# Intuition for opposite ring, equivalence of left and right modules, anti-involution.

Here is an excerpt from what I am reading.

One can similarly define right $A$-modules. A convenient way to give a formal definition of right module is as follows.

First, given a ring $A$, one defines $A^{op}$, the opposite ring, to be the abelian group $(A, +)$ equipped with an opposite multiplication $a$, $b \mapsto a \cdot_{op} b := ba$.

Then, a right $A$-module is, by definition, the same thing as a left $A^{op}$-module.

Remark. An anti-involution on a ring $A$ is a morphism $A \to A$, $a \mapsto a^*$ of abelian groups such that$$(ab)^* = b^* a^*, \quad (a^*)^* = a, \quad (1_A)^* = 1_A, \quad a, b \in A.$$An anti-involution on $A$ provides a ring isomorphism $A \overset{\sim}{\to} A^{op}$.

Transposition of matrices gives an example of a non-trivial anti-involution on $\text{M}_n(k)$, a noncommutative ring.

The identity map is an anti-involution on any commutative ring. Thus, we have $A^{op} \cong A$ for any commutative ring $A$ and hence the notions of left and right $A$-modules coincide in this case.

This is quite dense, and so I am wondering if anybody can help me unpack the following.

• What is the intuition behind the definition of opposite rings, and for working with them? Do they ever come up in practice?
• Why is a right $A$-module the same thing as a left $A^{op}$-module?
• What is the intuition behind the definition of opposite rings, and for working with them? Do they ever come up in practice? And why do they provide a ring isomorphism $A \overset{\sim}{\to} A^{op}$?
• What is the intuition behind $A^{op} \cong A$ for a commutative ring $A$ and having the notions of left and right $A$-modules coinciding?

Ugh, sorry for the large blocks of text. Thanks!

• The opposite ring is just what it says: it's the same thing except the multiplication is the other way around. One place this comes up is in the study of the Brauer group and central division algebras. For a commutative ring it doesn't matter what order you do the multiplication in, by definition. – Qiaochu Yuan Sep 5 '16 at 23:21
• Unpacking the definitions and doing a little leg work will go a long way here. For example, why is a right $A$-module the same thing as a left $A^{\rm op}$-module? Suppose $M$ is a right $A$-module with scalar multiplication given by the map $\mu:M \times A \longrightarrow M$. How can you define a left-sided scalar multiplication $\mu':A^{\rm op} \times M \longrightarrow M$? There really aren't that many possibilities. How about: $\mu'(a,m) = \mu(m, a)$? Can you verify that $\mu'$ satisfies all the axioms for a left scalar multiplication? – PeterJL Sep 6 '16 at 4:30
• Oh, and for down-to-earth examples of where we need to distinguish between left and right scalar multiplication, consider quaternionic "vector spaces". Technically, a vector space is a module over a field and the quaternions $\mathbb{H}$ aren't commutative (they form a skew-field, not a field.) You might gain some insight by simply looking at the spaces $\mathbb{H}^n$ and comparing what happens when you multiply by scalars on the left and right. – PeterJL Sep 6 '16 at 4:37

What is the intuition behind the definition of opposite rings, and for working with them? Do they ever come up in practice?

It's not clear why one would need any intuition to use them. You could say they are a very simple "construction" where you make a new ring out of a old one, but that view is not very fruitful.

There is a high-level, less accessible explanation. In category theory, you talk about objects and arrows between them (plus some axioms.) You might have guessed by now that there is in fact a notion of opposite category, and that's what happens when you take a category and point all arrows in the opposite direction.

Many things can be expressed as categories, and among those things are rings and partially ordered sets (or totally ordered sets if you prefer).

While viewing a partially ordered set as a category, the opposite category is just the reversed partial ordering. The opposite ring of a ring is just the opposite category of that ring viewed as a category.

Other than drawing this parallel between opposite ordering and opposite rings, I don't really have any further insight into what they are. Really they are most useful as a notational convenience.

The first place they arise naturally in a textbook on noncommutative algebra is probably while explaining the Artin-Wedderburn theorem. The way I remember it, no matter what setup you start out with, you eventually need to introduce the opposite ring of one of the rings in play. That's an example of using it for notational convenience.

Two more places they show up:

An abelian group $M$ is an $R, S$ bimodule iff it is a left $R\otimes S^{op}$ module.

If $R$ is a ring, then the ring of module endomorphisms $End(R_R)\cong R$, but $End(_RR)\cong R^{op}$.

Why is a right $A$-module the same thing as a left $A^{op}$-module?

You just check that $r\cdot m :=mr$ defines a left module structure on $M$. Without "opposite multiplication $\circ$", there is no way to prove that $(r\circ s)\cdot m=r\cdot(s\cdot m)$.

What is the intuition behind the definition of [anti-involutions], and for working with them? Do they ever come up in practice? And why do they provide a ring isomorphism $A\cong A^{op}$?

They come up in practice, for example, in the complex and quaternion conjugation maps. The first one is trivial since the complex numbers are commutative, but it's nontrivial for the quaternions. Additionally, the whole subject of $^\ast$-rings is devoted to the study of involutions like that.

As to why an anti-involution $f:A\to A$ yields an isomorphism $A\cong A^{op}$, I advise you to guess what the obvious candidate for a map is and then check to see that it's true.

What is the intuition behind $A^{op}\cong A$ for a commutative ring $A$ and having the notions of left and right $A$-modules coinciding?

A ring being isomorphic to its opposite ring just guarantees some left-right symmetry of the ring. For example, if $R$ is right Noetherian, $R^{op}$ is left Noetherian. If these two rings are isomorphic, then $R$ is Noetherian on both sides. If a ring is isomorphic to its opposite, then any one-sided condition that it has, it has on both sides.

The category of right modules and the category of left modules for a given ring can be quite different from each other. It could be, for example, that every left module admits a projective cover while there are right modules without projective covers. If, on the other hand, the two categories share the same properties, that is something special and is again a sort of 'symmetry' about the ring.