Cfrgtkky

I am reading “Anderson and Fuller - Rings and Categories of Modules (2nd edition, 1992)”
and I am trying to find a solution of exercise 8.3 on page 112, which says:

Let $M$ be a left $R$-module. Show that if $M$ cogenerates a generator, then $M$ is faithful.

So, $M$ cogenerates a generator, say $M$ cogenerates $X$ and $X$ is a generator, where $X$ is a left R-module.

(1) $M$ cogenerates $X$, by Corollary 8.13.(1) (page 109) the reject of $M$ in $X$ is zero, that is:

$$ Rej_X (M) = bigcap { ker(h) text{ | } h in Hom_R (X,M) } = 0 $$

(2) $X$ is a generator, this means that $X$ generates $_R R$, that is, $R$ as a left $R$-module.

Then, by Corollary 8.13.(2) (page 109) the trace of $X$ in $R$ equals $R$, that is:

$$ Tr_R (X) = sum { im(h) text{ | } h in Hom_R (X,R) } = R $$

(3) It follows that there are finitely many $f_i in Hom_R (X,R)$ and $x_i in X$, $i=1, 2, cdots ,n$ such that

$$1 = f_1 (x_1) + f_2 (x_2) + cdots + f_n(x_n)$$

( I do not know if we need this in the proof. )

(4) Proposition 8.22 on page 111 says that the annihilator of $M$ in $R$, that is:

$$ l_R (M) = { r in R text{ | } rM = 0 } $$

equals the reject of $M$ in $R$, that is:

$$ Rej_R (M) = bigcap { ker(h) text{ | } h in Hom_R (R,M) } $$

Thus

$$ l_R (M) = Rej_R (M) $$

(5) Now we have to show that M is faithful, that is, the annihilator is zero, that is,

$$ l_R (M) = Rej_R (M) = 0 $$

So, we have to prove that if $s in l_R (M) = Rej_R (M)$ then $s = 0$

Take $s in Rej_R (M) = cap { ker(h) text{ | } h in Hom_R (R,M) }$

Then, for all $ h in Hom_R (R,M)$ we have $h(s) = 0$

Also, because $ l_R (M) = Rej_R (M) = 0 $, we have $s in l_R (M)$

So, for all $m in M$ we have $sm = 0$

Take $x in X$

Then, for all $ h in Hom_R (X,M)$ we have $h(sx) = sh(x) = 0$, because $h(x) in M$

Thus, for all $ h in Hom_R (X,M)$, $sx in ker(h)$

Thus, by (1) above,

$sx in Rej_X (M) = cap { ker(h) text{ | } h in Hom_R (X,M) } = 0$

Therefore, $sx = 0$, for all $x in X$

So, $s in l_R (X) = Rej_R (X) = cap { ker(h) text{ | } h in Hom_R (R,X) } = 0$

Then, for all $ h in Hom_R (R,X)$ we have $h(s) = 0$

Now I am stuck and I do not know where I am heading to.
Can someone, please, help me to finish this proof ?

Saying that $M$ cogenerates a generator $G$ means that $G$ embeds in a direct product of copies of $M$. Indeed, since $operatorname{Rej}_G(M)=0$, if we take all homomorphisms $fcolon Gto M$, the natural map
$$
Gto M^{operatorname{Hom}_R(G,M)}
$$
defined by $xmapsto (f(x))_{finoperatorname{Hom}_R(G,M)}$ has zero kernel.

If $r$ annihilates $M$, then it also annihilates $G$. Since a generator is necessarily faithful, then $r=0$. Therefore $M$ is faithful as well.

Why is a generator $G$ faithful? Because $R$ (as a left $R$-module) is an epimorphic image of a direct sum of copies of $G$, thus any $r$ that annihilates $G$ has to annihilate $R$.

I found the answer in line of my argument in my original post,

and I am embarrassed because I was so close and I did not see it.

We proved $sx = 0$, for all $x in X$, and we have to show that $s=0$

In (3) we found
$$1 = f_1 (x_1) + f_2 (x_2) + cdots + f_n(x_n)$$
So
$$s = f_1 (s x_1) + f_2 (s x_2) + cdots + f_n(s x_n)$$

Then, for each $i=1, 2, cdots ,n$: $ s x_i=0 $, because $x_i in X$ thus

$$s = f_1 (0) + f_2 (0) + cdots + f_n(0) = 0$$

Now, we are ready, because thie proves that $M$ is faithful