# Abstract Jordan Decomposition different from usual Jordan Decomposition

It's known that if $L\subset gl(V)$, with $V$ finite dimensional, is a semisimple Lie algebra, then the abstract and usual Jordan decompositions in $L$ coincide. Is it possible to provide a counter-example if $L$ isn't semisimple?

Remark: The underlying field is algebraically closed of characteristic $0$ .

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The underlying field should be algebraically closed of characteristic 0 (otherwise the discussion gets more complicated). –  Jim Humphreys Apr 15 '11 at 22:17
You're right. I'll edit my post. Thank you! –  user14312 Apr 15 '11 at 22:54

Consider the subalgebra $\left(\begin{matrix} 0& a\\ 0& 0\end{matrix}\right)$ in $\mathfrak{gl}(2)$. This is abelian, so in abstract JD, every element is semi-simple, but these are nilpotent linear operators.