# Abstract Nonsense

## Endomorphism Ring of an Abelian Group

Point of Post: In this post we discuss the endomorphism ring of an abelian group as being a ‘typical ring’ in the sense that every ring is embeddable into the endomorphism ring of its underlying group structure.

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Motivation

This post will inaugurate a short run of posts developing certain classes of rings (endomorphism rings, rings of continuous functions, Boolean rings, ring of ring morphisms, matrix rings, group rings, etc.) to give us a firm set of examples. Our first class of rings will be endomorphism rings of abelian groups. Intuitively how one would think to create the endomorphism ring of an abelian group would be to note that for a given abelian group $G$ and endomorphisms $\phi,\psi:G\to G$ there are two ‘natural’ operations to perform on them, namely you can add them $\phi+\psi$ (in the sense that $(\phi+\psi)(g)=\phi(g)+\psi(g)$) and you can compose them $\phi\circ \psi$. It turns that these two operation on the set $\text{End}(G)$ of all endomorphisms on $G$ form a ring. That said, unlike the other aforementioned rings which clearly come up often in mathematics the endomorphism ring of an abelian group seems a bit…well, contrived. So, why would anyone really care about it at this stage in our studies? Well, the main reason why the endomorphism ring of an abelian group is interesting is it lets us formulate a unital ring theoretic analogy to the group theoretic Cayley’s theorem. Namely, we shall see that every ring $R$ naturally embeds into the endomorphism ring of its underlying group structure.

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Endomorphism Ring of an Abelian Group

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Let $G$ be an abelian group. Let $\text{End}(G)$ denote the set of all group endomorphism of $G$. Then if we define the sum $\phi+\psi$ of $\phi,\psi\in\text{End}(G)$ to be such that $(\phi+\psi)(g)=\phi(g)+\psi(g)$ for every $g\in G$ and multiplication $\circ$ to be the composition $\phi\circ \psi$ then $\left(\text{End}(G),+,\circ\right)$ forms a unital ring. To see this we note that evidently $\left(\text{End}(G),+\right)$ and $\left(\text{End}(G),\circ\right)$ form an abelian group (one can check this is where you need that $G$ is abelian) and monoid (with unit equal to the identity map $\text{id}_G$) respectively and thus it suffices to show distributivity. But, this is clear since, for example, given any $\phi,\psi,\eta\in\text{End}(G)$ and $g\in G$ we have that

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$\displaystyle \left(\phi\circ(\psi+\eta)\right)(g)=\phi\left(\left(\psi+\eta\right)(g)\right)=\phi\left(\psi(g)+\eta(g)\right)=\phi(\psi(g))+\phi(\eta(g))$

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For a specific example we consider the following:

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Theorem: Let $n\in\mathbb{N}\cup\{0\}$ then $\text{End}\left(\mathbb{Z}/n\mathbb{Z}\right)\cong\mathbb{Z}/n\mathbb{Z}$.

Proof: Define $f:\mathbb{Z}/n\mathbb{Z}\to\text{End}\left(\mathbb{Z}/n\mathbb{Z}\right)$ given by $\left(f(\overline{k})\right)(\overline{\ell})=\overline{k}\,\overline{\ell}=\overline{k\ell}$ where, for the sake of notational convenience, we denote $m+n\mathbb{Z}$ by $\overline{m}$. We first note that $f$ is well-defined (in the sense that $f(\overline{k})\in\text{End}(\mathbb{Z}/n\mathbb{Z}$) by right distributivity. To see that $f$ is a morphism we merely note that for any $\overline{k},\overline{\ell},\overline{m}\in\mathbb{Z}/n\mathbb{Z}$

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$\displaystyle \left(f(\overline{k}+\overline{\ell})\right)(\overline{m})=\left(\overline{k}+\overline{\ell}\right)=\overline{k}\overline{m}+\overline{\ell}\overline{m}=(f(\overline{k}))(\overline{m})+(f(\overline{\ell}))(\overline{m})$

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and

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$\left(f\left(\overline{k}\overline{\ell}\right)\right)(\overline{m})=(\overline{k}\overline{\ell})\overline{m}=\overline{k}(\overline{\ell}\overline{m})=\left((f(\overline{k}))\circ (f(\overline{\ell})\right)(\overline{m})$

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so that $f(\overline{k}+\overline{\ell})=f(\overline{k})+f(\overline{\ell})$ and $f(\overline{k}\overline{\ell})=f(\overline{k})\circ f(\overline{\ell})$ and so $f$ is a morphism as claimed. Clearly $f$ is injective since if $f(\overline{k})=f(\overline{\ell})$ then $\overline{k}=(f(\overline{k}))(\overline{1})=(f(\overline{\ell}))(\overline{1})=\overline{\ell}$. To see that $f$ is surjective we merely note that if $\phi\in\text{End}(\mathbb{Z}/n\mathbb{Z})$ then for any $\overline{k}$ we have that $\overline{k}=k\overline{1}$ and so by definition $\phi(\overline{k})=k\phi(\overline{1})$ and so in particular $\phi=f(\phi(\overline{1}))$. $\blacksquare$

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This example motivates the following unital ring-theoretic analogue of Cayley’s theorem:

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Theorem: Every unital ring $R$ is canonically isomorphic to a subring of $\text{End}(G)$ for some abelian group $G$.

Proof: Consider $\text{End}(R)$ (considering only $R$‘s underlying group structure) and define $f:R\to\text{End}(R)$ given by $(f(r))(s)=rs$. Using the ring structure similar to the previous theorem it’s easy to see that $f$ is a morphism.  Moreover, using the fact that $(f(r))(1)=(f(s))(1)$ for every $r,s\in R$ we see that $f$ is injective. Thus, we have that $R\cong \text{im }f$ and $\text{im }f$ is a subring of $R$. $\blacksquare$

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It should be seen equally as easy that the above in fact generalizes to

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Theorem: Let $R$ be a ring (not necessarily unital) with at least one non-zero divisor. Then, $R$ is canonically isomorphic to a subring of $\text{End}(R)$.

Proof: It evidently suffices to prove injectivity. To do this we merely note that $f(r)=f(s)$ implies $(r-s)t=0$ for all $t\in R$. But, if $t$ is the guaranteed non-zero divisor then $(r-s)t=0$ implies $r-s=0$ and so $r=s$ as claimed. $\blacksquare$

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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.

July 10, 2011 -

1. Thank you for this post. It has the information necessary to introduce very quickly the Endomorphism Ring without distracting examples

Comment by jocerfranquiz | August 12, 2011 | Reply

• You’re welcome, I’m glad it helped! Thank you for caring enough to respond!

Comment by Alex Youcis | August 12, 2011 | Reply

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