# In the ring $\mathbb Z[i]$ explain why our four units $\pm1$ and $\pm i$ divide every $u\in\mathbb Z[i]$.

In the ring $\mathbb Z[i]$ explain why our four units $\pm1$ and $\pm i$ divide every $u\in\mathbb Z[i]$.

This is obviously a elementary question, but Gaussian integers are relatively new to me. I found this exercise in the textbook, and my professor overlooked it, but i'm curious.

Is this basically the same thing as, for a lack of a better term, "normal" integers? As in, $\pm1$ divides everything?

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I understand, but what about $\pm i$? –  ground.clouds1 Nov 16 '12 at 3:09

Let

$$\,a+ib\in\Bbb Z[i]\Longrightarrow a+ib=i(b-ia)\Longrightarrow i\mid(a+ib)$$

and it's similar with the other ones.

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Hint $\$ Units $\rm\:u\mid 1\mid x.\$ Alternatively $\rm\ u\mid u(u^{-1}x) = (uu^{-1})x = x$

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As important as it is to understand why the units divide everything, it's also important to understand why those are the only units (as they are probably being referred to as "the units" in class).

Let $z=a+bi$. Then we can define the norm of $z$ to be $N(z)=|z|^2=z\overline z = a^2+b^2$. Note, some people call $|z|$ the norm, and I'm not entirely sure which is more standard. Hopefully no confusion will arise here.

The norm satisfies two important properties. First, $N(xy)=N(x)N(y)$, and second, if $z$ is a Gaussian integer, then $N(z)$ is an integer.

Because of the first property, if $z$ has an inverse in the Gaussian integers, then $1=N(1)=N(z z^{-1})=N(z)N(z^{-1})$, and so $N(z^{-1})=N(z)^{-1}$. The only numbers such that $x$ and $x^{-1}$ are both integers are $\pm 1$, and since the norm is always non-negative (being the sum of square of real numbers), we just have to solve the equation

$$N(z)=a^2+b^2=1.$$

Since every non-zero square of an integer is at least $1$, this has no solutions other than $z=\pm 1, \pm i$.

We are not done yet because we have only shown that if $z$ has an inverse then $N(z)=1$. We must check that each of these solutions is actually invertible. However, for any nonzero complex number, we have $z^{-1}=\frac{\overline z}{N(z)}$, so

$$(1)(1)=(-1)(-1)=(i)(-i)=1,$$

and so the inverses of these Gaussian integers are still integers.

By definition, an element $r$ of a ring is called a unit if there exists an $s$ such that $rs=sr=1$. In this case, $s$ is called the inverse of $r$, and is unique when it exists, so it is often written as $r^{-1}$. What we have shown so far is that the only units in $\mathbb Z[i]$ are $\pm 1, \pm i$.

As Gerry Myerson wrote, if $u$ is a unit, then for any $r$ in the ring, we can write $r=1r=(uu^{-1})r=u(u^{-1}r)$. Therefore, $r$ is divisible by $u$.

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Let $w$ be a unit (in any ring). By definition of unit, there's an $x$ in the ring such that $wx=xw=1$. Now let $r$ be any element of the ring. Then $r=(rx)w$, so $w$ divides $r$.

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