Past exam of the mathematics course of the University of Cambridge / 2019 / ib / Paper 3 / 11G (Groups, Rings and Modules)

The Eisenstein integers are $\mathbb{Z}[\omega ]$, where ${\omega }^2+\omega +1=0$. Their multiplicative algebraic norm isGiven $\alpha ,\beta \ne 0$, choose an Eisenstein integer $q$ nearest to the complex number $\alpha /\beta$. The hexagonal lattice has covering radius $1/\sqrt{3}$, soWith $r=\alpha -q\beta$, multiplicativity givesThus division with remainder decreases the nonnegative integer norm, proving that $\boxed{\mathbb{Z}[\omega ]\text{ is a Euclidean domain}}$.

Every Euclidean domain is a principal ideal domain: in a nonzero ideal choose an element of least positive Euclidean value and divide every other element by it; minimality forces every remainder to vanish. Every principal ideal domain is a unique factorization domain: the ascending-chain condition supplies factorizations into irreducibles, and Euclid lemma follows because every irreducible generates a prime ideal. Therefore

Factor the equation in the Eisenstein integers:Let $\pi =1-\omega$. Its norm is $3$, andso $3$ is a unit times ${\pi }^2$. Suppose $x\equiv 2(\mathrm{mod}3)$, say $x=3k+2$. Since $\omega =1-\pi$,The first term is divisible by ${\pi }^2$ and the second has discrete valuation exactly one, so $v_{\pi }(x+\omega )=1$. Conjugation gives $v_{\pi }(x+{\omega }^2)=1$ as well. ConsequentlyBut $v_{\pi }(y^3)=3v_{\pi }(y)$ is divisible by three, a contradiction. Hence