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I want to find the maximal and prime ideals of $\mathbb{Z}_4 \times \mathbb{Z_4}$ and $\mathbb{Z}_5 \times \mathbb{Z}_6$.

I figured out in a previous post that the ideals of $\mathbb{Z}/4 \times \mathbb{Z}/4$ are a total of $9$.

I know that a maximal ideal is a proper ideal which cannot be contained in a larger ideal. I've been able to find these for single rings like $\mathbb{Z_12}$, but I am unsure how they work for cross products.

If $M_1$ and $M_2$ are the maximal ideals of $R_1; R_2$ where $R_1,R_2$ are rings, then are the maximal ideals of $R_1 \times R_2$ are $M_1 \times M_2$?

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The maximal ideals of a product are $M_1\times R_2$ and $R_1\times M_2$. $M_1\times M_2$ is never a maximal ideal because the two proper ideals above contain it.

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  • $\begingroup$ Well, that makes a lot of sense. Thanks again. To note then, the maximal ideals of $\mathbb{Z}/4 \times \mathbb{Z}/4$ are simply $\mathbb{Z}/4 \times \langle 2 \rangle $ and $\langle 2 \rangle \times \mathbb{Z}/4$. For prime ideals, if our maximal ideals of $R_1 \times R_2$ have products by prime generators, are they considered $\textit{prime}$ ideals of $R_1 \times R_2$? $\endgroup$
    – alnoge
    Commented Mar 1, 2015 at 21:20
  • $\begingroup$ Again, it's better to ask new questions in new questions (not the comments), but I'll answer your question here. If $P_1$ is a prime ideal of $R_1$, and $P_2$ is a prime ideal of $R_2$, then $P_1\times P_2$ is not a prime ideal of $R_1\times R_2$. The reason is that since $P_1$ and $P_2$ are proper, let $a_1\in R_1\setminus P_1$ and $a_2\in R_2\setminus P_2$. Then $(a_1,0)\cdot (0,a_2)=(0,0)\in P_1\times P_2$, but neither factor is in the ideal. However, $P_1\times R_2$ and $R_1\times P_2$ are prime ideals. $\endgroup$
    – Michael Burr
    Commented Mar 1, 2015 at 21:32
  • $\begingroup$ Since I had included both prime and maximal ideals in my title of my post, I thought it would be okay to ask. Thank you for still answering and in the future I will make separate posts for questions. In addition, your explanation is nice and I understand it well. Thank you. $\endgroup$
    – alnoge
    Commented Mar 1, 2015 at 21:54
  • $\begingroup$ @MichaelBurr do you have a link where I can find a proof that the maximal and prime ideals of product of rings $R\times S$ are precisely of the form $M \times S$, $R \times M'$ for maximal ideals $M,M'$ in $R$ and $S$ and $P\times S$ and $R \times P'$ for prime ideals $P,P'$ in $R$ and $S$? $\endgroup$
    – no lemon no melon
    Commented Feb 17, 2021 at 3:58
  • $\begingroup$ @nolemonnomelon Hint: For anything not of this form, look at the product of $(r,0)$ and $(0,s)$ for appropriate $r$ and $s$. For the positive direction, just take quotients and look at the types of rings you get. $\endgroup$
    – Michael Burr
    Commented Feb 17, 2021 at 6:04

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