If $f(x)=0$ for all $f\in \operatorname{Hom}_R(U,V)$, then must $x=0$?

Question

I know that in normed vector spaces, if $f(x)=0$ for all functionals in the dual, then $x=0$. This follows quite easily using the norm. I've been studying ring and modules lately, and was wondering if the norm structure was necessary for this phenomen, or whether it is a more general characteristic of modules. I.e for right $R$ modules $U$ and $V$ such that $\operatorname{Hom}_R(U,V)\neq 0$ does $f(x)=0$ for all $f\in \operatorname{Hom}_R(U,V)$ imply that $x=0$? If not are there any restrictions we can put on $U$ or $V$ that will make this true?

Answer 1

The $R$-module $V$ is called a cogenerator if, for all $R$-modules $U$, $\operatorname{Hom}_R(U,V)=0$ implies $U=0$.

A necessary and sufficient condition for $V$ to be a cogenerator is that it contains a submodule isomorphic to the direct sum of the injective envelopes of the simple modules.

In particular, if $R$ is the ring $\mathbb{Z}$ of integers, a cogenerator must contain (as a direct summand) the group $\mathbb{Q}/\mathbb{Z}$.

The dual notion is that of a generator: a module $U$ is a generator if, for all modules $V$, $\operatorname{Hom}_R(U,V)=0$ implies $V=0$.

A module $U$ is a generator if and only if $R$ is a direct summand of $U^n$, for some integer $n$.

In the case $R$ is a field, every nonzero module (vector space) is both a generator and a cogenerator. There exist rings $R$ with the same property that are not fields (not even semisimple rings), for instance $\mathbb{Z}/p^2\mathbb{Z}$, where $p$ is prime.

One can relax the condition: $V$ cogenerates $U$ if $U$ embeds in a direct product of copies of $V$. This happens exactly when $f(x)=0$, for every $f\colon U\to V$ implies $x=0$. One direction is easy. For the other one, consider the map $$ \varphi\colon U\to V^{\operatorname{Hom}_R(U,V)} \qquad \varphi(x)=(f(x))_{f\in\operatorname{Hom}_R(U,V)} $$

Answer 2

This is false: if $m$ and $n$ are coprime, $\DeclareMathOperator\Hom{Hom}\;\Hom(\mathbf Z/m\mathbf Z,\mathbf Z/n\mathbf Z)=\{0\}$, yet $\;\mathbf Z/m\mathbf Z\ne \{0\}$.

Answer 3

Let $R=\mathbb{Z}$, $U=\mathbb{Z}/(6\mathbb{Z})=\{0,1,2,3,4,5\}$ and $V=\mathbb{Z}/(2\mathbb{Z})=\{0,1\}$.

Let's take $x=3\in U$.

Then for all $f:U\to V$, we must have $2f(3)=f(2\cdot 3)=f(0)=0$.

Notice that $x\mapsto [x]_2$ is a non-zero homomorphism from $U$ to $V$.