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$\mathbb{Z}[i] = \{a+bi | a,b\in\mathbb{Z}\}$

Show that if $N(t)$ is an ordinary prime in $\mathbb{Z}$ then $t$ is a Gaussian prime in $\mathbb{Z}[i]$ (we say that $t\in\mathbb{Z}[i]$ is a Gaussian prime if it has no non-trivial factorisations.)

An attempt:

$t = a + bi \implies N(t) = a^2 + b^2$, since $a^2 + b^2$ is prime $\forall x,y \in \mathbb{Z}$, $x$ does not divide $a$ or $b$ and $y$ does not divide $a$ or $b$. So since $t = a + bi$ no $x,y \in \mathbb{Z}$ divides $t$ and since all elements $s \in \mathbb{Z}[i]$ have integer coefficients can we conclude that no non trivial element factors $t$?

Warz
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2 Answers2

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If $t=uv$ in $\mathbb{Z}[i]$, then $N(t)=N(u)N(v)$. So $t$ composite in $\mathbb{Z}[i]$ implies $N(t)$ composite in $\mathbb{Z}$. (Note that $N(x)=1$ iff $x$ is a unit in $\mathbb{Z}[i]$, so one factorization is trivial iff the other is.)

Michael Weiss
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Key Idea $\ $ Multiplicative maps preserve multiplicative properties. We seek to pullback along the multiplicative norm map $N$ the property of being an $\rm\color{#c00}{atom}$ (irreducible), i.e. a nonunit which can't be nontrivially split: $\ x = y\ z\ \Rightarrow\ y\mid 1\:\ or\:\ z\mid 1,\ $ i.e. $\:y\:$ or $\:z\:$ is a unit. Follow your $N$ose!

$$ t = \alpha\beta\, \Rightarrow\, Nt = N\alpha\ N\beta\overset{\,\large{\rm\color{#c00}{ atom}}\ Nt} \Rightarrow N\alpha\mid1\ \,{\rm or}\,\ N\beta\mid1\, \color{#0a0}\Rightarrow\, \alpha\mid 1\,\ {\rm or}\,\ \beta\mid 1\, \Rightarrow\, {\rm \color{#c00}{atom}}\,\ t$$

where we $\rm\color{#0a0}{used}$ transitivity of "divides", $\ \alpha\mid \smash[b]{\underbrace{N\alpha}_{\large \alpha\,\alpha'}}\mid 1\,\color{#0a0}{\Rightarrow}\,\alpha\mid 1._{\phantom{I_{I_{I_I}}}}$

In fact much of the multiplicative structure of a number ring is reflected in its monoid of norms. For example, in many favorable contexts (e.g. Galois) a number ring enjoys unique factorization iff its monoid of norms does. For references (Bumby and Dade, Lettl, Coykendall) see my sci.math post on 19 Dec 2007.

Bill Dubuque
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