Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How would you show that the only units in $\mathbb{Z}[i] :=\{a + ib \, |\, a,b\in \mathbb{Z}\}$ are $1,\, -1, \, i, \, -i$?

share|improve this question

4 Answers 4

up vote 7 down vote accepted

Hint: Every element of $\mathbb Z[i]$ has a natural number norm, and $\nu(xy)=\nu(x)\nu(y)$. This imposes a restriction on $\nu(x)$ if $x$ is a unit.

share|improve this answer

$\mathbb{Z}[i]$ coincides with the lattice points in the complex plane. If you pick a lattice point outside the unit circle, its inverse lies inside. The only lattice point inside however is $0,$ which is not invertible. So any units must lie on the unit circle.

share|improve this answer
    
Note that the implied proof is essentially the same as the proof using multiplicativity of the norm, except here it is rephrased as a proof by contradiction, namely $\rm \alpha \beta = 1\:$ $\Rightarrow$ $\rm\:|\alpha||\beta| = 1.\:$ Therefore $\rm\:|\alpha|>1\:$ $\Rightarrow$ $\:|\beta|< 1,\:$ contradiction. Thus $\rm\:|\alpha| = 1.$ But that's a roundabout route, since once one has $|\alpha||\beta| = 1$ we can immediately deduce $|\alpha| = 1.\:$ –  Bill Dubuque May 22 '12 at 15:01
    
In any case, it's worth emphasizing that one needs to be very careful when applying "geometric" intuition. In particular, note that the above proof does depend crucially on multiplicativity, a fact that is, alas, hidden deeply in the presentation. It's easy to deceive oneself that way. Better to bring essential hypotheses to the fore. –  Bill Dubuque May 22 '12 at 15:02
    
@Bill, I completely agree that this proof alone would not be the most beneficial for a student new to the topic. Alex had already posted the canonical approach, and I just thought it was a 2nd view worth mentioning. –  Ragib Zaman May 23 '12 at 0:43
    
@RagibZaman +1 Nice answer. –  user38268 May 23 '12 at 2:06

Here is an elementary proof that requires no knowledge of field theory (the field-theoretic proof requires not only the proof of multiplicativity of the norm, but also the proof of multiplicativity (and well-definedness!) of conjugation, a subtlety that is too frequently overlooked).

Let $\rm \ c = (a\!+\!b{\it i})(a\!-\!b{\it i}) = a^2 + b^2\in\mathbb Z.\:$ Then by rationalizing denominators we deduce

$$\rm \frac{1}{a\!+\!b{\it i}} = \frac{a\!-\!b\:\!{\it i}}{c}\in \mathbb Z[{\it i}\:]\:\Rightarrow\:\frac{a}c,\frac{b}c\in\mathbb Z\:\Rightarrow \frac{a\!+\!b{\it i}}c\in \mathbb Z[{\it i}\:]\:\Rightarrow\frac{a\!-\!b{\it i}}{c}\frac{a\!+\!b{\it i}}c = \frac{c}{c^2} = \frac{1}c\in\mathbb Z[{\it i}\:]$$

Thus $\rm\:1/c \in \mathbb Z[{\it i\:}]\cap\mathbb Q = \mathbb Z\:$ $\Rightarrow$ $\rm\: 1 = c = a^2 \!+\! b^2,\:$ so $\rm\:a = \pm 1, b = 0\:$ or $\rm\:a = 0, b = \pm 1.\ \ $ QED

Alternatively, more slickly via gcds: the gcd $\rm\:\!(a,b)\! =\! d\:\!|\!\:a,\!b,\:$ so $\rm\:d^2\:\!|\!\:a^2,b^2\Rightarrow\:d^2\:\!|\:c\!=\! a^2\!+\!b^2,\:$ so

$$\rm \frac{1}{a\!+\!b{\it i}} = \frac{a\!-\!b\:\!{\it i}}{c}\in \mathbb Z[{\it i}\:]\:\Rightarrow\:c \:|\:a,b\:\Rightarrow\: c\:|\:(a,b)\!=\!d\:\Rightarrow\: d^2\:\!|\:c\:|\:d\:\Rightarrow\: d = 1 = c\quad QED$$

share|improve this answer

The ring $\mathbb{Z}[i]$ is a subset of $\mathbb{C}$. Any nonzero $a + bi$ is invertible in $\mathbb{C}$, and the inverse is

$$(a + bi)^{-1} = \frac{a}{a^2 + b^2} - \frac{b}{a^2 + b^2}i$$

Thus nonzero $a + bi \in \mathbb{Z}[i]$ is invertible in $\mathbb{Z}[i]$ if and only if $\frac{a}{a^2 + b^2} \in \mathbb{Z}$ and $\frac{b}{a^2 + b^2} \in \mathbb{Z}$.

Use this to show that $a + bi \in \mathbb{Z}[i]$ is invertible if and only if $a + bi = \pm1$ or $a + bi = \pm i$. To do this, treat the cases where $a = 0$ or $b = 0$ first. What happens when $a \neq 0$ and $b \neq 0$?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.