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 \b[this <module> is not cyclic], over every <field> $K$.

Solution

Past exam of the algebra course of the University of Oxford / 2007 / A / 1 / 2 / Solution

 Solution (source code)