What are the prime and maximal ideals of the cartesian product of commutative rings?

Question

Let $R$ and $S$ be two commutative rings. Describe the prime and maximal ideals of $R\times S$ in terms of the prime and maximal ideals of $R$ and $S$.

I am asked to solve this question. Under the influence of this question and the hint given by Michael Burr I'm now trying to prove that all maximal and prime ideals are of the form $P \times S$ or $R\times P$ or $M\times S$ or $R\times M$ for $P,M$ a prime and maximal ideal respectively. I was wondering if my ideas are correct, or if there were any holes in my proof ? ._.

I am able to deduce that all ideals of $R\times S$ must be of the form $I\times J$ for $I,J$ ideals of $R$ and $S$ respectively, from here. (If $I$ is an ideal of $R$ and $J$ is an ideal of $S$ then $I\times J$ is an ideal of $R\times S$, and if $I$ is an ideal of $R\times S$ then there are ideals of $R$ and $S$ such that $I=I_R \times I_S$).

From Corollary 14 of chapter 7 of Dummit and Foote:

Assume $R$ is commutative. Every maximal ideal of $R$ is a prime ideal.

As mentioned in the comments if $P_1,P_2$ are proper prime ideals, then $P_1 \times P_2$ cannot be a prime ideal because given $r \not\in P_1$ and $s \not \in P_2$ then $(r,0)\times(0,s) = (0,0) \in P_1\times P_2$ but $(r,0),(0,s) \not\in P_1 \times P_2$. Since maximal ideals are prime ideals, given $M_1, M_2$ maximal/prime proper ideals, then $M_1 \times M_2$ is not prime, and thus not maximal. A two for one...

If the above is correct, both ideals cannot be proper, thus only one ideal must be the whole ring $R$ or $S$, $R\times S$ would not be proper and prime and maximal ideals are proper ideals. (All prime and maximal ideals must be of the form $P \times S$ or $R\times P$ or $M\times S$ or $R\times M$).

Following through to the next step in the hint I try taking quotients (to show that they are actually maximal or prime ideals):

$P \times S$ or $R\times P$ must be prime ideals for $P$ a prime ideal:

${R\times S}/{R\times P} = \{(r,s)+R\times P | r\in R \text{ and } s\in S\}$

If $s \in P$ then $(r,s) + R\times P = 0 + R\times P$

suppose $(r,s),(r',s')\not\in R\times P$ and $[(r,s)+R\times P]\times [(r',s')+ R\times P] = (r\cdot r', s\cdot s') + R\times P = 0 + R\times P$

then $s,s' \not\in P$ and $s\cdot s' \in P$, but $P$ is prime, contradiction. ${R\times S}/{R\times P}$ has no zero divisors. I guess that makes ${R\times S}/{R\times P}$ into an integral domain if it is assumed that $R,S$ have identity $1\not= 0$... thus $R\times P$ is a prime ideal, similarly $P\times S$ is prime.

I'm not so sure about a two for one in this case,

Is there a way to show that $R\times S / M\times S$ or $R\times S / R\times M$ is a field?

I think the latter would be maximal by a direct proof?

Suppose there is an ideal $M\times S \subset I \subset R$ then $I$ must be of the form $J\times S$ for some ideal $J$ of $R$ where $M \subset J \subset R$. Contradiction because $M$ is maximal. Thus $M\times S$ is maximal in $R\times S$ and similarly $R\times M$ is maximal in $R\times S$. Does this work?

Answer 1

In general, if $I \subset R$ and $J \subset S$ are left ideals of rings $R$ and $S$, then the product of the quotient maps

$$ p_I \times p_J \colon R \times S \to R/I \times S/J $$

has in its kernel the precisely the elements $(x,y)$ such that $x \in I, y \in J$. By the first isomorphism theorem, we get

$$ \frac{R \times S}{I \times J} \simeq R/I \times S/J.\tag{1} $$

Now, a product $A \times B$ of two non-zero rings $A$ and $B$ aways has zero divisors: taking $a \in A$ and $b \in B$ both non-zero, we get $(0,0) = (a,0)(0,b)$.

Hence if $I \times J$ is prime (resp. maximal), one of the factors in the right hand side of $(1)$ must be trivial, and the other must be a domain (resp. field). This translates to saying that one of the ideals must be the whole ring and the other prime (resp. maximal).