Rings Isomorphisms I study ring theory, and I wondered about the next question that I asked myself for
practice:
Let $\mathcal{R}_1, \mathcal{R}_2$ be ring, and let $G_1, G_2$ be their additive groups, and $M_1, M_2$ be their multiplicative monoids.
Suppose there are $f_G: G_1\to G_2$ and $f_M:M_1\to M_2$ isomorphisms ($G_1\cong G_2, M_1\cong M_2$).
Is it true that $\mathcal{R}_1 \cong \mathcal{R}_2?$
My motivation to proving this: $G_1=M_1,G_2=M_2$ as sets, by adding binary operetions they induces the rings $\mathcal{R}_1$ and $\mathcal{R}_2$, but the problem is $f_G$ and $f_M$ not necessarily induce ring homomorphism $f_{\mathcal{R}}$.
Thanks in advance.
 A: The answer is no, but nontrivial in the finite case.
In the infinite case, let $F$ be a field, and let $R=F[x,y]$, $S=F[x]$, the rings of polynomials in one and two (commuting) variables. The additive groups of both are isomorphic to a direct sum of countably many copies of $F$, and the multiplicative structure of nonzero elements is isomorphic to the direct product of multiplicative group of $F$ (the group of units), and a direct sum of $\aleph_0|F|$ copies of the free monoid of rank $1$ (this is given by the irreducible polynomials; that is, the free monoid generated by the irreducible polynomials; since both rings are UFDs). You can then adjoin the $0$ to both with no problems to get the full isomorphism of multiplicative monoids. But $S$ is a PID (every ideal is principal), and $R$ is not, so they cannot be isomorphic as rings.
In the finite case, in this math.overflow question that asked precisely this question, Keith Kearnes gives an explicit example of two finite unital rings with isomorphic additive and multiplicative structures but that are no isomorphic. He mentions that if you allow non-unital rings, then examples exist with an $8$-element ring-without-unity.
