Is there a name for an algebraic structure on two copies of $A$

Question

Let $A$ be an associative ring with identity.

On $A\times A$ there is a usual ring structure defined componentwisely, i.e. $(a, b)+(a', b'):=(a+a', b+b')$ and $(a, b)(a', b'):=(aa', bb').$

There is another way to define a ring structure on $A\times A$ as follows:

$(a, b)+(a', b'):=(a+a', b+b')$ and $(a, b)(a', b'):=(aa'+ba'+ab', bb')$.

In general, these two structures would not be isomorphism.

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I am wondering is there a name for this ring structure? Any reference on it would be welcome!

Answer 1

Those two rings actually are isomorphic! The second ring is $A[x]/(x(x-1))$. Letting $I=(x)$ and $J=(x-1)$, so $I+J=(1)=A$, then by the Chinese Remainder Theorem, $$ A[x]/(x(x-1))=A[x]/(I\cap J)\cong A[x]/I\oplus A[x]/J\cong A\oplus A. $$ The isomorphism is $p(x)\mapsto (p(0),p(1))$. This is a surjective ring homomorphism from $A[x]$ to $A\oplus A$. The kernel is $(x(x-1))$, since $p(0)=0$ if and only if $x$ divides $p(x)$ and $p(1)=0$ if and only if $x-1$ divided $p(x)$. Therefore, this mapping induces an isomorphism $A[x]/(x(x-1))\to A\oplus A$. Concretely, $ax+b\mapsto (b,a+b)$.

Therefore, letting $\star$ be the skewed multiplication on $A\times A$, then $A\oplus A$ is isomrphic to $(A\times A,\star)$ via the isomorphism $$ \phi:(A\times A,\star)\to A\oplus A,\\ \phi(a,b) = (b,a+b). $$ To check that this is indeed an isomorphism, note $$ \phi(a,b)\phi(c,d) = (b,a+b)(d,c+d)=(bd,(a+b)(c+d)),\\ \phi((a,b)\star (c,d)) = \phi(ac+ad+bc,bd)=(bd,ac+ad+bc+bd). $$

Answer 2

If $R$ is a ring (possibly without identity) and $S$ is a ring with identity such that $R$ is an $S,S$ bimodule, then $R\times S$ with the product you described is known as the Dorroh extension of $R$ by $S$. In the case I have described, it is a ring with identity $(0,1_S)$ which contains an ideal which is ring isomorphic to $R$.

You could let $S$ not have identity too, probably, you just wouldn't have an identity in the resulting ring. Quite often you see this with $S=\mathbb Z$.

As Mike Earnest has shown in the other solution, it does turn out that if $R=S$ you don't really get anything novel. The solution even works for noncommutative rings since $x$ and $x-1$ are always central elements of $R[x]$, even when $R$ is noncommutative.