Does there exist a non-trivial commutative binary operation on $\Bbb C$ that distributes over both multiplication and addition?

In other words, if our operation is denoted by $\odot$, then I want the following to hold:

  1. $a \odot (b \cdot c) = a \odot b \cdot a \odot c$
  2. $a \odot (b + c) = a \odot b + a \odot c$
  3. $a \odot b = b \odot a$

All of the things I can find so far distribute over either multiplication or addition, but not both. Alternatively, is there a proof that no such operation can exist?

I wasn't sure if this question was too elementary for MO, so I'm trying here first. This question is obliquely related to the following other questions I've asked on MO and here:


One such operation is $a\odot b=0$ for all $a,b$. I claim this is the only such operation. Indeed, we have $$a\odot c=a\odot(1\cdot c)=(a\odot 1)\cdot (a\odot c).$$ Taking $c=1$ gives that $a\odot 1$ must be either $0$ or $1$ for each $a$. But if $a\odot 1=1$, then $(a+a)\odot 1=2$, which is impossible. So in fact $a\odot 1=0$ for all $a$, and now the equation above tells us $a\odot c=0$ for all $c$ as well.

This argument uses only the fact that $\odot$ distributes over multiplication on the left and $\odot$ distributes over addition on the right. With slight modification, it applies equally well with $\mathbb{C}$ replaced by any ring in which $2$ is not a zero divisor. Note that in arbitrary rings, there can be other such operations $\odot$. For instance, in a Boolean ring (in which $a\cdot a=a$ for all $a$), $a\odot b=a\cdot b$ is such an operation.

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    $\begingroup$ To clarify: if $a\cdot c=0$ for all $c$ then in particular $a\cdot 1=0$, otherwise if $a\cdot c\ne 0$ for some $c$ then the equation forces $a\cdot1=1$. $\endgroup$ – whacka Aug 3 '15 at 20:53
  • $\begingroup$ Thanks, this is obviously the right answer and I've accepted it. I have a question about MSE etiquette though: I'm now interested in operations that match my original description, but that can be be undefined for 1. Do I post that as a separate question, or edit this one, or...? $\endgroup$ – Mike Battaglia Aug 3 '15 at 21:04
  • $\begingroup$ This isn't just me being difficult BTW, but is actually related to the original problem I was trying to solve with the Dirichlet ring. If you mix the Dirichlet and ordinary convolution rings together, you get an interesting 3-operation structure similar to the thing I'm asking about, except for some difficulty in how Dirichlet convolution works with the ordinary convolution identity. The situation is solvable there by leaving certain things undefined, and leads to a pretty nice structure as a result, so I'm curious if it's solvable here too. $\endgroup$ – Mike Battaglia Aug 3 '15 at 21:08
  • $\begingroup$ @MikeBattaglia: You should post that as a new question. $\endgroup$ – Eric Wofsey Aug 3 '15 at 21:40
  • $\begingroup$ Doesn't matter anyway, here's a stronger result not depending on any multiplicative identity. 2[a@(bc)] = a@(bc) + a@(bc) = (a+a) @ (bc) = [(a+a) @ b] * [(a+a) @ c] = [(a@b) + (a@b)] * [(a@c) + (a@c)] = (a@b)(a@c) + (a@b)(a@c) + (a@b)(a@c) + (a@b)(a@c) = a@(bc) + a@(bc) + a@(bc) + a@(bc) = 4[a@(b*c)] So the existence of two types of distributivity is, itself, bad. I need to go over that Dirichlet ring again, as I've clearly screwed something up here. $\endgroup$ – Mike Battaglia Aug 3 '15 at 22:52

To add to Eric's answer above, it doesn't even require the existence of a multiplicative identity to shoot this down. See below:

$2[a\odot(b \cdot c)] \\ = a\odot(b \cdot c) + a\odot(b \cdot c) \\ = (a+a) \odot (b \cdot c) \\ = [(a+a) \odot b] \cdot [(a+a) \odot c] \\ = [(a \odot b) + (a \odot b)] \cdot [(a \odot c) + (a \odot c)] \\ = (a \odot b)(a \odot c) + (a \odot b)(a \odot c) + (a \odot b)(a \odot c) + (a \odot b)(a \odot c) \\ = a \odot (b \cdot c) + a \odot (b \cdot c) + a \odot (b \cdot c) + a \odot (b \cdot c) \\ = 4[a \odot (b \cdot c)]$

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    $\begingroup$ Told you it might end up being more trivial than you may think. :-) $\endgroup$ – Asaf Karagila Aug 3 '15 at 22:59

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