Past exam of the algebra course of the University of Oxford / 2010 / A / 1

An $R$-module $M$ is a finitely generated module if there are $m_1,\dots ,m_s\in M$ such that every element of $M$ has the form $r_1{m}_1+\cdots +r_s{m}_s$ with $r_i\in R$. Equivalently, $M=R{m}_1+\cdots +R{m}_s$, or $M$ is a quotient module of $R^s$.

The structure theorem for finitely generated modules over a principal ideal domain applies because every Euclidean domain is a principal ideal domain. In its cyclic decomposition form, every finitely generated module has a decompositionwhere the $d_i$ are nonzero nonunits. The nonnegative integer $r$ and the invariant factors of a finitely generated module $d_i$ are unique, with each $d_i$ determined up to multiplication by a unit. The zero module corresponds to no summands. Equivalently, factor the $d_i$ into powers of prime elements and use the Chinese remainder theorem to obtain a direct sum of cyclic prime-power torsion modules and a free summand.

Give $V$ a $k[t]$-module structure by $t\cdot v=\alpha v$. A finite vector space basis generates it as a module. It is a torsion module: the Cayley-Hamilton theorem gives a nonzero polynomial annihilating all of $V$. Since $k[t]$ is a Euclidean domain, cyclic decomposition giveswith monic nonconstant invariant factors of a linear operator $f_j$ and no free summand. On $k[t]/(f_j)$, use the basis $1,t,\dots ,t^{\mathrm{deg}{f}_j-1}$. Multiplication by $t$ then has the companion matrix $C(f_j)$, because the relation $f_j(t)=0$ expresses the last power in terms of earlier ones. Concatenating these bases yieldsthe rational canonical form of $\alpha$. In particular, its minimal polynomial is $f_s$ and its characteristic polynomial is ${\prod }_j{f}_j$.

The first block is the companion matrix of $t-3$. The last block is the companion matrix of $f(t)=t^3-3t^2+9t-27=(t-3)(t^2+9)$. Since $t-3$ divides $f$, these are the ordered invariant factors of a linear operator. ThusThe minimal polynomial is the largest invariant factor, and the characteristic polynomial is their product.

The minimal polynomial factors over $\mathbb{C}$ as $(t-3)(t-3i)(t+3i)$ with distinct roots. The diagonalizability criterion using the minimal polynomial therefore makes $\alpha$ diagonalizable. Its characteristic polynomial gives multiplicities two, one and one, so its Jordan normal form isup to reordering the one-dimensional Jordan blocks.

For a two-by-two matrix over a principal ideal domain, its first Smith normal form invariant is the greatest common divisor of its entries, while the product of its two invariants is its determinant, up to a unit. Hereso the entries generate the unit ideal in the Gaussian integers. The first invariant is therefore $1$. AlsoHence a valid Smith normal form isMultiplying the second diagonal entry by a unit $\pm 1,\pm i$ gives an equivalent normal form. The divisibility condition holds because $1$ divides every Gaussian integer.