Internal product of rings is homomorphic image of direct product

75 Views Asked by At

Let $R$ be a ring (commutative with $1$) and $A,B \subseteq R$ subrings (both still having the same $1$). Then we can consider the outer direct product of the rings $A \times B$ with elementwise addition/multiplication. We can also consider the inner product $$ AB= \{ \sum_{i=0}^n a_i b_i \; | \; n \in \mathbb{N}, a_i \in A ,b_i \in B \} \subseteq R $$ with addition/multiplication in $R$ (which is the smallest subring of $R$ containing both $A$ and $B$).

Is the ring $AB$ a homomorphic image of the ring $A \times B$? Is it true and if so, is there a reference for ist? If it is not true, what counterexample do we have?

If $A \subseteq B$ this is the case as $\phi \colon A\times B \to AB=B,(a,b) \mapsto b$ is an epimorphism. But what about the general case? The obvious map $(a,b) \mapsto ab$ is not a ring-homomorphism and I could not find any other map which would work.

1

There are 1 best solutions below

3
On BEST ANSWER

Suppose that there exists $f\colon A\times B\to AB$; then $f(1,1)=1$; set $x=f(1,0)$ and $y=f(0,1)$. Then $xy=f(1,0)f(0,1)=f(0,0)=0$, but also $x+y=1$, $x^2=x$ and $y^2=y$.

Take your favorite ring with no nontrivial idempotents (a field, for instance) and having at least two distinct subrings.