Is it true that any prime ideal of a finite direct product ring $R=\prod_{i=1}^nR_i$ is of the form $P=\prod_{i=1}^nI_i$, where $I_j$ is a prime ideal of $R_j$ for some $j$ and $I_i=R_i$, for $i\neq j$?

Any ideal of the form above is prime in $R$. Indeed, if $X=\prod_{i=1}^nA_i$ and $Y=\prod_{i=1}^nB_i$ are ideals of $R$ (where $A_i$'s and $B_i$'s are ideals of $R_i$) such that $XY\subseteq P$, then either $A_j$ or $B_j$ is a subset of a prime ideal $I_j$, for some $j$. Hence either $X$ or $Y$ is a subset of $P$.

Any help/suggestion would be appreciated!

  • $\begingroup$ Yes. In your ring there are $n$ central orthogonal idempotents $e_1,\ldots, e_n$ adding up to $1$, and at most one of these can be outside of $P$, because if $e_i, e_j \notin P$ with $i\neq j$ then $0 = e_i.e_j \in P$ gives a contradiction. On the other hand, if all of the $e_i$ lie in $P$ then $1 = e_1+\ldots+e_n \in P$ which is impossible since prime ideals are proper. So, exactly one $e_j$ lies outside $P$, and $1 - e_j$, which equals the sum of all the other ones, lies in $P$. It follows that $P$ has the form you want. $\endgroup$ – Konstantin Ardakov Oct 9 '17 at 9:59

If I am not mistaken, you've proven ideals of that form are prime, and now you are headed for the converse. Let $I$ be a prime ideal of $R$.

I assume you already know the ideal structure of finite products of rings, so you can see why the ideal must be of the form $I=\prod I_i$ for some ideals $I_i\lhd R_i$. Then $R/I=(\prod R_i)/(\prod I_i)\cong\prod R_i/I_i$.

Now, in order for this product to be a prime ring (which is exactly what a quotient by a prime ideal gives you) it is necessary for all factors to be zero except one, and the one that is nonzero must also be a prime ring.

The first condition implies there is a $j$ such that $R_k=I_k$ for all $k\neq j$, and the second condtion implies $I_j$ is a prime ideal of $R_j$.

A quick note about KonstantinArdakov's comment-solution: it's fine, except for a small danger of readers misunderstanding part of the logic.

As we know, the general definition of prime ideals says that $AB\subseteq P$ implies $A\subseteq P$ or $B\subseteq P$ for any two ideals $A,B$ of $R$. That is, it's the ideal-wise version of the element-wise condition for primality. So something needs to be said about why $e_ie_j=0$ implies one of the two is in $P$.

Now, given that at least one of $a$ and $b$ is central, we can conclude that $(a)(b)=\{0\}\subseteq P$, and it follows that $a\in P$ or $b\in P$, so everything works as hoped.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.