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The Jacobson radical is $J(R)={\bigcap }_{L}L$, where $L$ runs over the maximal right ideals. Every proper right ideal $I$ lies in such an $L$: the union of a chain of proper right ideals containing $I$ is a right ideal and remains proper because it omits $1$. Thus Zorn lemma applies.

The right-module form of Nakayama lemma says that a finitely generated module $M$ with $MJ=M$ is zero. More generally, if $N$ is a submodule and $M=N+MJ$, then $M=N$. To prove the first statement, suppose $M\ne 0$ and take a generating set $m_1,\dots ,m_s$ of minimal size. Since $MJ=M$, there are $a_i\in J$ withThe unit criterion for the Jacobson radical makes $1-a_s$ a unit. Hence $m_s$ is generated by the other $m_i$, a contradiction. Apply this result to the quotient module $M/N$ to obtain the general form.

For an injective module endomorphism $f$ of an Artinian module $M$, the descending chain condition givesfor some $n$. Given $x\in M$, write $f^n(x)=f^{n+1}(y)$; injectivity of $f^n$ gives $x=f(y)$. Thus $f$ is surjective, as asserted by Artinian modules are co-Hopfian.

Now $R$ is a semilocal ring and $\overline{V}=V/VJ$ is a finitely generated module over the right Artinian ring $\overline{R}=R/J$. A finite direct sum of copies of $\overline{R}$, and hence its quotient module $\overline{V}$, is an Artinian module. The module endomorphism $\alpha$ induces $\overline{\alpha }$ on $\overline{V}$, and the given inverse-image condition says precisely that $\ker \overline{\alpha }=0$. The preceding argument makes $\overline{\alpha }$ surjective, soThe quotient module $W=V/\alpha (V)$ is finitely generated and satisfies $WJ=W$. Applying Nakayama lemma yields $W=0$. Therefore $\alpha$ is surjective, by surjectivity from injectivity modulo the Jacobson radical.

An Artinian module is an $A$-module satisfying the descending chain condition on submodules: every chain $M_0\supseteq M_1\supseteq \cdots$ is eventually constant. An Artinian ring is a ring that is Artinian as a module over itself, so its descending chains of ideals stabilize.

We first prove the closure properties needed for finite generation. A submodule of an Artinian module is Artinian, because any descending chain in it is also a chain in the original module. A quotient module is Artinian, because inverse images turn a descending chain in the quotient into one upstairs. Conversely, consider a short exact sequencewith $N$ and $Q$ Artinian. For a descending chain $(M_i)$, the chains $M_i\cap N$ and $\pi (M_i)$ both stabilize. Choose $r$ beyond both stabilization indices. If $i\ge r$ and $x\in M_i$, equality of the images gives $y\in M_{i+1}$ with $\pi (y)=\pi (x)$. Thenso $x\in M_{i+1}$. Hence $M_i=M_{i+1}$. This proves the extension property of Artinian modules in a short exact sequence.

Apply this property inductively to $0\to A^{n-1}\to A^n\to A\to 0$. Every finite direct sum $A^n$ is Artinian when $A$ is an Artinian ring. If $m_1,\dots ,m_n$ generate $M$, the map $(a_i)\mapsto \sum a_i{m}_i$ is a surjection $A^n\twoheadrightarrow M$. The quotient property therefore provesThis argument also covers the zero module; it does not require a structure theorem for Artinian rings.

For a counterexample without finite generation, take any field $k$ and the moduleThe ring $k$ is Artinian, since its only ideals are $0$ and $k$. But $M_i⊋M_{i+1}$ for every $i$, since $e_i\notin M_{i+1}$. Thus an infinite-dimensional vector space over a field provides the requested non-Artinian module.

Finally, let $f:M\to M$ be an injective endomorphism with $M$ Artinian. Its image chain stabilizes, so $f^{r}M=f^{r+1}M$ for some $r$. Given $x\in M$, choose $y$ with $f^{r}x=f^{r+1}y=f^r(fy)$. Since $f^r$ is injective, $x=fy$. ThereforeThis is precisely the assertion that Artinian modules are co-Hopfian.

For a finitely generated module $V$ on the right over a semilocal ring, the displayed condition makes the induced module endomorphism on $V/VJ$ injective. This quotient is an Artinian module, so Artinian modules are co-Hopfian makes that map surjective. The remaining quotient module $V/\alpha (V)$ equals its product with $J$, and Nakayama lemma makes it zero.