# Is there a name for, or notable structure that uses, weird “distributive laws” such as $a\times(b+c)=b\times a+c\times a$?

Consider the following "multiplication" over "addition": $$a \times (b + c)$$ The distributive law in common notion is the left distributive law: $$a \times (b + c) = a \times b + a \times c$$ But what if?: $$a \times (b + c) = b \times a + c \times a$$ $$a \times (b + c) = a \times b + c \times a$$ $$a \times (b + c) = b \times a + a \times c$$ Are there any name for these? I guess they're "anti", "exo", and "endo" distributive, respectively.
Are there any notable algebraic structure with any of these laws?

• I know this isn't what you're looking for, but any abelian structure that has a distributive law will also satisfy these distributive laws. – Carl Schildkraut Oct 14 '18 at 7:54
• Ad hoc, I'd call them skew-distributivity. – Wlod AA Oct 3 '19 at 3:25

My comment was a bit wrong, and too brief, so I'm going to expand this into a partial answer.

Suppose you have a set $$R$$ with operations $$+$$ and $$\times$$ such that $$(R, +)$$ is a group, and $$\times$$ left-endodistributes over $$+$$. Let $$0$$ be the identity under $$+$$.

Fix $$a, b \in R$$. Then $$a \times 0 = a \times (0 + 0) = 0 \times a + a \times 0 \implies 0 \times a = 0.$$ We also have $$0 = 0 \times (a + 0) = a \times 0 + 0 \times 0 = a \times 0,$$ and $$a \times b = a \times (b + 0) = b \times a + a \times 0 = b \times a.$$

Therefore, $$\times$$ is commutative, and therefore distributes over $$+$$.

The same can be done with left-exodistributivity.

For left-antidistributivity, consider first $$0 \times 0 = 0 \times (0 + 0) = 0 \times 0 + 0 \times 0 \implies 0 \times 0 = 0.$$ Next, note that $$0 \times a = 0 \times (a + 0) = a \times 0 + 0 \times 0 = a \times 0.$$ Then, $$(0 + a) \times (0 + a) = (0 + a) \times 0 + (0 + a) \times a = 0 \times 0 + 0 \times a + a \times 0 + a \times a,$$ which, when combined with the above identity, simplifies to $$0 \times a + 0 \times a = 0$$. But then, $$0 = 0 \times a + 0 \times a = a \times (0 + 0) = a \times 0 = 0 \times a.$$ Finally, this again gives us that $$\times$$ is commutative and distributes over $$+$$, as $$a \times (b + 0) = b \times a + 0 \times a = b \times a.$$

While this doesn't mean that left anti/endo/exo-distributivity properties are of no interest, it does mean that, in order to avoid "trivial" examples (i.e. ones where $$\times$$ distributes), we have to sacrifice a fair amount of structure of the additive magma, which means the result is not going to be as "ring-like" as you might have hoped.