For what values of $a\in \mathbb{Z}[i]$ do $(2,1)$ and $(2+i,a)$ form a basis of $\mathbb{Z}[i]^2$?

Question

For what values of $a\in \mathbb{Z}[i]$ do $(2,1)$ and $(2+i,a)$ form a basis of $\mathbb{Z}[i]^2$?

My original thought is to let $a=1+i$. But $(2,1)$ and $(2+i,1+i)$ don't span $\mathbb{Z}[i]^2$.

Answer 1

The elements $(2,1),(2+i,a) \in \mathbb{Z}[i]^2$ form a basis if and only if the matrix $$\begin{pmatrix} 2 & 2+i \\ 1 & a \end{pmatrix}$$ has unit determinant in $\mathbb{Z}[i]$, in other words, if and only if $$2a-(2+i) \in \{ 1,i,{-1},{-i} \}$$ Thus the possible values of $a$ are $$\frac{3+i}{2}, \quad 1+i, \quad \frac{1+2i}{2}, \quad 1$$ Of course, not all of these are elements of $\mathbb{Z}[i]$...

Answer 2

I beg your pardon, but $(1,1)$ and $(2+i,1+i)$ are a basis of $\mathbf Z[i]^2$. The condition for $(a,b)$ and $(c,d)$ to be a basis of $R^2$ ($R$ a commutative ring) is the determinant of these vectors is a unit in $R$. The units in $\mathbf Z[i]$ are $\{1,-1,i,-i\}$.

For your initial question, the condition is $$2a-2-i\in\{1,-1,i,-i\},\enspace\text{i.e.}\quad a\in\Bigl\{1,1+i,\frac{1+i}2,\frac{3+i}2\Bigr\}.$$

Answer 3

If you look at $\begin{pmatrix}2\\1\end{pmatrix}$ and $\begin{pmatrix}2+i\\ a\end{pmatrix}$ as a basis spanning $\mathbb{Q}(i)$, then for them to be a basis for $\mathbb{Z}[i]$, of course $a$ must be in $\mathbb{Z}[i]$, but the determinant of their matrix must be a unit. The determinant is

$\det\begin{pmatrix}2&2+i\\1&a\end{pmatrix}=2a-(2+i)$. We need this to be a unit, so either $1, i, -1$ or $-i$. It cannot be $\pm 1$, so it must be $\pm i$. Then this gives $a=1+i$ or $1$. Why did you not believe that $a=1+i$ did not work?