Decide if quotient module is free module

abstract-algebramodules

Let $U=\{\lambda\cdot (-3+i):\lambda \in \mathbb{Z}\}$ a sub-module of the $\mathbb{Z}$-module $\mathbb{Z}[i]$ (Gaussian integers). Is $\mathbb{Z}[i]/U$ a free module?

My attempt:

It's pretty obvious that $\mathbb{Z}[i]/U$ is a straight line in the complex plane and that's very familiar with quotient spaces from linear algebra. So $\{-3+i\}$ is a basis in $U$. Now I have to complete this basis to a basis in $\mathbb{Z}[i]$, right?

I could just choose $\{\{-3+i\},\{i\}\}$, because they're linearly independent.

But I could in principal do the same for $V=\{\lambda\cdot(6+2i ):\lambda\in \mathbb{Z}\}$ but I was told that $\mathbb{Z}[i]/V$ is not a free module. I think this is because $(-3+i) + V \in \mathbb{Z}[i]/V$ and
$$\mathbb{Z}\times (\mathbb{Z}[i]/V) \longrightarrow \mathbb{Z}[i]/V, \quad (\lambda,x+V)\mapsto \lambda\cdot x +V$$

So choosing $\lambda=2$, we have:
$$(2,(-3+i)+V) = (-6-2i)+V=V$$
and $V$ is in that sense the neutral Element $0_{\mathbb{Z}[i]/V}$? I my thinking correct?

So it seems like I need to check something else to decide if $\mathbb{Z}[i]/U$ is a free module?

Best Answer

Claim: $\mathbb Z[\mathrm i]/U\cong \mathbb Z$ as $\mathbb Z$-modules.

Proof. One shows that the map $\phi:\mathbb Z[\mathrm i]/U\to\mathbb Z, a+b\mathrm i+U\mapsto a+3b$ is an isomorphism of $\mathbb Z$-modules, that is

  • $\phi$ is well-defined (one shows that if $a+b\mathrm i+U=c+d\mathrm i+U$, then $a+3b=c+3d$);
  • $\phi$ is a $\mathbb Z$-module (this is straightforward);
  • $\phi$ is surjective (as for $a\in\mathbb Z$, we have $\phi(a+U)=a$);
  • $\phi$ is injective (since $\ker\phi=0$).$\quad\tiny\blacksquare$

Since $\mathbb Z$ is obviously $\mathbb Z$-free (having $B=\{1\}$ as a basis), the module $\mathbb Z[\mathrm i]/U$ is as well.

To see what $\phi$ does intuitively, let us take a look at the following graph.

intuition

Each element of $\mathbb Z[\mathrm i]/U$ corresponds to one of the depicted lines (intersected with $\mathbb Z[\mathrm i]$). The map $\phi$ now sends each such line to its intersection point with the $x$-axis.

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