Proof about ideals of cartesian product of rings

Question

I'm trying to prove the following theorem:

Let $R$ and $S$ be rings. Then every ideal of $R\times S$ is of the form $A \times B$ where $A \subset R$ and $B \subset R$ are ideals.

Here's my proof:

Let $I \subset R \times S$. By definition (in the book) of an ideal of a ring $I$ is an ideal of $R\times S$ iff:

1. $I$ is an additive subgroup of $R\times S$
2. for every $x \in R\times S$ then $xI \subset I$ and $Ix \subset I$.

Since $I$ is an additive subgroup of $R\times S$, every element of $I$ is also an element of $R\times S$, that implies that every element of $I$ has the form of $A\times B$ where $A \subset R$ and $B \subset S$.

$$ I = \{(a,b) \text{ | } a \in A, b \in B\} $$

It remains to show that $A$ and $B$ being ideals is a necessary condition for $I$ to be an ideal of $R\times S$.

In order for the additive subgroup condition be true for $I$, we need that $A$ and $B$ are also additive subgroups of $R$ and $S$ respectively.

Now, for the second condition let $(r,s)$ be an arbitrary element of $R\times S$. Therefore $$ (r,s)I = \{(ra,sb)\text{ | }a \in A, b \in B\} $$ To guarantee that $(ra,sb)I \subset I$ we need to make sure that $ra \in A$ and $sb \in B$.

We can't say anything about $ra$ and $sb$ since $A$ and $B$ are only additive subgroups of $R$ by now.

If we let $A$ and $B$ be ideals of $R$ and $S$ respectively, we'll have $rA \subset A \implies \forall a \in A$: $ra \in A$. Same for $B$. Therefore $A$ and $B$ need to be ideals of $R$ and $S$ respectively in order for $I$ be an ideal of $R\times S$.

Remark: The right-hand side of the proof of $I$ being an ideal was omitted since it uses the same argument of the left-hand side.

Can someone please check my proof? Sorry if I was too detailed in the proof, but I'm trying to be careful in order to not make any silly mistakes.

Any help or constructive critics about the proof are highly appreciated. Thanks!

Answer 1

If (r,s) is in the ideal I in RxS, then (1,0)*(r,s) = (r,0) is in I, and similarly (0,s) is in I. I think this should help.

Answer 2

Your argument has several problems.

You say that because $I$ is a subgroup of $R\times S$ that means that every element has the form $A\times B$ with $A\subseteq R$ and $B\subset S$. Well, this isn't quite right: $A\times B$ is a subset of $R\times S$, and the objects in $I$ are not subsets of $A\times B$, they are elements of $R\times S$.

Now, it is true that there exist subsets $A\subseteq R$ and $B\subseteq S$ such that $I\subseteq A\times B$ (for one thing, you could take $A=R$ and $B=S$; but you can be more precise; you can take $A=\{r\in R\mid \text{there exists }s\in S\text{ such that }(r,s)\in I\}$ and similarly with $B$). But it is not enough to show that these sets are ideals of $R$ and of $S$, because you haven't shown that $I$ is equal to this $A\times B$, just that it is contained in it.

You just asert that $I=\{(a,b)\mid a\in A,b\in B\}$; that is, that $I=A\times B$. But this is precisely what you are supposed to prove, and you have not done so.

Note that not every subgroup of $R\times S$ is of the desired form. For example, if $R=S=\mathbb{Z}$, then the subset $\{(n,n)\mid n\in\mathbb{Z}\}$ is a subgroup of $R\times S$, but it is not of the form $A\times B$ for some $A,B\subseteq \mathbb{Z}$. There is something special about ideals that makes this work, because it doesn't work just for subgroups.

Finally, $A$ and $B$ being ideals of $R$ and $S$ "is necessary for $I$ to be an ideal of $R\times S$" also does not establish that $I$ is of the desired form, namely the collection of all (not just some) elements of the form $(a,b)$ with $a\in A$ and $b\in B$.