I'm wondering if there is a structure that builds on the ring axioms analogously to how rings build on the group axioms.
For example, $(X, +)$ is a group and $(X, +, \times)$ is a ring with identity. Is there a structure $(X,+,\times,\wedge)$ that adds another operator and additional constraints to the ring structure? For example:
- $\times$ is commutative
- $\wedge$ is associative
- $\wedge$ is distributive across $\times$
The first thing I thought of as a possible example is the ring $(\mathbb{Z},+, \times)$ with exponentiation as the third operator. But of course exponentiation is not associative and it only distributes on the right over multiplication.
I don't see this kind of structure in eg. Grillet, Abstract Algebra, or in the Wikipedia list of algebraic structures. Or perhaps I'm not recognizing some common structure as being essentially the kind of thing I'm asking about?
To understand how different types of structures relate to each other, category theory is a very useful and powerful tool. This answer makes heavy use of it.
Rings are monoid objects in the monoidal category $(\mathbf{Ab},\otimes)$ of abelian groups with the tensor product. So one way of answering the question is to take a monoidal category of ring-like objects and consider the monoid objects therein.
There are several ways to do this, but $(\mathbf{Ring},\otimes)$ (with the tensor product of rings) is not a very interesting choice*. Much more interesting are plethories. In particular, have a look at the paper Plethystic algebra by Borger-Wieland. Here, one considers the monoidal category of $k$-$k$-birings ($\mathbf{BiRing}_k,\odot_k)$, and monoids inside of it are called plethories over $k$. An important example is the ring $\Lambda$ of symmetric functions in countably many variables. The following quote confirms that plethories answer your question.
Just as a k-algebra is exactly the structure that knows how to act on a k-module, a k-plethory is the structure that knows how to act on a commutative k-algebra.
*Exercise: $\mathbf{Mon}(\mathbf{Ring},\otimes) \cong \mathbf{CRing}$