Let $K$ be an algebraic closed field and $M_n(K)$ the $K$-algebra of all matrices $n\times n$ over $K$. If $L$ and $M$ are two commutative isomorphic subalgebras of $M_n(K)$ it is true that there exista a regular matrix $S\in M_n(K)$ such that $SLS^{-1}=M$. That is the isomorphism is inner?
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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 $M_n(k)$. |
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