Ring isomorphism given by $\varphi: R \to \text{end}_R(M), a \mapsto \lambda_a$

Question

Let $R$ be a commutative Ring with $1$ and $M$ a $R$-Module. $$\varphi: \begin{cases}R & \longrightarrow \text{end}_R(M) \\ a & \longmapsto \lambda_a \end{cases} $$ is a Ringisomorphism for $M=R$

My approach: First to clarify, $\lambda_a: M \to M, x \mapsto ax$ is the homothetic mapping.

I managed to show that $\varphi$ is a Ringmorphism with $\varphi(1_R)=\lambda_{1_R}=1_{\text{end}_R(M)}=id_M$

Now I am stuck with the 2nd part. I was told that the easiest way to complete this exercise is to find the inverse mapping and show that the two composition yield the identity mapping (on the respective set)

At some point I was given the following mapping $$\xi : \begin{cases}\text{end}_R(R) & \longrightarrow R \\ \delta & \longmapsto \delta (1) \end{cases} $$

I yet fail to understand the intuition behind this mapping, how one comes up with such an idea and ontop of that, why it works. Here are my calculations:

Let $x \in R$ be arbitrary $$\varphi(\xi(\delta(x)))=\varphi(\delta(1))=\lambda_{\delta(1)} $$ My $x \in R$ seems to have 'vanished' which is clearly a bad calculation on my end. So I suppose I have to do the calculation like that: $$\varphi(\xi (\delta(x)))=\varphi(\delta(1)(x))=\lambda_{\delta(1)(x)}\overset{?}=\lambda_{\delta(1)}(x)=\delta(1)x =x \delta(1) \\ = \delta(x) = id_{\text{end}_R(R)}\tag{*}\delta(x)$$

After the answers provided below I understand that the above calculation does hold, but there is clearly some magic (and with magic I mean bad mathematics performed by me) going on at the step indicated with ?. It seems that my argument is always getting 'eaten up' or ends up in places where I can no longer work with it.

Furthermore let $a \in R$ be arbitrary: $$\xi(\varphi(a))=\xi(\lambda_a)=\lambda_a(1)=a\cdot 1=a=id_M(a) $$ Which I am okay with.

Could someone please enlighten me with some insight regarding this exercise? Especially in the calculation marked with (*) I am hopelessly lost (because of a mapping defined by a mapping through a mapping .....)

Answer 1

Extending the comments:

• Ring structure on $\operatorname{End}_R(M)$ (sketch)

The composition $\circ$ is nothing but $(\phi\circ\psi)(m):=\phi(\psi(m))\in M$, for all $\phi, \psi\in \operatorname{End}_R(M)$ and $m\in M$. It is clearly associative with unit $1_{\operatorname{End}_R(M)}$ given by $1_{\operatorname{End}_R(M)}: m\mapsto 1(m):=m$.

• Isomorphism $\lambda: R\rightarrow \operatorname{End}_R(R)$

We want to show that

$$\lambda\circ\Psi = 1,~~~ \Psi\circ\lambda = 1,$$

where $\Psi: \operatorname{End}_R(R)\rightarrow R$ is given by $$\Psi(\varphi):=\varphi(1),~~ \forall \varphi\in \operatorname{End}_R(R)$$ and identities are $1:\operatorname{End}_R(R)\rightarrow \operatorname{End}_R(R)$, and $1: R\rightarrow R$, respectively.

Let us prove that $\Psi$ is a ring homomorphism, to start with. We have

$$\Psi(\rho\circ\psi):=(\rho\circ\psi)(1)=\rho(\psi(1))=(*)=\rho(1)\psi(1)=\Psi(\rho)\Psi(\psi), $$ where the equality $(*)$ follows as both $\rho$ and $\psi$ are $R$-linear: more explicitly $$\rho(\psi(1))=\rho(1\psi(1))=\rho(1)\psi(1).$$

Let us prove $\lambda\circ\Psi = 1$; for all $\varphi\in \operatorname{End}_R(R)$ we have $\lambda(\Psi(\varphi)):r\mapsto \varphi(1)r=\varphi(r)$, for all $r\in R$ as, by definition, $\varphi$ is $R$-linear. This is equivalent to $\lambda(\Psi(\varphi))=\varphi$, as claimed.

Now, for $\Psi\circ\lambda = 1$, one can write $\Psi(\lambda)=\lambda(1)$, where the r.h.s. is the $R$-linear map $\lambda(1):r\mapsto 1r=r$, for all $r\in R$, i.e. the identity map on $R$. This is what we had to prove.

Answer 2

I think you're forgetting a tiny thing here: observe $\;\delta\;$ is an $\;R$- homomorphism, and thus

$$\delta(x)=\delta(x\cdot1)=x\cdot\delta(1)\;,\;\;\forall\,x\in R$$

I think this solves the whole conundrum.

Answer 3

My impression is that you're overcomplicating things.

For $a\in R$, $\lambda_a\colon M\to M$ is an $R$-endomorphism of $M$

This is just a simple verification.

$\varphi$ is a ring homomorphism

Indeed, $\varphi(ab)=\lambda_{ab}$ and this is the map $$ x\mapsto \lambda_{ab}(x)=(ab)x=a(bx)= \lambda_a(\lambda_b(x))=(\lambda_a\circ\lambda_b)(x) $$ so $\lambda_{ab}=\lambda_a\circ\lambda_b$ or $\varphi(ab)=\varphi(a)\circ\varphi(b)$.

Similarly, $\lambda_{a+b}$ is the map $$ x\mapsto\lambda_{a+b}(x)=(a+b)x=ax+bx=\lambda_a(x)+\lambda_b(x) =(\lambda_a+\lambda_b)(x) $$ and so $\varphi(a+b)=\varphi(a)+\varphi(b)$.

Of course $\varphi(1)=\lambda_1$ is the identity on $M$.

The kernel of $\varphi$ is the annihilator of $M$

Easy: $\varphi(a)=0$ if and only if $aM=0$.

As a consequence, if $M$ is faithful, $\varphi$ is injective. Note that $R$ is faithful as $R$-module, because only $0$ annihilates $1$.

If $M=R$, then $\varphi$ is surjective

Let $f\colon R\to R$ be an $R$-endomorphism and set $a=f(1)$. Then, for $x\in R$, we have $$ f(x)=f(x1)=xf(1)=xa=ax=\lambda_a(x) $$ so $f=\lambda_a=\varphi(a)$.

The inverse map is just $f\mapsto f(1)$.