Let R be a ring with 1 and M a left R-module

Proof:
Let Ann(N)={r in R: rn=0 for all n in N} be the annihilator of N in R.
We want to show that Ann(N) is a two-sided ideal of R.
First, I show that Ann(N) is a subring of R. For any r,s in Ann(N), we have
rn=0, sn=0 for any n in N. Then (r+s)n=rn+sn=0+0=0. This implies that r+s
is in Ann(N). (rs)n=r(sn)=r*0=0. This implies that rs is in Ann(N). Similarly,
(sr)n=s(rn)=s*0=0. So sr is in Ann(N). Thus Ann(N) is closed under addition
and multiplication and thus is a subring of R.
Second, I show that Ann(N) is a two-sided ideal of R.
For any r in Ann(N), s in R, for any n in N, we have (sr)n=s(rn)=s*0=0, then
sr is in Ann(N). For (rs)n, since N is a submodule of M and M is a left R-module,
then sn is in N for any s in R and n in N. So (rs)n=r(sn)=r*0=0. Thus rs is in
Ann(N). Therefore, Ann(N) is a two-sided ideal of R.