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How much of ring theory arises simply from applying group-theoretic results to the additive structure, or semigroup/monoid-theoretic results to the multiplicative structure? To be more precise, can anyone describe how much rigidity is imposed on the structure of rings by the distributive axiom which "connects" addition and multiplication? What are some examples of results, say, where distributivity really plays a key role?

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    $\begingroup$ Every ring (commutative or otherwise) embeds into the endomorphism ring of its additive group. But this is just a fancy way of stating the distributive law. $\endgroup$ – Zhen Lin Jul 10 '12 at 7:06
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How much of ring theory arises simply from applying group-theoretic results to the additive structure, or semigroup/monoid-theoretic results to the multiplicative structure?

Almost nothing. And in any case this would not be called ring theory. For example, a finite subgroup of the multiplicative group of a field is cyclic. This belongs to group theory, not to field or ring theory, although it is crucial for field theory.

To be more precise, can anyone describe how much rigidity is imposed on the structure of rings by the distributive axiom which "connects" addition and multiplication? What are some examples of results, say, where distributivity really plays a key role?

There is no field $K$ with $K^* \cong \mathbb{Z}$.

This kind of rigidity arises everywhere. Just take any interesting result on rings and you will see that it makes no sense at all to try to prove this for pairs of (abelian group,monoid). So, just open a textbook and you will see lots of evidence that distributivity plays a key role.

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    $\begingroup$ I don't get your point "a finite ... is cyclic. This belongs to group theory". Are you suggesting that for a "non-distributive ring", a structure $X$ with additive group structure on $X$ and unrelated multiplicative group structure on $X\setminus\{0\}$ (both commutative), any finite subgroup of the multiplicative group would be cyclic? I think I can easily give a counter-example. $\endgroup$ – Marc van Leeuwen Jul 10 '12 at 10:14
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One example of where the distributivity axiom adds additional properties to ring-like structures is the following:

The ring axioms with the multiplicative identity axiom and without the commutativity of addition axiom can already prove the commutativity of addition.

This requires the distributive property.

(1 + 1)(x + y) = (1 + 1)(x) + (1 + 1)(y) = x + x + y + y

(1 + 1)(x + y) = 1(x + y) + 1(x + y) = x + y + x + y

so $x + y = y + x$.

So distributivity forces the commutativity of addition in the case that the ring has a multiplicative identity.

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    $\begingroup$ Shorter and more generally: If $M$ is an $R$-module, then $(M,+)$ is forced to be commutative. Namely, multiplication with $-1$ is linear, but on the other hand it is inversion with respect to $+$. But the inversion of a group is a homomorphism iff the group is abelian. $\endgroup$ – Martin Brandenburg Jul 10 '12 at 6:58

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