The annihilator of a module is an ideal Let $R$ be a ring and let $M$ be an $R$-module. Define the annihilator of $M$ to be the following subset of $R$:
$Ann_R(M) =\{r\in R \mid rm = 0_M  \text{ for all } m\in M \}$
Prove that $Ann_R(M)$ is an ideal of $R$.
please help as much as you can a full answer is greatly appreciated 
Thanks
 A: We show that $\text{Ann}_R(M) = \{r \in R: rm = 0 \, \text{for all} \, m \in M\}$ is in fact a two-sided ideal of $R$.  To accomplish this, two assertions must be established:
1.)  $\text{Ann}_R(M)$ is an abelian group under the ring operation "$+$" defined in $R$;
and
2.)  $ra \in \text{Ann}_R(M)$ and $ar \in \text{Ann}_R(M)$ for any $a \in \text{Ann}_R(M)$ and $r \in R$.
We show (1) via the following proposition, which is valid for any group $G$:
Proposition:  Let $G$ be a group, an $S \subseteq G$ an subset of $G$; then
$S$ is a subgroup of $G$ if and only if $rs^{-1} \in S$ for all $r, s \in S$.
Proof of Proposition:  Clearly, if $S$ is a subgroup of $G$, $rs^{-1} \in S$ for all $r, s \in S$; thus we merely need prove the "if" direction.  Pick any $r \in S$; then $e = rr^{-1} \in S$, where $e \in G$ is the identity element.  Since $e \in S$, we have $r^{-1} = er^{-1} \in S$.  Finally, for $s, r \in S$, since $r^{-1} \in S$, we have $sr = s(r^{-1})^{-1} \in S$.  $S$ thus contains the identity of $G$, the inverse of each of its elements, and is closed under the group operation of $G$; these facts confirm that $S$ is a subgroup of $G$.  QED
To apply this proposition to the case at hand, we need to show that for $a, b \in \text{Ann}_R(M)$ we have $a - b \in \text{Ann}_R(M)$; but this is easy:  for such $a, b$ and $m \in M$ we have $(a - b)m = am - bm = 0 - 0 = 0$.  So by our proposition, $\text{Ann}_R(M)$ forms an abelian group under "$+$".
Now if $a \in \text{Ann}_R(M)$ and $r \in R$, for all $m \in M$ we have $(ra)m = r(am) = 0$, whence $\text{Ann}_R(M)$ is a left ideal in $R$.  Likewise, $(ar)m = a(rm) = 0$, 
since $rm \in M$.  $\text{Ann}_R(M)$ is thus a right ideal as well, hence a two-sided ideal.  And we are done.  QED
Nota Bene:  Using the above proposition, proved in the contexts of general groups, to establish that $\text{Ann}_R(M)$ is a subgroup of $R$ under the operation "$+$" is perhaps overkill, but it does serve to illustrate why many authors define ideals $I$ to be closed under the operation $a - b$: $a - b \in I$ for all $a, b \in I$ forces $0 = b - b \in I$, whence $-b = 0 - b \in I$, whence $a + b = a - (-b) \in I$.  On the other hand, merely assuming $I$ is closed under "$+$" does not necessarily imply $0, -a \in I$.  So requiring $a - b \in I$ gives a concise way of positing that $I$ be an abelian group under "$+$".  I have seen many authors take this approach.
Hope this helps.  Cheerio,
and as always, 
Fiat Lux!!!
A: Use the definition of ideal and the definition of the set $\operatorname{Ann}_R(M)$ to check that $\operatorname{Ann}_R(M)$ is an ideal.
Hint 1: if $a, b \in \operatorname{Ann}_R(M)$, show that $a+b \in \operatorname{Ann}_R(M)$.
Hint 2: for $a \in \operatorname{Ann}_R(M)$ and $r \in R$, show that $ra \in \operatorname{Ann}_R(M)$.
