J where J is a prime ideal of S. Since the prime ideals of Z are pZ for prime p, the prime ideals of \mathbb{Z}\oplus \mathbb{Z} are p \mathbb{Z} \oplus \mathbb{Z}.

BUT, we have (\mathbb{Z} \oplus \mathbb{Z})/(p \mathbb{Z} \oplus \mathbb{Z})\cong (\mathbb{Z}/p\mathbb{Z}) \oplus (\mathbb{Z}/\mathbb{Z})\cong \mathbb{Z}/p \mathbb{Z} which is a field.

Either I'm missing something simple, or there are no primes that are not also maximal.

 

[morphism]

Don't forget \mathbb Z \oplus p\mathbb Z and the two nonmaximal primes 0 \oplus \mathbb Z and \mathbb Z \oplus 0.