1
$\begingroup$

Here is my attempted proof of the question in title:

It is given that $R$ is a commutative, unital ring. Assume that $F$ is faithful and $FM = 0$, for some $R$-module $M$ and $\alpha:M\to N$. We have an obvious exact sequence: $$0\to\ker\alpha\to M\to \Im\alpha\to 0.$$ Since $F$ is exact, it preserves images and kernels. That is, $F(\ker\alpha) = \ker(F\alpha)$ and $F(\Im\alpha) = \Im(F\alpha).$ Hence, applying $F$ to the given exact sequence above yields: $$0\to\ker(F\alpha)\to 0=FM\to\Im(F\alpha)\to 0 .$$ This means that $\ker(F\alpha) = \Im(F\alpha) =0$, because the corresponding homomorphisms must be injective. Hence, $F\alpha = 0\Rightarrow \alpha = 0$. But this gives an exact sequence $\ker\alpha = 0\to M\to 0 =\Im\alpha$, which then forces $M$ to be zero.

Is there a problem in my proof? If there isn't, is it unnecessarily long?

Thanks in advance.

  • 0
    what is the category on which $R$ is defined – 2017-02-12
  • 0
    @TsemoAristide $R$ is a commutative ring with a unit. Edited now. – 2017-02-12

1 Answers 1

2

Suppose that $F$ is defined on $R$-modules and $R$ has a unit. $F(M)=0$ and $F$ is faithful implies that $F(id_M)=F(2id_M)=0$ and $2id_M=id_M$since $F$ is faithful, we deduce $2x=x$ and $x=0$ for every $x\in M$.

You also have let $0_M:M\rightarrow M$ the zero map, $F(0_M)=F(Id_M)$ this implies that $0_M=Id_M$ since $F$ is faithhful.