Finitely generated left $A$-modules, over a division algebra $A$

Question

Let $A$ be a division algebra over $k$, and let $M$ be a finitely generated left $A$-module. Show that $A^n \cong M$ for some natural number $n$.

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My attempts: Assume that $M$ is generated by a minimal set $\{x_1, x_2, \cdots, x_n\}$. We have an $A$-module homomorphism $\varphi: A^n \longrightarrow M$ ($(a_1, a_2, \cdots, a_n) \mapsto a_1x_1+ a_2x_2+ \cdots+ a_nx_n$). Why is this an $A$-module homomorphism? If one can show that it is homomorphism, then clearly it is surjective.

Let $(0, 0, \cdots, 0) \neq (a_1, a_2, \cdots, a_n) \in \ker(\varphi)$, such that $a_1x_1+ a_2x_2+ \cdots+ a_nx_n=0$. Note that at least one of $a_i$'s are non-zero, without loose of generality suppose that $a_n\neq0$. Then we have $x_n=-a_n^{-1}a_1x_1- \cdots - a_n^{-1}a_{n-1}x_{n-1}$, so $\{x_2, \cdots, x_n\}$ is a generating set, which contradicts the minimality of the gererating set.

Is my proof correct? Is there any need for our algebra to be finitely generated? Are some of the assumptions extra?

Answer 1

It's all good. Your $\varphi$ is an $A$-module homomorphism because for all $a$, $a_1,\cdots,a_n \in A$, \begin{align*} \varphi(a \cdot (a_1,\cdots,a_n)) &=\varphi(aa_1,\cdots,aa_n) \\ &= aa_1x_1+ \cdots+ aa_nx_n \\ &=a(a_1x_1+\cdots+a_nx_n) \\ &=a \varphi(a_1,\cdots,a_n) \end{align*}