Take $R$ to be a commutative ring and $A \in R_{n} $ and the homomorphism $f: R^{n} \rightarrow R^{n}$ defined by the equation $f(B) = AB $ be surjective.
How do I show that this implies $f$ is an isomorphism (as in it's invertible)?
I have established that since it's surjective, it obviously has a right inverse but what about the left inverse?
On
One way to see this is by the theory of determinant. Namely, since $x \mapsto Ax$ is surjective, we can find a matrix $B$ such that $AB = I$. Then applying determinant we obtain that $\det (A) \det (B) = 1$ and thus $\det (A)$ is a unit in $R$. Then we have $adj (A) \cdot A = \det (A) \cdot I$ and we can write $\frac{1}{\det (A)}adj (A) \cdot A = I$ and so $A$ has a left inverse.
See https://en.wikipedia.org/wiki/Adjugate_matrix for $adj(A)$.
To define a map between $R^n$ and $R^n$ it is enough to determine what the map does to a basis. For example, we could consider the standard basis $e_1,\dots,e_n$, where $e_i$, as a vector, has a $1$ in the $i$th spot and $0$'s elsewhere. Since $f$ is surjective, for each $i$, we can choose a $b_i$ such that $f(b_i)=e_i$. Now we can define a map $g:R^n \to R^n$ by defining $g(e_i)=b_i$ and extending by linearity. Extending by linearity means if we are given $r \in R^n$, then, writing $r$ (uniquely) as $r=\displaystyle \sum^n_{i=1} r_ie_i$ where the $r_i \in R$, we then define $g(r)=\displaystyle \sum^n_{i=1} r_ig(e_i)=\displaystyle \sum^n_{i=1} r_ib_i$.
The maps $f$ and $g$ are inverses by construction.