# prime ideals in a product of rings

• Jun 28th 2009, 07:30 AM
Biscaim
prime ideals in a product of rings
Let $\displaystyle B=A_{1}\times\cdots\times A_{n}$ be a product of rings. Any prime ideal of B is of the form $\displaystyle A_{1}\times\cdots\times I_{i}\times\cdots \times A_{n}$, where $\displaystyle i\in\{1,\cdots,n\}$ and $\displaystyle I_{i}$ is a prime ideal of $\displaystyle A_{i}$.

• Jun 28th 2009, 10:57 AM
NonCommAlg
Quote:

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

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

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

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

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

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

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