Past exam of the algebra course of the University of Oxford / 2007 / A / 1

A left module over the unital $K$-algebra $A$ is a $K$-vector space $V$ with a $K$-bilinear action $A\times V\to V$ satisfying $(ab)v=a(bv)$ and $1v=v$. For a vector $v$, the map $A\to V$, $a\mapsto av$, is an $A$-module homomorphism. Its kernel is $\mathrm{Ann}(v)$; explicitly this is a linear subspace and $b(av)=(ba)v=0$ for $a\in \mathrm{Ann}(v)$, so it is a left ideal.

If $V=Av$, this homomorphism is surjective, and the module isomorphism theorem gives $V\cong A/\mathrm{Ann}(v)$. Conversely the quotient by a left ideal is generated by $1+I$. Thus cyclic modules are exactly the quotients $A/I$ by left ideals.

For the first example set $e=\mathrm{diag}(1,1,0)$. Right multiplication by $e$ keeps the first two columns and zeroes the third. Every matrix with third column zero equals its product with $e$. Hence $V=Ae$ is cyclic, generated by $e$.

For a matrix $v$ supported in its first row and any upper triangular matrix $a$, multiplication gives $av=a_{11}v$. Thus $Av=Kv$ has dimension at most one for every $v$, whereas the entire module has dimension three. Hence this module is not cyclic, over every field $K$.

Coordinatewise multiplication gives $V=A(1,1,0)$, so the third module is cyclic.

A submodule of a cyclic module need not be cyclic. In the second example the first-row space is a left ideal and hence a submodule of the regular module $A$. The regular module is cyclic, generated by $1_A$, whereas its first-row submodule has just been shown not to be cyclic. Thus the proposed inheritance property fails even for a finite-dimensional algebra.