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I have asked this on Math Stack Exchange but without answers:

The usual ring operations on $\mathbb{Z}$ can be defined via polynomials in $\mathbb{Z}[a,b]$ (when viewing $a,b$ as variables):

Addition: $(a,b) \mapsto a+b \in \mathbb{Z}[a,b]$

Multiplication: $(a,b) \mapsto a \cdot b \in \mathbb{Z}[a,b]$

There are more ring structures on $\mathbb{Z}$ that can be described with polynomials, e.g. $(\mathbb{Z},P_A,P_M)$ with $P_A(a,b) = a+b-1$ and $P_M(a,b)=a+b-ab$. Here the neutral element w.r.t. $P_A$ is $1$ and those w.r.t. $P_M$ is $0$.

When we don't require the ring to have a neutral element w.r.t. multiplication, we could even take $P_A(a,b)=a+b$ and $P_M(a,b)=0$.

My question is: Can one characterize the pairs $(P_A,P_M)$ of polynomials, such that $(\mathbb{Z},P_A,P_M)$ is a ring? How does this change when we

  • dont' require a neutral element w.r.t. $P_M$?
  • want to get a commutative ring?
  • want to get an integral domain?
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  • $\begingroup$ The second example is not essentially different from the first; you can always assume WLOG that $0$ is the additive identity by translating appropriately. $\endgroup$
    – Qiaochu Yuan
    Commented Jul 22, 2016 at 20:45
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    $\begingroup$ If my scribbling is right, every example with quadratic functions is given by $P_A(a,b)=K+a+b$, $P_M(a,b)=K(KL-1)+KL(x+y)+Lxy$, for constants $K$ and $L$ --- assuming commutativity of addition, associativity of both operations, distributivity, an additive identity and additive inverses. Commutativity of multiplication is then implied. There is a multiplicative identity if and only if $L=\pm 1$. $\endgroup$
    – Steven Landsburg
    Commented Jul 22, 2016 at 21:02

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We can start by looking at $P_A$. The requirement that it be a commutative group implies that for any $a$, $P_A(a, x)$ is a bijective function - so it must be of the form $x + f(a)$ or $-x + f(a)$ for some $f(a)$. By similar consideration on $b$, we can conclude that $P_A(a, b) = a + b + K$ or $P_A(a, b) = -a - b + K$ for some constant $K$. In the latter case, $P_A(P_A(a, b), c) = P_A(-a - b + K, c) = a + b - c + K$, while $P_A(a, P_A(b, c)) = -a + b + c + K$, so this case is eliminated.

We therefore have that $P_A(a, b) = a + b + K$. By translation, as Qiaochu says above, we can take that $P_A(a, b) = a + b$ WLOG. The interesting question then lies in multiplication.

Distributivity says that for $a$ constant, $P_M(a, b)$ is $\mathbb{Z}$-linear in $b$, so $P_M(a, b) = f(a)b$ for some function $f$. Similarly, $P_M(a, b) = a g(b)$. By dividing, we can see that there must be some constant $L$ with $P_M(a, b) = L a b$.

Therefore, the answer to your question is that all are of the form Steven Landsburg gave above, and that all of them are equivalent to the usual addition and multiplication on some ideal of $\mathbb{Z}$ (that ideal being the ideal of numbers divisible by $|L|$.

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  • $\begingroup$ That's a nice point about the ring structure being transported across a bijection $\mathbb{Z} \to K+L\mathbb{Z}$ (assuming the more general case Steven gave, with $K\not=0$). $\endgroup$
    – David Roberts ♦
    Commented Jul 27, 2016 at 6:30
  • $\begingroup$ As a note: the map would have to be $\mathbb{Z} \rightarrow L(K + \mathbb{Z})$. Otherwise, addition doesn't work. Completely unrelatedly, I've been playing around with trying to generalize this proof for more rings, including the reals. Mostly, it comes down to proving the same statement for addition, and then the same proof works for multiplication. However, I don't see a nice way to generally prove the same statement for addition. $\endgroup$
    – user44191
    Commented Aug 3, 2016 at 18:17
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Your polynomial $a+b-ab$ is more or less what is used for the formal group associated to $\mathbb G_m$, since the "identity" of a (1-dimensional) formal group is taken to be $0$. I don't know to what extent there's a theory of "formal rings", but my recollection is that the only polynoimals that give a characteristic 0 formal group law (with $0$ as identity) are $X+Y$ (for the additive group) and $X+Y+XY$. Here a formal group of $R$ is a power series $F\in R[\![X,Y,]\!]$ satisfying various axioms:

  • $F(0,0)=0$
  • $F(X+Y) = X+Y+\text{higher order terms}$
  • $F(X,F(Y,Z))=F(F(X,Y),Z)$

One could also impose $F(X,Y)=F(Y,X)$, but it turns out that this is automatic unless $R$ has zero divisors and nil-potents. There are lots of interesting power series with these properties, but as I mentioned, only two polynomials.

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