Commutative binary operations on $\Bbb C$ that distribute over both multiplication and addition 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:


*

*$a \odot (b \cdot c) = a \odot b \cdot a \odot c$

*$a \odot (b + c) = a \odot b + a \odot c$

*$a \odot b = b \odot a$


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:


*

*A particularly “natural” algebraic structure with three commutative, pairwise-distributive operations

*Translations AND dilations of infinite series

*Discrete “difference” equations that involve changes in both shift and scale
 A: To add to Eric's answer above, it doesn't even require the existence of a multiplicative identity to shoot this down. See below:
$2[a\odot(b \cdot c)] \\
= a\odot(b \cdot c) + a\odot(b \cdot c) \\
= (a+a) \odot (b \cdot c) \\
= [(a+a) \odot b] \cdot [(a+a) \odot c] \\
= [(a \odot b) + (a \odot b)] \cdot [(a \odot c) + (a \odot c)] \\
= (a \odot b)(a \odot c) + (a \odot b)(a \odot c) + (a \odot b)(a \odot c) + (a \odot b)(a \odot c) \\
= a \odot (b \cdot c) + a \odot (b \cdot c) + a \odot (b \cdot c) + a \odot (b \cdot c) \\
= 4[a \odot (b \cdot c)]$
A: 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.
