Which associative and commutative operations are defined for any commutative ring? In a commutative (unital) ring $R$, the binary operation $x*_{abc}y:=axy+b(x+y)+c$ can be defined for arbitrary $a,b,c\in R$. This operation is obviously commutative. It is associative iff $b(b-1)=ac$, because $(x*_{abc}y)*_{abc}z-x*_{abc}(y*_{abc}z)=(b(b-1)-ac)(z-x)$.
Both addition ($x+y=x*_{010}y$) and multiplication ($xy=x*_{100}y$) are special cases of this operation, as is the or operation $1-(1-x)(1-y)=x*_{-110}y$. The operation $\operatorname{min}(x,y)$ is not a special case of this operation, but it is only defined for ordered rings.

Does the operation $x*_{abc}y$ with $b(b-1)=ac$ cover all possible associative and commutative binary operations that are defined for any (general) commutative ring?


One idea how this could be proved is to show that a binary operation, which can be defined for any commutative ring, can be written in the form $x*y=\sum_{ij}a_{ij}x^iy^j$. Then show that the operation won't be associative if the highest exponent is bigger than one, and then the rest should be easy.
Edit This strategy fails, because operations in rings are not limited to addition and multiplication, but can also be defined with the help of an "if $x\in R^*$ then $f(x)$ else $g(x)$" construct, where $R^*$ is the group of units of $R$. This construct is not natural in the sense of category theory, because ring homomorphisms usually don't preserve non-units. If we insist that the operation should be natural, then $a,b,c\in R$ must be replaced by $a,b,c\in \mathbb Z$, and Martin Brandenburg's answer shows that all natural associative and commutative operations are covered by this. If we don't restrict ourselves to natural operations, then it's easy to give a counterexample to the question:
 Let $u(x):=$"if $x\in R^*$ then $1$ else $0$", and define $x*y:=u(xy)$. This binary operation is associative and commutative, but not covered by the operation $x*_{abc}y$.

Appendix The condition $b(b-1)=ac$ becomes clearer, if we note that we can get associative operations from existing associative operations $*$ with the help of an injective transformation $\phi$ via $(x,y)\to\phi^{-1}(\phi(x)*\phi(y))$. Obviously $xy$, $x+y$ and $(x,y)\to c$ are associative. $\phi_{ab}(x)=ax+b$ is injective if $a$ is not a zero-divisor.
$$\phi_{ab}^{-1}(\phi_{ab}(x)\phi_{ab}(y))=((ax+b)(ay+b)-b)/a=axy+b(x+y)+b(b-1)/a$$ $$\phi_{ab}^{-1}(\phi_{ab}(x)+\phi_{ab}(y))=((ax+b)+(ay+b)-b)/a=x+y+b/a$$ Hence we see that no genuinely new associative and commutative operations have been found here. Even the counterexample can be seen as a slight generalization of this construction with a pseudo-inverse $\psi$ instead of $\phi^{-1}$ via $(x,y)\to\psi(\ \phi(\psi(\phi(x)))\ *\ \phi(\psi(\phi(y)))\ )$ where $\phi(\psi(x))$ satisfies $\phi(\psi(x*y))=\phi(\psi(x))*\phi(\psi(y))$ and $\phi(\psi(\phi(\psi(x))))=\phi(\psi(x))$.
 A: As argued in the comments, it is natural to require that the operation is natural ;) in the sense that $*$ is a natural transformation, i.e. for every homomorphism of commutative rings $R \to S$ the diagram
$$\begin{array}{c} R \times R & \xrightarrow{*} & R \\ \downarrow && \downarrow \\ S \times S & \xrightarrow{*} & S \end{array}$$
commutes. Then, using the universal property of the polynomial ring $\mathbb{Z}[x,y]$, it follows that natural, commutative and associative operations on commutative rings correspond to $p \in \mathbb{Z}[x,y]$ with $p(x,y)=p(y,x)$ and $p(x,p(y,z))=p(p(x,y),z)$. The polynomials with the property $p(x,p(y,z))=p(p(x,y),z)$ have been classified in: Cohn, Universal algebra (1965), p. 168, Exercise 2; see also Bergman, Clark in Automorphism class group of the category of rings, Prop 3.1 for a correction. The polynomials are:
$x$, $y$, $axy + b(x+y) + c$, $ayx+b(x+y)+c$, where $a,b,c \in \mathbb{Z}$ with $ac=b(b-1)$. (Unfortunately, I haven't been able to solve this exercise, so hopefully somebody can post his solution.) The only polynomials here which satisfy $p(x,y)=p(y,x)$ are clearly $axy + b(x+y) + c$.
