prime ideals in a product of rings

Quote:

Originally Posted by Biscaim

Let $B=A_{1}\times\cdots\times A_{n}$ be a product of rings. Any prime ideal of B is of the form $A_{1}\times\cdots\times I_{i}\times\cdots \times A_{n}$ , where $i\in\{1,\cdots,n\}$ and $I_{i}$ is a prime ideal of $A_{i}$ .

Thanks in advance.

you should always mention what kind of rings do you have? for example, are they commutative? are they unitary? since you didn't mention that, i'll assume that your rings are commutative.

we first need an important fact:

Fact: every ideal of $B$ is in the form $I=I_1 \times \cdots I_n,$ where each $I_j$ is an ideal of $A_j.$

Proof: it's clear that $I$ is an ideal of $B.$ conversely, suppose $I$ is any ideal of $B.$ for any $1 \leq j \leq n$ consider the map $I \overset{\iota} \longrightarrow B \overset{\pi_j} \longrightarrow A_j,$ where $\iota$ and $\pi_j$ are the inclusion and the projection maps

respectively. let $I_j=\pi_j \iota(I).$ see that $I_j$ is an ideal of $A_j$ and $I=I_1 \times \cdots \times I_n.$ Q.E.D.

now let $I=I_1 \times \cdots \times I_n$ be any ideal of $B.$ we have $\frac{B}{I} \cong \frac{A_1}{I_1} \times \cdots \times \frac{A_n}{I_n}.$ we know that $I$ is prime iff $\frac{B}{I}$ is a domain. now suppose $I_i \neq A_i, \ I_j \neq A_j,$ for some $i \neq j.$ choose $a_i \in A-I_i$ and

$a_j \in A_j - I_j.$ let $x=(0, \cdots, 0, a_i + I_i, 0 , \cdots , 0)$ and $y=(0, \cdots, 0, a_j + I_j, 0 , \cdots , 0).$ then $xy=0 \in \frac{A_1}{I_1} \times \cdots \times \frac{A_n}{I_n}$ and $x \neq 0, y \neq 0.$ so in this case $\frac{B}{I}$ is not a domain. thus in order for $\frac{B}{I}$

to be a domain, we must have $I_j=A_j$ for all but one $j,$ which we'll call it $i.$ then $I=A_1 \times \cdots \times I_i \times \cdots \times A_n$ and $\frac{B}{I} \cong \frac{A_i}{I_i}.$ clearly $I$ is a prime ideal of $B$ iff $I_i$ is a prime ideal of $A_i. \ \Box$