TITLE: Prime ideals in a finite direct product of rings
QUESTION [19 upvotes]: Let $S=\prod_{i=1}^{n}{R_i}$ where each $R_i$ is a commutative ring with identity. The prime ideals of $S$ are of the form $\prod_{i=1}^{n}{P_i}$ where for some $j$, $P_j$ is a prime ideal of $R_j$ and for $i\neq j$, $P_i=R_i$.

REPLY [8 votes]: Let $R_{1}$ and $R_{2}$ be two commutative rings with unity.
Let $P$ be a prime ideal in $S=R_{1} \times R_{2}$.
Let $ P= P_{1} \times P_{2}$ where $P_{1}$ is a ideal in $R_{1}$ and $P_{2}$ is a ideal in $R_{2}$.
Then $S/P \simeq R_{1}/P_{1}  \times R_{2}/P_{2} $.
In the whole quotient ring we have $(\hat 1,0)(0,\hat 1)=(0,0)$, hence one of $\hat 1$ it must be zero in the summand quotient ring which means $R_1/P_1 =0$ or $R_2/P_2 =0$. And the whole quotient ring which is integral domain will isomorphism to the non zero summand ring, which will turn to be a integral ring, hence its factor will be a prime ideal.
Therefore only one of the  $P_{1}$ or $P_{2}$ is a prime ideal and other should be corresponding ring.