Past exam of the mathematics course of the University of Cambridge / 2006 / ib / Paper 2 / 2E (Groups, Rings and Modules)

A Euclidean domain is a commutative integral domain $R$ with a function $\delta :R\setminus \{0\}\to {\mathbb{Z}}_{\ge 0}$ such that, for $a,b\in R$ with $b\ne 0$, there are $q,r\in R$ satisfying $a=qb+r$, where either $r=0$ or $\delta (r)<\delta (b)$. One may also impose the usual monotonicity $\delta (a)\le \delta (ab)$ for nonzero $a,b$; the example below has it.

The integers form a Euclidean domain that is not a field. Use $\delta (a)=|a|$. Integer division by $|b|$ supplies a remainder with $0\le r<|b|$, and changing the sign of the quotient handles negative $b$. The integers are an integral domain, while $2$ has no multiplicative inverse in them. This verifies both the Euclidean property and the failure to be a field.

The structure theorem for finitely generated modules over a principal ideal domain applies because every Euclidean domain is a principal ideal domain. In invariant-factor form, every finitely generated module $M$ over such a domain $R$ has a decompositionwhere the $d_i$ are nonzero nonunits, $r\ge 0$, and either list of summands may be empty. The free rank $r$ and the invariant factors $d_i$ are unique up to multiplication of each $d_i$ by a unit. Equivalently the torsion part is a direct sum of cyclic modules with prime-power annihilators. The nonunit convention omits zero summands $R/(1)$.

Change the generators to $a=m_1$, $b=m_3-m_2$ and $c=m_2$. This is an invertible integer change of generators, since $m_1=a$, $m_2=c$, $m_3=b+c$. The relations become $2b=0$ and $4c=0$, with no relation involving $a$. ConsequentlyThis is already the invariant-factor order required by the structure theorem for finitely generated modules over a principal ideal domain, because $2\mid 4$. The inverse change of generators shows that there are no hidden extra relations.