Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2009 / 1

The zero polynomial belongs to $I_v$. If $f,g\in I_v$, linearity gives $(f-g)(T)v=0$. If $a\in K[X]$ and $f\in I_v$, polynomial evaluation obeys $(af)(T)=a(T)f(T)$, so $(af)(T)v=0$. Thus $I_v$ is an ideal of the polynomial ring $K[X]$.

The $n+1$ vectors $v,Tv,\dots ,T^{n}v$ are linearly dependent in the $n$-dimensional vector space. Their dependence supplies a nonzero polynomial of degree at most $n$ in $I_v$. Hence this ideal is nonzero. A polynomial ring over a field is a Euclidean domain under degree, and every ideal in a Euclidean domain is principal. It follows that $I_v=(m_v)$ for a nonzero polynomial $m_v$, which can be scaled uniquely to be monic.

For completeness, uniqueness follows because any two generators divide each other and hence differ by a nonzero constant; the monic normalization forces that constant to be one. Since $m_v$ divides the polynomial of degree at most $n$ constructed above, its degree is at most $n$. The generator characterization givesThis is the minimal polynomial of a vector under a linear operator. If $v=0$, the ideal is the whole ring and the unique monic generator is the constant polynomial one.