Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 3 / 11E (Groups, Rings and Modules)

A Euclidean domain is an integral domain $R$ with a function $\nu :R\setminus \{0\}\to \mathbb{N}$ such that for each $a\in R$ and $b\ne 0$ there are $q,r\in R$ with $a=qb+r$ and either $r=0$ or $\nu (r)<\nu (b)$. Some conventions add a multiplicative monotonicity condition; it is unnecessary for this implication.

Let $I$ be a nonzero ideal. Choose $d\in I\setminus \{0\}$ minimizing $\nu (d)$, possible because the values lie in the nonnegative integers. For $a\in I$, division by $d$ gives $a=qd+r$, with $r=a-qd\in I$. A nonzero remainder would contradict minimality, so $r=0$ and $a\in (d)$. Since $d\in I$, the reverse inclusion also holds. The zero ideal is $(0)$. Thus

Reducing coefficients modulo $p$ gives the ring isomorphismIf $d=\mathrm{deg}f>0$, its monic leading coefficient remains one modulo $p$, so $\mathrm{deg}\bar{f}=d$. Polynomial division gives a unique representative of degree less than $d$: existence follows by division, and the difference of two such representatives cannot be a nonzero multiple of a degree-$d$ monic polynomial. There are $p$ choices for each of the $d$ coefficients, provingIrreducibility of $\bar{f}$ is not needed for finiteness; without it the quotient ring need not be a field.

Write the Gaussian integer $\alpha =a+bi$, with $a,b\in \mathbb{Z}$, and set $R=\mathbb{Z}[\alpha ]$. Complex conjugation preserves $R$, because $\bar{\alpha }=2a-\alpha$. Choose a nonzero $\beta \in P$. Then $N=\beta \bar{\beta }$ is a positive integer in $P$. It cannot be one because $P$ is a proper prime ideal. Factoring $N$ into rational primes and applying primality repeatedly gives a prime $p\in P$.

The element $\alpha$ satisfies the monic polynomial $F(t)=t^2-2at+a^2+b^2$. Evaluation induces a ring homomorphism that is a surjectionThe source has $p^2$ elements by part (b), soIf $b=0$, use $F(t)=t-a$ instead and get at most $p$ elements. Moreover $R/P$ is an integral domain, so its finiteness makes it a field. This proves the finite residue-field theorem for Gaussian integer orders without incorrectly assuming that every Gaussian integer order is a principal ideal domain.