Prime Ideals of direct sum of Z and Z

1. Feb 3, 2012

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.

2. Feb 3, 2012

mathwonk

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

notice the zero ideal is not prime, why not?

3. Feb 3, 2012

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 $\langle (4,6) \rangle \subset \langle (2,3) \rangle$ .

4. Feb 3, 2012

micromass

Staff Emeritus
We have

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

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

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

5. Feb 4, 2012

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.

6. Feb 9, 2012

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.

7. Feb 12, 2012

morphism

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