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How can we find all invertible elements of $\mathbb Z[i]=\{a+bi | a,b \in \mathbb Z\} \subseteq \mathbb C$?

The idea is: let $u=a+bi$ be an invertible element and let $v=c+di$ be the inverse element of $u$ over $\mathbb Z[i]$. This implies that $uv=1$, but how do i continue from here on out? Are there any more elegant solutions to this question?

And shouldn't be every element invertible in $\mathbb Z[i]$ since $\mathbb C$ is an integral domain? I appreciate your help, thank you!

null1
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  • For an element of $\mathbb{Z}[i]$ to be invertible, it needs to have an inverse in $\mathbb{Z}[i]$, so every element of $\mathbb{C}\setminus{0}$ being invertible has nothing to do with this (and this is a stronger property than just being an integral domain). – Thorgott Jan 11 '20 at 13:45

3 Answers3

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The point is invertible in $\Bbb{Z}[i]$ so $u=a+ib $ and its inverse in $\Bbb{C}$ (this one always exists if $u\neq 0$), say $v=c+id $, both belong to $\Bbb{Z}[i]$. The clean way is to use the norm as in @Somerandommathematician answer. Let’s try the brute force way.

One has

$$\begin{align} c=&{a\over a^2+b^2}\\ d=&-{b\over a^2+b^2} \end{align}$$

And we want both $c$ and $d$ to belong to $\Bbb{Z}$. This means

$$a^2+b^2|a,b$$

Because $a,b\lt a^2+b^2$ this can only happen iff $a^2+b^2=1$ and we’re left with

$$(a,b)\in \{(1,0),(-1,0),(0,1),(0,-1)\}$$

And we have proven that

$$\Bbb{Z}[i]^{\times}=\{1,-1,i,-i\}$$

marwalix
  • 17,045
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No, the element, say, $2=2+0\cdot i$ has no inverse with integer coefficients, i.e., of the form $a+bi\in \Bbb Z[i]$. The invertible elements of this ring form a group, the group of units. We have $$ \Bbb Z[i]^{\times}=\{\pm 1,\pm i\}. $$

Reference: This duplicate

Find all units in the ring Z[i] = { a+bi : a,b ϵ Z }

Dietrich Burde
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The units are just $1,-1,i-i.$

We know this because if $u=a+bi$ is an unit, then the function $$N\colon \mathbb{Z}[i]\rightarrow \mathbb{N}$$ $$a+bi\mapsto a^2+b^2$$ applied to $u$, $N(u)$ must be $1.$

Somerandommathematician
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  • Could you please elaborate what the function $N: \mathbb Z[i] \rightarrow \mathbb N$ is or what it is used for? Thank you. – null1 Jan 11 '20 at 13:59
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    It's an euclidean function, often called the norm function (for obvious reasons). With it, the ring $(\mathbb{Z},N)$ turns out to be an euclidean domain. – Somerandommathematician Jan 11 '20 at 14:04