Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2004 / 2 / ii

Suppose $m$ has no repeated irreducible factor. For every polynomial $p$ and integer $n\ge 1$, its monic greatest common divisors satisfy $\gcd (p^n,m)=\gcd (p,m)$, since raising $p$ to a power changes no membership of an irreducible factor of the square-free $m$. This is the square-free minimal polynomial criterion for polynomial kernel stabilization. The kernel identity proved above givesThus (ii) implies (iii) over every field.

Over $\mathbb{C}$, the distinct irreducible factors of $m$ are linear. The primary decomposition theorem therefore reads $V={⨁}_j\ker (T-{\lambda }_{j}I)$, a direct sum of eigenspaces. Taking a basis in each summand gives an eigenbasis, proving (ii) implies (i) over the complex numbers.

Over $\mathbb{R}$, take $T=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. Its minimal polynomial is $X^2+1$, which is an irreducible polynomial over $\mathbb{R}$ and has no repeated factor. But it has no real eigenvalues, so no real eigenbasis. This is the required counterexample. Over a general field, the phrase “no repeated factors” here refers to irreducible factors in $K[X]$, rather than a prior assumption that those factors split.