Ring, Matrix, and Endomorphism

Before start every ring $R$ has 1 but does not need to be commutative. Also all modules are left $R$-module.

Ring

Definition

$I\subset R$ is two sided ideal if $I$ is an abelian subgroup of $R$, and $ri\in I$ and $ir\in I$ for all $r\in R$ and $i\in I$.

Remark

Matrix ring has a one-sided ideal. However matrix ring $M_n(R)$ has no non-trivial proper ideal, because there are elementary matrices that ideal have $E_{ij}$ for all $i,j=1,\dots ,n$. In this context $E_{ij}$ has only one 1 in $ij$ position, otherwise $0$.

Matrix and Endomorphism

When we have a ring $R$, we can use the matrix $M_n(R)$.

Also, we can consider the $R$-module $M$ with endomorphism ring $En{d}_R(M):=\left\{f:M\to M\right\}$ with

$$\{\begin{array}{l}f+g(m)=f(m)+g(m)\\ f\cdot g(m)=(f\circ g)(m)\end{array}.$$

For the free module with rank $n$ case

If $R$ is a field, then $M_n(R){\cong }_{\text{Ring}}En{d}_R{R}^n$. However for general ring $R$, it does hold, i.e. $M_n(R){≆}_{\text{Ring}}En{d}_R(R^n)$.

Easy case $n=1$

We can easily see that it does not hold for $n=1$. Let $r_0\in R$ and $f_{r_0}:R\to R$ as $r\mapsto r_{0}r$. It is matrix map so that $f_{r_0}\in M_1(R)=R$ as ring multiplication. However, it is not a left module map because

$$f_{r_0}(r{r}^{\prime })=r{r}^{\prime }f(1)=r{r}^{\prime }{r}_0\ne r_{0}r{r}^{\prime }.$$

Note that if $g_{r_0}:R\to R$ as $r\mapsto r{r}_0$, $g_{r_0}\in En{d}_R(R)$, but still $R\to En{d}_R(R)$ such that $r_0\mapsto g_{r_0}$ is not a ring homomorphism. That’s because

$$g_{r_0{r}_1}(r)=r{r}_0{r}_1\ne g_{r_0}{g}_{r_1}(r)=r{r}_1{r}_0.$$

Fixing

We can fix it with $R^{op}$, where $R^{op}$ is opposite ring with multiplication $r{\cdot }_{op}{r}^{\prime }=r^{\prime }r$. Let $\varphi :R^{op}\to En{d}_R(R)$, such that $r\mapsto g_r$. It is a ring homomorphism as following.

$$\varphi (r_1{\cdot }_{op}{r}_2)=g_{r_2{r}_1}=g_{r_1}\cdot g_{r_2}=\varphi (r_1)\varphi (r_2).$$

Then $R^{op}{\cong }_{\text{Ring}}En{d}_R(R)$.

Arbitary $n$

Let $E_j$ and $F_i$ be a $R$-module where $i,j=1,\dots ,n$ and ${\phi }_{ij}:E_j\to F_i$.

Then using direct sum, we can build a module homomorphism.

$$\phi :{\oplus }_{i=1}^n{E}_i\to {\oplus }_{j=1}^n{F}_j$$

and

$${\phi }_{ij}:E_j\to F_i.$$

Let be a ring map $M:En{d}_R(E,F)\to M_n(R^n)$ such that $\phi \mapsto \left({\phi }_{ij}\right)$. So consider $E^n=F^n$, then the map $M$ is a ring isomorphism.

If $E^n={\oplus }_{i=1}^{n}E$ and $E=R$ as a left $R$-module, then $En{d}_R(R^n)=R^{op}$. Therefore, $En{d}_R(R^n){\cong }_{\text{Ring}}{M}_n(R^{op})$.

Exercise