I'm trying to solve an exercise which asks me to determine for what values of $a\in \mathbb{Z}[i]$ $(2,1)$ and $(2+i,a)$ form a basis of $\mathbb{Z}[i]^2$ (where we're considering $\mathbb{Z}[i]^2$ as a module over $\mathbb{Z}[i]$).
I see that someone has asked the same question before:
For what values of $a\in \mathbb{Z}[i]$ do $(2,1)$ and $(2+i,a)$ form a basis of $\mathbb{Z}[i]^2$?
but this hasn't helped me. The answers given here rely on the fact that $(2,1),(2+i,a)$ form a basis if and only if the matrix $$\begin{pmatrix} 2 & 2+i \\ 1 & a \end{pmatrix}$$ has unit determinant in $\mathbb{Z}[i]$.
I've never seen this result (or anything of the sort before) and have no idea how I'd go about proving it.
Does anyone know of an alternative way to do this, or even the name of the result if it's a standard one or just some idea how you'd go about proving it?
I'd really appreciate any help you could offer.
Left multiplication by the matrix $A := \begin{pmatrix} 2 & 2+i \\ 1 & a \end{pmatrix}$ defines a $\mathbb{Z}[i]$-linear map $\varphi: \mathbb{Z}[i]^2 \to \mathbb{Z}[i]^2$ that sends the standard basis $\pmatrix{ 1\\ 0}, \pmatrix{ 0\\ 1}$ to $\pmatrix{ 2\\ 1}, \pmatrix{ 2+i\\ a}$. So the vectors $\pmatrix{ 2\\ 1}, \pmatrix{ 2+i\\ a}$ form a basis iff $\varphi$ is invertible iff $A$ is invertible iff $\det(A)$ is a unit in $\mathbb{Z}[i]$. (This last implication can be proved using the standard cofactor formula for the inverse of a matrix.)