Semisimple endomorphism: Equivalent definitions? Let $V$ be a finite-dimensional vector space over a field $k$.
There are two definitions of semisimple endomorphisms I am aware of.

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*We call a linear endomorphism $\phi: V \rightarrow V$ semisimple if the endomorphism $$\tilde{\phi}: \overline{k} \otimes_{k} V \rightarrow \overline{k}\otimes_k V;\lambda \otimes v \mapsto \lambda \otimes \phi(v)$$ is diagonalizable. Here, $\overline{k}$ denotes the algebraic closure of $k$.


*We call a linear endomorphism $\phi: V \rightarrow V$ semisimple if every $\phi$-invariant linear subspace $U \subset V$ has a $\phi$-invariant complement.
Are these two definitions equivalent?
 A: These are not equivalent in general.  Definition 1 always implies definition 2, but the converse holds only if the base field $k$ is perfect.
To understand these implications better, let me take a more abstract perspective.  Let us consider a vector space $V$ together with an endomorphism $\phi$ as a $k[x]$-module, with $x$ acting as $\phi$. Definition 2 then just says that $V$ is a semisimple $k[x]$-module.  By the classification of finitely generated modules over a PID, such a $V$ is always isomorphic to a direct sum of modules of the form $k[x]/(f^n)$ where $f$ is some irreducible polynomial.  Also, $V$ is semisimple iff each of these direct summands is semisimple, and $k[x]/(f^n)$ is semisimple iff $n=1$.  (When $n=1$, $k[x]/(f)$ is in fact simple since $f$ is irreducible, and if $n>1$ then the submodule of $k[x]/(f^n)$ generated by $f$ does not have a complement.)
We can give a similar description of definition 1.  First, $\tilde{\phi}$ is diagonalizable iff $\overline{k}\otimes_k V$ splits as a direct sum of $\overline{k}[x]$-modules of the form $\overline{k}[x]/(x-\lambda)$ (these being the subspaces generated by a single eigenvector).  Since $\overline{k}$ is algebraically closed, the only irreducible polynomials over $\overline{k}$ are the linear polynomials, so the only simple $\overline{k}[x]$-modules are those of the form $\overline{k}[x]/(x-\lambda)$.  So, definition 1 is equivalent to saying that $\overline{k}\otimes_k V$ is a direct sum of simple $\overline{k}[x]$-modules, or equivalently that it is a semisimple $\overline{k}[x]$-module.
Now how do these definitions relate?  Well, when we tensor with $\overline{k}$, $k[x]/(f^n)$ turns into $\overline{k}[x]/(f^n)$, which then splits as a direct sum of modules of the form $\overline{k}[x]/((x-\lambda)^m)$ where the $\lambda$'s are the roots of $f^n$ over $\overline{k}$ and the $m$'s are their multiplicities.  All of these exponents $m$ are multiples of $n$, and in particular if $n>1$ then the $m$'s are also greater than $1$.  So, if $V$ is not a semisimple $k[x]$-module, then $\overline{k}\otimes_k V$ is not a semisimple $\overline{k}[x]$-module either, i.e. definition 1 implies definition 2.  For the converse, we need to know that an irreducible polynomial $f$ over $k$ always has distinct roots over $\overline{k}$, so that $\overline{k}[x]/(f)$ is a semisimple $\overline{k}[x]$-module.  This happens exactly when $k$ is perfect.
For an explicit counterexample if $k$ is not perfect, suppose $k$ has characteristic $p$ and $a\in k$ has no $p$th root.  Let $V=k[x]/(x^p-a)$ with $\phi$ acting by multiplication by $x$.  Then $\phi$ satisfies definition 2, as there are no nontrivial proper $\phi$-invariant subspaces of $V$ (since $x^p-a$ is irreducible over $k$).  However, $\tilde{\phi}$ is not diagonalizable: if $b\in\overline{k}$ is the $p$th root of $a$, then $\overline{k}\otimes_k V\cong \overline{k}[x]/((x-b)^p)$ and the only eigenvalue of $\tilde{\phi}$ is $b$ but the eigenspace is only $1$-dimensional (spanned by $(x-b)^{p-1}$).
