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 1


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



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 $f^{-1}(1)$


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


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




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

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


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

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

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

  • $\begingroup$ 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 $S \to S.$ $\endgroup$
    – Surojit
    Aug 18, 2019 at 10:30
  • $\begingroup$ @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). $\endgroup$ Aug 18, 2019 at 12:11
  • $\begingroup$ 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. $\endgroup$
    – Surojit
    Aug 18, 2019 at 13:36
  • $\begingroup$ 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 $(S, +, \ast).$ $\endgroup$
    – Surojit
    Aug 18, 2019 at 13:40
  • $\begingroup$ @Surojit This is known as transport of structure $\endgroup$ Aug 18, 2019 at 14:23

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