# Different ring structures on the product of rings

Let $$(R_1, +_1, \circ_1)$$ and $$(R_2, +_2, \circ_2)$$ be two rings. Let us consider the set $$R = R_1 \times R_2$$. We know it has a ring structure given by the following:

$$(x,y) + (x^\prime, y^\prime) = (x+_1 x^\prime, y +_2 y^\prime)$$ and $$(x,y)\circ(x^\prime, y^\prime) = (x\circ_1 x^\prime, y \circ_2 y^\prime)$$

Question: I am looking for different ring structures on $$R.$$ Any help will be appreciated.

## 1 Answer

In general if you have a bijective map $$f$$ of a ring $$S$$ to itself then you can define two new operations

$$x+_fy:=f^{-1}(f(x)+f(y))$$

$$x*_fy:=f^{-1}(f(x)f(y))$$

In this case $$(S,+_f, *_f)$$ is a ring, where the neutral element with respect to $$+_f$$ is $$f^{-1}(0)$$ and the neutral element with respect to $$*_f$$ is $$f^{-1}(1)$$

Example:

If you choose $$f: S\to S$$ such that

$$f(x):=-x+1$$

you have that the inverse is $$g(x)=-x+1=f(x)$$ and in this case we have

$$g(0)=1$$ and $$g(1)=-1+1=0$$

So $$1$$ is the new neutral element with respect to $$+_f$$ and $$0$$ is neutral element with respect to $$*_f$$ on the new ring $$(S,+_f,*_f)$$. The new operations are

$$x+_fy=f(-x-y+2)=x+y-1$$

$$x*_fy=f((-x+1)(-y+1))=$$

$$f(xy-x-y+1)=x+y-xy$$

The problem is that $$f: (S,+_f,*_f)\to (S,+,*)$$ is an isomorphism of rings, so the two structures are equal.

In your case if you want a new structure on $$R$$, equal to the initial structure up to isomorphism, you can consider two bijective maps $$f:R_1\to R_1$$ and $$g:R_2\to R_2$$, and in this case you have that

$$(R, +_{(f,g)}, *_{(f,g)})$$ is a new ring but it is isomorphic to the initial ring $$(R,+,*)$$

If you want a different structure you can consider a generalization of semi-direct products for Rings:

If you have a morphism

$$\psi: R_2\to Aut((R_1,+,*))$$

then you have that

$$(a,b)+^\sim(c,d)=(a+\psi(b)(c), b+d)$$

and

$$(a,b)*^\sim(c,d)=(a\psi(b)(c), bd)$$

are two operation on $$R_1\times R_2$$ such that

$$(R_1\times R_2, +^\sim,*^\sim)$$ is a ring different from $$(R,+,*)$$

• For so your particular example $f(x) = -x+1$ we get the ring structure on $S$ given by $x+_f y = x+y-1$ and $x \ast_f y = x+y-xy.$ Also, we know that any map $S \times S \to S$ can be thought as a map $S \to Hom(S, S)$. So, I would like know is this function $f$ induces any ring homomorphism from $S \to S.$ – Surojit Aug 18 at 10:30
• @Surojit it is not true, the map $S\times S \to S$ must be bilinear to get that it can be thought as a map S\to hom(S,S). – Federico Fallucca Aug 18 at 12:11
• So, if $f: S \to S$ is a bijection such that $f \circ f= id$, then the map $f : (S, +, \ast) \to (S, +_f, \ast_f)$ is a ring homomorphism. – Surojit Aug 18 at 13:36
• Also, this map $f$ is a ring isomorphism too. Thus, it implies that the rings $(S, +, \ast)$ and $(S, +_f, \ast_f)$ are isomorphic rings. But I need to construct some ring $(S, +^\prime, \ast^\prime)$ such that it is not isomorphic to $(S, +, \ast).$ – Surojit Aug 18 at 13:40
• @Surojit This is known as transport of structure – Bill Dubuque Aug 18 at 14:23