# Prime Ideals, Maximal Ideals

• Mar 3rd 2011, 06:07 PM
matt.qmar
Prime Ideals, Maximal Ideals
Is every prime ideal of $Z \oplus Z$ 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 $Z \oplus Z$ has zero divisors, so we are not in an integral domain, and it is also infinite.

Any help much appreciated!
Thanks.
• Mar 3rd 2011, 06:08 PM
matt.qmar
$Z$ referring to the integers... sorry, I am not sure how to type-face it as bold.
• Mar 3rd 2011, 06:54 PM
Tinyboss
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.
• Mar 3rd 2011, 07:45 PM
Ackbeet
Note that $\mathbb{Z}$ is \mathbb{Z}.
• Mar 3rd 2011, 10:41 PM
matt.qmar
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.
• Mar 4th 2011, 06:07 AM
Ackbeet
You're quite welcome for the formatting tip. Cheers!
• Mar 4th 2011, 08:47 AM
kira
• Mar 4th 2011, 10:17 AM
matt.qmar
Hi... I am a little unsure of this result...

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

$\{0\} \oplus \mathbb{Z}$ is a prime ideal of $\mathbb{Z} \oplus \mathbb{Z}$

but

$\{0\} \oplus \mathbb{Z}$ is (properly) contained in $2\mathbb{Z} \oplus \mathbb{Z}$, also an ideal of $\mathbb{Z} \oplus \mathbb{Z}$, so $\{0\} \oplus \mathbb{Z}$ is not maximal...

so I think there is an issue here?
• Mar 7th 2011, 08:02 AM
topspin1617
Quote:

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.

$\{0\} \oplus \mathbb{Z}$ is a prime ideal of $\mathbb{Z} \oplus \mathbb{Z}$

but

$\{0\} \oplus \mathbb{Z}$ is (properly) contained in $2\mathbb{Z} \oplus \mathbb{Z}$, also an ideal of $\mathbb{Z} \oplus \mathbb{Z}$, so $\{0\} \oplus \mathbb{Z}$ is not maximal...

so I think there is an issue here?

No, you are correct.

It is true that $I\subseteq\bigoplus_{i=1}^n R_n$ is an ideal if and only if $I=\bigoplus_{i=1}^n I_n$, each $I_k$ an ideal of $R_k$. Also, $I$ is prime implies each $I_k$ is prime in $R_k$ or equals $R_k$; however, the converse is not necessarily true (e.g., $2\mathbb{Z}\oplus (0)\notin \mathrm{Spec}(\mathbb{Z}\oplus\mathbb{Z})$).

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

In this case we have $\mathrm{Spec}(\mathbb{Z}\oplus\mathbb{Z})=\{\mathb b{Z}\oplus p\mathbb{Z}\}\cup \{p\mathbb{Z}\oplus\mathbb{Z}\}\cup \{\mathbb{Z}\oplus (0),(0) \oplus\mathbb{Z}\}$ (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.