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Suppose R, S are two rings (with identity elements in both).

Suppose further I've already proven the following -

Claim #1: any ideal in RxS is necessarily of the form $I_1 \times I_2$ where $I_1, I_2$ are ideals in R, S respectively.

I'd like to prove the following assertion: K is a maximal ideal in $R\times S$ iff it is of the form $I_1 \times I_2$ and at least one of $I_1, I_2$ is a maximal ideal (in R, S respectively).

I reasoned as follows, would appreciate feedback if this is right:

Assume $K$ is a maximal ideal in $R\times S$, so by Claim #1 above it is of the form $I_1 x I_2$ (where $I_1, I_2$ are ideals in $R, S$ respectively).

Assume now that $I_1, I_2$ are not maximal ideals, then we can find $I'_1, I'_2$ such that $I_1 \subset I'_1 \lhd R, I_1 \subset I'_1 \lhd S$, where $I'_1, I'_2$ are (larger) non-trivial ideals in $R, S$ respectively.

By the above Claim #1 again, we know that $I'_1 \times I'_2$ is an ideal of $RxS$, and further we can see that $I_1 \times I_2 \subset I'_1 \times I'_2$ since of course every $(r, s) \in I_1 \times I_2$ also has $(r, s) \in I'_1 \times I'_2$.

Also $I'_1 \times I'_2$ is not all of $R \times S$ since neither $I'_1 = R$ nor $I'_2 = S$ (therefore in particular $1 \notin I'_1$ for example).

Is this right? the proof I provide that $I'_1\times I'_2$ is not all of $R\times S$ and the last step where we show containment of the ideals is legit? I've seen other answers that make use of fancier homomorphism theorem based arguments so left me a bit uneasy.

rschwieb
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giorgio
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    $I_1=I_2={0}$ is a maximal ideal in $F_2$ (the field of two elements), but $I_1\times I_2\subset F_2\times F_2$ is not maximal. So you can see the hypothesis of "at least one" is not going to work. – rschwieb May 07 '24 at 01:10
  • @rschwieb aha, thank you. So I believe I can easily amend my argument a bit to show that we get a similar contradiction to one I mention above in the case that even one of $I_1, I_2$ are not maximal? and then the conclusion is they both need to be maximal? – giorgio May 07 '24 at 01:46
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    In the example I gave they are both maximal in their respective rings. Stop guessing and start thinking. You've basically guessed all cases except the right one. – rschwieb May 07 '24 at 02:48
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    The correct characterization of maximal ideals $I_1 \times I_2$ of $R \times S$ is that either $I_1$ is a maximal ideal of $R$ and $I_2=S$ or $I_1=R$ and $I_2$ is a maximal ideal of $S$. More generally, in a finite direct product $R_1 \times R_2 \times ... \times R_n$, maximal ideals are of the form $I_1 \times I_2 \times ... \times I_n$ where $I_i=R_i$ for all but a single index $i$ and $I_i$ is a maximal ideal of $R_i$ for the unique index $i$ where $I_i \ne R_i$. A similar statement is not true for infinite direct products, however (think of ultrafilters). – Geoffrey Trang May 07 '24 at 13:50
  • It looks good to me! – Marcus K. Johnson May 09 '24 at 11:18

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