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So I made a mistake in assuming every subring of $R \times S$ is equal to some $A \times B$ for some $A\subset R, B\subset S$ for one of my homeworks.

The grader mention this is a common mistake, hence I was wondering if some one could explain the mistake to me, or better, provide a counter-example to illustrate.

Any insight is deeply appreciated.

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    Obviously there must be some mistake in the line of reasoning that led you to believe that every subring of $R\times S$ is equal to some $A\times B.$ But how can you expect anyone to explain the mistake in a line of reasoning ***that you have not revealed***? Now, if you just assumed something for no particular reason, then ***that*** was your mistake and there's nothing to explain. In a mathematical proof, each step should have a reason. – 2017-02-03
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    The answers given below already address my issue. – 2017-02-03
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    @some1fromhell It's also good to know that all the *ideals* (resp right ideals, left ideals) of the product *do* turn out to be products of ideals (resp right ideals, left ideals) in their respective factor. But that depends on the extra properties of (one/two-sided) ideals. – 2017-02-03

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The problem occurs already at the level of vector spaces over a field $F$. If you take the $2$-dimensional vector space $V = F \times F$, the diagonal $D = \{ (a, a) : a \in F \}$ is not of the form $A \times B$, as $D$ intersects $F \times \{0\}$ and $\{0\} \times F $ trivially.

Now look at the very same example as a ring.

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    Hmm, I could be wrong. But I want to see if my understanding is right. So basically If $S = R$, then lets view it as a vector space (for sake of explaining) Claiming that every subspace of $R^2$ is of the form $A\times B$ is like saying every subspace is 2-D. But as you pointed out, this is clearly not the case. The Diagonal sub-space is 1 dimensional. – 2017-02-03
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    @some1fromhell, yes, sticking to the real case, there are only $4$ possible subspaces of the form $A \times B$, namely $\{0\} \times \{0\}$, $\mathbb{R} \times \{0\}$, $\{0\} \times \mathbb{R}$, $\mathbb{R} \times \mathbb{R}$, but then infinitely many other subspaces $\{ (a, \lambda a) : a \in \mathbb{R}\}$ not of this form, for $0 \ne \lambda \in \mathbb{R}$. – 2017-02-03
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    I see. Thank you, it does clear some of my misconceptions. – 2017-02-03
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Here's a counterexample:

Consider the ring $\mathbb{Z} \times \mathbb{Z}$. It has a subring generated by the element $(1,1)$. The intersection of this subring with each of the factors is just $\{(0,0)\}$, but it is not the trivial ring so is not the product of hte trivial subrings of each factor.

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You were probably confusing subrings with ideals. An ideal $K$ of $R\times S$ is of the form $K=I\times J$ with $I$ an ideal of $R$ and $J$ an ideal of $S$. Indeed, if we consider $$ I=\{r\in R:(r,0)\in K\}, \qquad J=\{s\in S:(0,s)\in K\}, $$ we can show $K=I\times J$.

Suppose $(r,s)\in K$; then $(r,0)=(r,s)(1,0)\in K$, so $r\in I$; similarly, $(0,s)=(r,s)(0,1)\in K$, so $s\in J$. Hence $K\subseteq I\times J$.

If $r\in I$ and $s\in J$, then $(r,s)=(r,0)+(0,s)\in K$; therefore $I\times J\subseteq K$.

If you look closely, you'll notice that actually right ideal has been used; it's easy to modify the proof for left ideals.

For subrings this trick doesn't work and, indeed, $\Delta_R=\{(r,r):r\in R\}$ is a subring of $R\times R$ which is not of the form $A\times B$.

More generally, the minimal subring $P$ of $R\times S$, that is, the subring generated by $(1,1)$, is of the form $A\times B$

Suppose $A$ and $B$ are subrings of $R$ and $S$ respectively and $P=A\times B$. Then, for every integers $m$ and $n$, $(m1,n1)\in P$, so $$ (m1,n1)=k(1,1) $$ for some integer $k$. Just take $m$ and $n$ such that the greatest common divisor of the characteristics of $R$ and $S$ doesn't divide $m-n$ and you get a contradiction, provided the characteristics are not coprime.

We have proved that for any pair of rings (with nonzero identity) $R$ and $S$ having coprime characteristics, there exists a subring of $R\times S$ which is not of the form $A\times B$, for any choice of subrings $A$ of $R$ and $B$ of $S$.