TITLE: Conjugation between commutative subalgebras of a matrix algebra?
QUESTION [9 upvotes]: Let $K$ be an algebraically closed field and $M_n(K)$ the $K$-algebra of all matrices $n\times n$ over $K$. If $L$ and $M$ are two isomorphic commutative subalgebras of $M_n(K)$, it is true that there exists a regular matrix $S\in M_n(K)$ such that $SLS^{-1}=M$. That is, the isomorphism can be chosen to be inner?

REPLY [10 votes]: Two isomorphic subalgebra of $M_n(K)$ do not need to be conjugated. The following example is taken from Exercise 161 of my web site http://www.umpa.ens-lyon.fr/~serre/DPF/exobis.pdf
Set $n=p+q$ with $q>p>0$. Then define  $\mathcal A$ as the subset of $M_n(k)$ made of the matrices with block form
$$\left(\begin{array}{cc} 0_p & 0_{p\times q} \\\\ A & 0_q \end{array}\right).$$
Likewise, ${\cal B}$ is made of the matrices
$$\left(\begin{array}{cc} 0_q & 0_{q\times p} \\\\ B & 0_p \end{array}\right).$$
Both $\cal A$ and $\cal B$ are subalgebras of $M_n(k)$, with dimension $pq$ and the property that $MN=0_n$ for every two elements (of the same algebra). They are obviously isomorphic, because the algebra structure is trivial. But ${\cal A}$ and $\cal B$ are not conjugated in $M_n(k)$. However $\cal B$ is conjugated to ${\cal A}^T$ in $M_n(k)$.