Prove that $a+ib$ is prime in $\Bbb Z[i]$, if $a^2+b^2$ is prime in $\Bbb Z$.

Question

Prove that $a+ib$ is prime in $\Bbb Z[i]$, if $a^2+b^2$ is prime in $\Bbb Z$.

My Try: We can easily show that $\Bbb Z[i]$ is a FD but how can we show that $\Bbb Z[i]$ is a UFD. Because if we can show that $\Bbb Z[i]$ is a UFD then we can use the fact that every irreducible element is prime...

Answer 1

The fact that $\Bbb Z[i]$ is a UFD follows from the stronger fact that it is an Euclidean domain. Note that given any complex number $z$, we can find $w=a+ib\in\Bbb Z[i]$ such that $|z-w|^2 \leqslant \frac{1}{2}$. You can prove this by dividing the complex plane into the grid determined by $\Bbb Z[i]$. Then $z$ must fall inside (or on) some square with integer coordinates and area $1$. If you subdivide this into four smaller squares, they have diagonal $\sqrt{1/2}$, hence $z$ is at most this distance away from the corner which has integer coordinates. Squaring gives the result.

Using the above suppose you have $w,z$ two elements of $\Bbb Z[i]$ (two Gaussian integers) which we can assume are nonzero. Then we know there is a Gaussian integer $q$ with $|wz^{-1}-q|^2\leqslant \frac 1 2 $, which gives $|w-zq|^2\leqslant |z|^2/2<|z|^2$.

Call $r=w-zq$, then note we can write $w=qz+r$, and $Nr<Nz$ as we wanted, i.e. we have proven that given $w,z$ Gaussian integers there are a Gaussian integers $q,r$ such that $w=qz+r$ and $Nr<Nz$. This is precisely the statement that $\Bbb Z[i]$ is Euclidean with the norm $Nz=|z|^2$.

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Alternatively, note $\Bbb Z[i]$ as a $\Bbb Z$-module is free with basis $1,i$, and the ideal generated by $w=a+ib$ is generated by $a+ib,-b+ia$ as a $\Bbb Z$-module. The matrix for this generators is $A=\begin{pmatrix}a&-b\\b&a\end{pmatrix}$. Then $\det A = a^2+b^2$, so if we carry this to its normal form it will have two nonzero invariant factors. Hence $\Bbb Z[i]/(w)$ will have $a^2+b^2=p$ elements. This means the quotient ring is no other than $\Bbb Z/p\Bbb Z$, hence the ideal $(w)$ is maximal.

Answer 2

A sketch of a direct proof (not using $\,\Bbb Z[i]$ is a UFD). Suppose $\,w = a+bi\,$ and $\,ww' = p\,$ is prime. We show $\,\Bbb Z[i]/w \cong \Bbb Z/p\,$ is a domain, so $\,w\,$ is prime. $ $ Consider the natural image of $\,\Bbb Z\,$ in the quotient ring $\, h:\,\Bbb Z\to \Bbb Z[i]/w\,$ via $\,n\mapsto n+0\,i\pmod{\!w}.\,$ Note $\,p = a^2\!+b^2$ is $\rm\color{#c00}{coprime}$ to $\color{#c00}{a,b}$.

$\ \ h$ is onto: $\ a\!+\!bi\equiv 0 \equiv p\,\Rightarrow\, i\equiv -a/b \,\Rightarrow\, c\!+\!di\equiv c-da/b\ $ (note $\,\color{#c00}{p\nmid b}\,\Rightarrow b^{-1}$ exists mod $p$)

$\ \ker h = p\Bbb Z\:$ by $\,w\mid n \iff p=ww'\!\mid nw'\!=na\!-\!nbi\iff p\mid na,nb\color{#c00}{\iff} p\mid n$

Therefore the First Isomorphism Theorem yields $\ \Bbb Z[i]/w = {\rm im}\ h \,\cong\, \Bbb Z/\ker h = \Bbb Z/p$

Answer 3

Show the contrapositive.

Assume that $a+ib$ is not a prime in $\mathbb{Z}[i]$. Then there exists non-unit elements $z,w\in\mathbb{Z}[i]$ such that $zw=a+ib$.

Consider the norm $N$ on $\mathbb{Z}[i]$ given by $N(x+iy)=x^2+y^2$ (so this is basically square of the modulus of any complex number). Then norm is clearly multiplicative, that is, $N(nm)=N(n)N(m)$ for any $n,m\in \mathbb{Z}[i]$.

Now in $\mathbb{Z}[i]$ an element has norm $1$ if and only if it is a unit (an element has norm $1$ if and only if it is on a unit circle on complex plane). Also, since $zw=a+ib$, we know that $N(a+ib)=a^2+b^2=N(z)N(w)$. Since we know that $z,w$ are non-units, then $N(z),N(w)$ are also non-units in $\mathbb{Z}$. Hence $a^2+b^2$ is not a prime in $\mathbb{Z}$.