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 $\phi(\psi(\phi(\psi(x))))=\phi(\psi(x))$.

  • $\begingroup$ By "operation" you probably mean a operation which is defined for all rings simultanously, without any case distinction. This may be formalized using the notion of naturality in category theory: You require that for every ring homomorphism $R \to S$ the obvious diagram with the objects $R^2,S^2,R,S$ commutes. Now for $R=\mathbb{Z}[x,y]$ we can surely write $x*y=\sum_{ij} a_{ij} x^i y^j$ for some numbers $i,j$. Since for two elements $a,b$ of a commutative ring $R$ there is a (unique) hom $\mathbb{Z}[x,y] \to R$ mapping $x \mapsto a$, $y \mapsto b$, we see that $a*b=\sum a_{ij} a^i b^j$ in $R$. $\endgroup$ – Martin Brandenburg Nov 1 '14 at 13:26
  • $\begingroup$ That being said, it is clear that you want to classify those $p \in \mathbb{Z}[x,y]$ such that $p(x,p(y,z))=p(p(x,y),z)$ and $p(x,y)=p(y,x)$, i.e. the non-counital cocommutative coalgebra structures on the ring $\mathbb{Z}[x]$. The counital coalgebra structures are classified in Prop 3.1 in Bergman, Clarke, The automorphism class group of the category of rings. $\endgroup$ – Martin Brandenburg Nov 1 '14 at 13:33
  • $\begingroup$ @MartinBrandenburg I think I already found a counterexample to my question. Take the unary operation $u(x)$ which maps any element of the group of units to $1$, and all other elements to $0$. Then $u(xy)$ is an associative and commutative binary operation not covered by $x_{abc}y$. By "operation", I mean that only the ring structure is used, and not an order or field structure, which wouldn't be available in a general ring. But I don't put any special requirements on $a, b, c$, like being simultaneously defined for all rings. (I downloaded you reference and will read it now.) $\endgroup$ – Thomas Klimpel Nov 1 '14 at 13:48
  • $\begingroup$ Yes, this is an example, but it is not natural (in the sense of category theory). You really have to put some compatibility condition on the operations defined for various rings, because otherwise you may for example define it as $xy$ for one "half" of the rings, and $x+y$ for other half. Or take $\min(x,y)$ for ordered rings, and $x+y$ for non-ordered rings. $\endgroup$ – Martin Brandenburg Nov 1 '14 at 13:52
  • $\begingroup$ In your addendum, how do you know that $(ax+b)*(ay+b)$ is in the range of $\phi$, so that $\phi^{-1}$ is defined there? $/a$ is not well-defined in an arbitrary ring... $\endgroup$ – Thomas Andrews Nov 29 '14 at 1:50

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 $axy + b(x+y) + c$.

  • $\begingroup$ It should be possible to define operations in rings with the help of an "if $x\in R^*$ then $f(x)$ else $g(x)$" construct, where $R^*$ is the group of units of $R$. You argued that this construct is not natural in the sense of category theory, but I fail to see why it should not be natural, as long as $f(x)$ and $g(x)$ are natural. The property of being an element of the group of units seems to be invariant under (unitary) ring homomorphism to me. $\endgroup$ – Thomas Klimpel Nov 1 '14 at 18:26
  • $\begingroup$ Yes, ring homomorphisms preserve units, but they usually don't preserve non-units. $\endgroup$ – Martin Brandenburg Nov 1 '14 at 18:31
  • $\begingroup$ Maybe the "if $x\in R^*$ then $f(x)$ else $g(x)$" can't be interpreted as a natural transformation, but before we can even ask whether an operation is a natural transformation, we have to fix the source and target functor. This was the crucial point why it took me so long to come to terms with adjoint functors: I kept thinking about natural transformations without taking care that the source and target functor are as crucial as the family of morphisms itself. Your diagram looks as if the target functor is an arbitrary $\mathcal F$, and the source is $\mathcal F \times \mathcal F$. Correct? $\endgroup$ – Thomas Klimpel Nov 1 '14 at 18:32
  • $\begingroup$ A natural (binary) operation on commutative rings is a natural transformation $U \times U \to U$, where $U$ is the forgetful functor $\mathsf{CRing} \to \mathsf{Set}$. Using this abstract definition, we see the correspondence to $U(\mathbb{Z}[x,y])$ again: $U$ is represented by $\mathbb{Z}[x]$, hence $U \times U$ is represented by the coproduct $\mathbb{Z}[x] \coprod \mathbb{Z}[x] \cong \mathbb{Z}[x,y]$. Hence, the Yoneda Lemma implies $\hom(U \times U,U) \cong U(\mathbb{Z}[x,y])$. $\endgroup$ – Martin Brandenburg Nov 1 '14 at 18:34
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    $\begingroup$ Wow, you really taught me some category theory today. The fact that ring homomorphisms don't preserve non-units is so typical, yet I still fall into the "law of excluded middle" trap. And the lesson about natural transformations and representation of the forgetful functor was most welcome, since "theoretically I mastered that material some time ago", but this "practical application" was still challenging for me. It took me more than an hour and some repetition of material which I "believed to master", before I got the warm fuzzy feeling that I understood what you told me... $\endgroup$ – Thomas Klimpel Nov 2 '14 at 0:33

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