Proving $R$-modules $Ra$ and $R/\mathrm{Ann}_R(a)$ are isomorphic.

Question

I am solving an exercise problem which aims to prove $R$-modules $Ra$ and $R/\mathrm{Ann}_R(a)$ are isomorphic. This exercise problem is broken down to steps.

Let $R$ be a ring and $M$ be a (left) $R$-module. Let $a\in M$. Define the annihilator of $a$ in $R$ by $\mathrm{Ann}_R(a):=\{ r\in R\mid ra=0 \}.$

Show that the submodule of $M$ generated by $a$ is $Ra:= \{ra\mid r\in R\}$.

Show that $\mathrm{Ann}_R(a)$ is a left ideal of $R$.

Show that the $R$-modules $Ra$ and $R/\mathrm{Ann}_R(a)$ are isomorphic. Show, by example, that $\mathrm{Ann}_R(a)$ need not be a two-sided ideal in $R$.

I proved $\mathrm{Ann}_R(a)$ is a left ideal of $R$. However unable to prove the first and the third one. Can anyone give me hints or any ideas? Thanks!

Answer 1

I am assuming that your ring $R$ has a unit and all modules are unital ($1\cdot m = m$).

Let $N$ be a submodule of $M$ containing $a$. Then $ra\in N$ by definition for every $r\in R$. Hence $Ra\subseteq N$. On the other hand $Ra$ is a submodule of $M$ and it contains $a$, since $a = 1\cdot a\in Ra$.

For the third part. Consider a morphism $\phi:R\rightarrow Ra$ given by $\phi(r) = ra$. This is a morphism of left $R$-modules. It is also surjective. Now you can verify that its kernel is $\mathrm{Ann}_R(a)$. Thus by isomorphism theorem we derive that $\phi$ induces an isomorphism

$$R/\mathrm{Ann}_R(a)\cong Ra$$

Consider a matrix ring $R = \mathbb{M}_{2\times 2}(\mathbb{C})$. Pick $M=\mathbb{C}^2$ as a set of two-dimensional vectors on which $R$ acts by matrix multiplication from the left. Pick a vector $a = (1,0)\in M$. Then $\mathrm{Ann}_R(a)$ is a set of matrices in $R$ with first column equal to zero. Check that this is not a two-sided ideal in $R$.

Answer 2

Let $R\to Ra$ be the map sending $r\mapsto ra$. You can check that this is an $R$-module homomorphism. This map is surjective since every element of $Ra$ is of the form $ra$ for some $r\in R$ by definition. The kernel of this map is precisely $Ann_R(a)$, so by the first isomorphism theorem we get $R/Ann_R(a)\cong Ra$.

You can also define the isomorphism explicitly, but that can be deduced from the proof of the isomorphism theorem.