Defining additions in the ring of integers Is it possible to describe all possible ways in which one can define additions in the set of integers to give it a structure of ring when the multiplication is same as the usual multiplication ?
If the addition is same as the usual addition then one can easily describe all possible multiplications. What can we say when multiplication is usual ? 
 A: I assert the multiplicative structure of the integers is completely characterized by these three properties:


*

*There are exactly two units: the square roots of unity

*A unique factorization domain

*Has countably infinitely many prime elements


So any ring with these three properties is isomorphic to the integers with non-standard addition, and conversely any non-standard addition defined on the integers gives a ring with these three properties.
Some examples of rings with these properties are:


*

*Any polynomial ring over the integers

*Any polynomial ring over the finite field of 3 elements

*Any polynomial ring over the ring of integers in an imaginary quadratic number field of class number 1 that is neither $\mathbb{Q}(i)$ nor $\mathbb{Q}(\sqrt{-3})$.

A: As Hurkly mentioned, this amounts to finding rings whose multiplicative monoid is isomorphic to that of the ring of integers, which is freely generated by the primes, and by the Conway prime $\,-1,$ of order $\,2$. Thus one seeks UFDs with two units and countably many primes. It is easy to find examples among well-known examples of UFDs (polynomial rings, number fields, etc).
Many properties of domains are purely multiplicative so can be described in terms of monoid structure. Let R be a domain with fraction field K. Let R* and K* be the 
multiplicative groups of units of R and K respectively. Then 
G(R), the divisibility group of R, is the factor group K*/R*. 


*

*R is a UFD $\iff$ G(R) $\:\rm\cong \mathbb Z^{\,I}\:$ is a sum of copies of $\rm\:\mathbb Z\:.$

*R is a gcd-domain $\iff$ G(R) is lattice-ordered (lub{x,y} exists) 

*R is a valuation domain $\iff$ G(R) is linearly ordered 

*R is a Riesz domain $\iff$ G(R) is a Riesz group, i.e. 
an ordered group satisfying the Riesz interpolation property: if  $\rm\:a,b \le c,d\:$  then  $\rm\:a,b \le x \le c,d\:$  for some $\rm\:x\:.\:$ A domain $\rm\:R\:$ is  Riesz if every element is primal, i.e. $\rm\:A\:|\:BC\ \Rightarrow\ A = bc,\ b\:|\:B,\ c\:|\:C,\:$ for some $\rm b,c\in R.$  
For more on divisibility groups see the following surveys: 
J.L. Mott. Groups of divisibility: A unifying concept for 
integral domains and partially ordered groups, Mathematics 
and its Applications, no. 48, 1989, pp. 80-104. 
J.L. Mott. The group of divisibility and its applications, 
Conference on Commutative Algebra (Univ. Kansas, Lawrence, Kan., 
1972), Springer, Berlin, 1973, pp. 194-208. Lecture Notes in Math., 
Vol. 311. MR 49 #2712 
