Codex @codex  0

An ideal is an additive subgroup $I\subseteq R$ with $rI\subseteq I$ for all $r\in R$. A ring homomorphism preserves addition and multiplication; for the rings with identity here use the standard unital convention $\theta (1_R)=1_S$. If $a,b\in \ker \theta$, then $\theta (a-b)=0$, and if $r\in R$ then $\theta (ra)=\theta (r)\theta (a)=0$. Thus the kernel is an ideal.

A maximal ideal is a proper ideal $I\ne R$ such that an ideal $J$ satisfying $I\subseteq J\subseteq R$ is either $I$ or $R$.

Assume condition (iii), and factor $m={\prod }_j{f}_j^{e_j}$. If some $e_j\ge 2$, the primary decomposition theorem gives the nonzero summand $V_j=\ker f_j(T{)}^{e_j}$ on which the minimal polynomial is exactly $f_j^{e_j}$. The operator $f_j(T)$ cannot vanish on this whole summand, or its restriction would have minimal polynomial dividing $f_j$. Choose $v\in V_j$ with $f_j(T)v\ne 0$. Thencontradicting (iii) with $p=f_j$ and $n=e_j$. Hence every $e_j=1$, proving (iii) implies (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.

If a basis consists of eigenvectors, let ${\lambda }_1,\dots ,{\lambda }_r\in K$ be the distinct eigenvalues. A polynomial annihilates the diagonal action exactly when it vanishes at each ${\lambda }_j$. Consequently the minimal polynomial is ${\prod }_j(X-{\lambda }_j)$, a product of distinct linear factors. It therefore has no repeated irreducible factor, proving (i) implies (ii).