# Maximal Ideals of direct products

Maximal Ideals of $R\times S$ are either of the form $A \times S$, where $A$ is maximal in $R$, or of the form $R\times B$, where $B$ is maximal in $S$.

I started by assuming $U$ is maximal in $R \times S$, then $(R\times S)/U$ is a field, which means $((r,s)+U)((r',s')+U)=1+U$, where $(r,s)$,$(r',s') \in R\times S$. Then we get $(rr'-1,ss'-1)\in U$. After this, I have no idea where to go, and how to use the assumption that $U$ is maximal.

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What can you say about the images of $(1,0)$ and $(0,1)$ in $(R\times S) / U$? –  JSchlather Mar 29 '12 at 2:41
Your statement is more generally true of prime ideals of $R \times S$. Not just maximal ideals. –  Rankeya Mar 29 '12 at 3:25
@Rankeya Every maximal ideal of a commutative ring is a prime ideal. So we are taling the same thing. –  Shannon Mar 29 '12 at 14:39
No we are not, because every prime ideal of a ring is not maximal. –  Rankeya Mar 29 '12 at 21:54

Jacob's comment pretty much says it all, but here's the details:

Suppose $U$ contained neither all of $R=R\times \{0\}$ nor $S=\{0\}\times S$. Then we could pick nonzero elements $(r,0) \in R$ and $(0,s)\in S$, neither of which is in $U$, and then $$(r+U)(s+U)=rs+U=(0,0)+U$$ But of course a field does not contain zero-divisors. So $U$ must contain all of $R$ or all of $S$. Using the fact an ideal is maximal if and only if the quotient ring is a field, you can now easily show that the ideal $U$ is of the required form.

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I am confused by your denotion. R is a ring, but you let R = Rx{0} (which is ordered pairs). Again,r is an element in R, U is direct product, is this you mean ((r,0)+U) instead of r+U? –  Shannon Mar 29 '12 at 14:59
@Shannon Yes, this is a common abuse of notation. I am using $r$ to denote both the element of $R$ and the element $(r,0)$ of $R\times \{0\}$ (which is a subset of $R\times S$). $r+U$ denotes the corresponding residue class in $(R\times S)/U$. –  Brett Frankel Mar 29 '12 at 15:59

Prove that if $R_1, \dots, R_n$ are rings, then the ideals of

$$R_1 \times \dots \times R_n$$

are all of the form

$$\mathcal{I}=\mathcal{I}_1 \times \dots \times \mathcal{I}_n$$

where $\mathcal{I}_i \subseteq R_i$ is an ideal. [Hint: consider what happens when you multiply the ideal on the left by $(1,0,\dots,0), (0,1,\dots,0),\dots$.]

Suppose that $\mathcal{I}$ is such that more than one of the $\mathcal{I}_i$'s is different from $R_i$; then we can replace one of these $\mathcal{I}_i$'s with $R_i$ and get an ideal properly containing $\mathcal{I}$. Hence a maximal ideal has all $\mathcal{I}_i$'s but one equal to $R_i$. It is clear that the one $\mathcal{I}_i$ with $\mathcal{I}_i \neq R_i$ is also maximal.

Note: it is no longer true for infinite products of rings. For instance $\prod_{i=1}^\infty \mathbb{Z}/2\mathbb{Z}$ has uncountably many maximal ideals but only countably many maximal ideals of the form just described.

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