# Abstract Nonsense

## Opposite Ring and Antihomomorphism

Point of Post: In this post we discuss the notion of opposite rings and antihomomorphisms and prove some basic theorems regrading them.

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Motivation

There is a natural way to take non-commutative rings and create new rings out of them. In particular, you take the ring and just pretend the multiplication has its order reversed.  In other words if we have a ring whose multiplication is denoted $\ast$ then we can create a new multiplication called, say, $\ast'$ given by $a\ast' b=b\ast a$. What seems to be true then is that while $\text{id}_R:(R,+,\ast)\to (R,+,\ast')$ is not a homomorphism it is something that happens quite often a morphism which reverses order–the come up often when something takes “inverses”.

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Opposite Rings and Antihomomorphisms

Let $(R,+,\ast)$ be a ring, define the opposite ring denoted $R^\text{op}$ to be the ring $(R,+,\ast_\text{op})$ where $a\ast_\text{op}b=b\ast a$. It’s easy to see that $R^\text{op}$ is a ring.

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An obvious question is as to whether $R\cong R^\text{op}$. “Usually” this is true, not in the sense that in the scheme of everything it happens more often than not, but in the sense that it’s quite difficult to create rings which aren’t isomorphic to their opposite ring. Probably the easiest example is if one takes $\text{End}\left(\mathbb{Z}\times\mathbb{Q}\right)$ (endomorphism ring as an abelian group) but it’s not in my interest to prove this.

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We define an antihomomomorphism between two rings $R$ and $S$ to be a map $R\to S$ such that the induced map $R\to S^\text{op}$ is a homomorphism.  In other words, it’s a map $f:R\to S$ which is a group homomorphism and for which $f(ab)=f(b)f(a)$.

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References:

1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

2. Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.