1
$\begingroup$

If $f(x)$ and $g(x)$ are real functions and $g$ is even, so is $f(g(x))$. Even functions are also closed under addition. I noticed that these are similar properties to those of an ideal of a ring, except composition doesn't distribute over addition in general so the ring structure isn't there. In abstract algebra are rings without the distributive property studied? If so, are homomorphisms related to sets closed under addition that absorb the multiplication operation, like they are for normal rings?

$\endgroup$
1
$\begingroup$

This question investigates another algebraic object that acts like a ring without a distributive law.

The main point of that thread is to discover another operation so that the distributive law does hold. This is different from your question, but does get to the heart of what a ring is. A ring is both an abelian group under addition, and a semigroup (a group without identity or inverses) under multiplication (where addition and multiplication are just the titles given to the two binary operations). This additive and multiplicative structure doesn't interact with eachother in any way except through the distributive law. In this way, the distributive law "links" the additive and multiplicative parts of the ring. So, a ring without this distributive law is just two binary operations that share an underlying set of elements, but aren't required to interact with eachother in any meaningful way.

$\endgroup$
  • $\begingroup$ Hmm. I suppose you're right, the interaction between the operations is not interesting. Is there any operation that composition does interact with in an interesting way? $\endgroup$ – Vik78 May 24 '16 at 4:43
  • $\begingroup$ @Vik78 That'd probably be a better overall question, and probably worth submitting a new question about $\endgroup$ – Mark May 24 '16 at 5:40
  • $\begingroup$ @Vik78 I did just find this wikipedia page that talks about rings with a third binary operation known as "composition" that coincides with function composition in polynomial rings. This isn't really what you want, but the page mentions in boolean rings that function composition can coincide with multiplication, which may be interesting for you. $\endgroup$ – Mark May 24 '16 at 5:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.