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Distributivity is inherent in the definitions of rings, boolean algebras, etc.

But why distributivity specifically?

Let me qualify that question a bit and try to get to the bottom of what I mean by "why". Obviously given a set $A$ with two binary operations $\ast,\circ :A\times A\rightarrow A$, if we want to study both $\ast$ and $\circ$ together with the hopes that will complement each other, then we will need some kind of relation between them; distributivity clearly fills this role.

But why not another property? For example, the property that $$(x\ast y)\circ (z\ast w)= (x\circ y)\ast (z\circ w)$$

Now, I already have some motivating factors for why distributivity instead of something else:

  • Many of the most important structures that we deal with have distributivity holding between the relevant operations, whether it be fields like $\mathbb{Q},\mathbb{R},\mathbb{C}$, or some rings like $\mathbb{Z}$ or $\operatorname{End}(G)$ (where $G$ is an abelian group).
  • It is extremely simple while not being symmetric (the non-symmetry of the definition is important, as it helps to distinguish between the two operations, and makes the cases in which we do have a symmetry, like in lattices, all the more important and indicative of important ideas, like duality)
  • Geometrically, looking at the case of addition and multiplication in, say, $\mathbb{R}$, we can interpret distributivity as the additivity of area, e.g. given a rectangle $R$ with side lengths $a$ and $b$, if $R_1$ and $R_2$ are subrectangles partitioning $R$ with side lengths $a,c$ and $a,d$, then we would expect the area of $R$ to be equal to the sum of the areas of $R_1$ and $R_2$. I.e. $a(c+d)=ac+ad$.

I suppose I'm hoping for a more algebraic motivation that does not depend (largely) on any of the above arguments; I just don't find that they are "enough" of a motivation without already having worked through something like ring or field theory where distributivity is indispensable.

For example, I find a nice motivation for the associativity of a binary operation $\ast:A^2\rightarrow A$ as representing that $\ast$ is, when regarded as $A$ acting on itself by right multiplication through $\ast$, compatible with itself (e.g. given a group $G$, there is the group action $G\times G \rightarrow G$ given by $(g,h)=g\ast h$, and the fact that we have compatibility of this action is exactly the fact that $\ast$ is associative). Similarly, commutativity says that there is no distinction between $A$ acting on itself on the right or left by $\ast$, it simply acts.

(It is worth noting, I think, that the commutative diagram characterizing distributivity is quite large and ugly, unlike for associativity, which is a nice commutative square, or commutativity, which is a commutative triangle. However, the same can be said for the diagram for inverses.)

Hayden
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1 Answers1

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Let's give "left multiplication by $a \in A$" a name, $l_a: A \times A \to A$.

Now the usual distributive property, $a\cdot (b+c) = a\cdot b + a\cdot c$, is equivalent to saying that, for each $a \in A$,

$$l_a(b+c) = l_a(b) + l_a(c);$$

we've got a family of homomorphisms of $(A, +)$.

In this sense, it seems to me that requiring multiplication to induce a homomorphism is as "compatible" as we can ask $(A, \cdot)$ to be with $(A, +)$.

Sidenote: I've always wanted a shirt that says "respect addition" and has the distributive law on it.

pjs36
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  • Yeah, I've noticed this as well, but whether it is a homomorphism depends on the context a bit more (for example, when considering unital rings multiplication by an element does not preserve the identity, and so would not be a unital ring homomorphism). I suppose there isn't really any other algebraic information aside from "preserves addition", so thanks for your input! – Hayden Feb 21 '15 at 02:19
  • @Hayden I had a lot of hesitation with that part. At first I wanted to say "automorphism", but we could have zero-divisors, etc. I guess if our ring doesn't have 1, then the best we can hope for would be a semigroup homomorphism (if such a thing exists); in either case, it seems like the situation is "as good as it gets"! – pjs36 Feb 21 '15 at 02:24
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    It would form a group homomorphism (in the context of rings, unital or not), but this is the best we can get in general. Indeed, it does not preserve products in general, though if the element $a$ is an idempotent (i.e. $a^2=a$) and $a$ commutes with every element of the ring, then we would have $a(bc)=a^2(bc)=(ab)(ac)$. Indeed, this would always be the case in something like a Boolean ring, but still the map would only give a rng (or pseudo-ring) homomorphism. – Hayden Feb 21 '15 at 02:34
  • But agreed, it's as good as it gets! – Hayden Feb 21 '15 at 02:35
  • @Hayden You've clearly thought quite a bit about it, I bet you could craft a very good answer to your question, – pjs36 Feb 21 '15 at 02:41
  • haha, thank you. I've been thinking about the problem for a few days, but I wanted to try and see how other people thought about it. My hope was to get some interesting perspectives that shed some light on why distributivity happens to fit so well into the mathematical universe. – Hayden Feb 21 '15 at 03:10