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?
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$\begingroup$ Doesn't this answer the question (positively)? en.wikipedia.org/wiki/Skolem%E2%80%93Noether_theorem $\endgroup$– darij grinbergCommented Dec 30, 2011 at 14:43
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1$\begingroup$ Is this homework? $\endgroup$– Bruce WestburyCommented Dec 30, 2011 at 14:46
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8$\begingroup$ Take $n=4$. Let $L$ be the diagonal matrices whose upper left $2\times 2$-block and whose lower right $2\times 2$-block are scalar multiples of identity matrices. Let $M$ be the diagonal matrices whose upper left $3\times 3$-block is a scalar multiple of the identity matrix. $\endgroup$– darij grinbergCommented Dec 30, 2011 at 15:49
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2$\begingroup$ If you aggree my answer, feel free to accept it. $\endgroup$– Denis SerreCommented Dec 30, 2011 at 21:30
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1$\begingroup$ An example was already given, but there are also examples with (commutative) semisimple algebras. Namely, in $4\times 4$ matrices, there are two non-conjugate isomorphic $K$-subalgebras isomorphic to $K\times K$, namely the subalgebras of diagonal matrices $(a_1,a_2,a_3,a_4)$ satisfying $a_1=a_2=a_3$ on the one hand, and $(a_1,a_3)=(a_2,a_4)$ on the other hand. $\endgroup$– YCorCommented Oct 4, 2019 at 18:45
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1 Answer
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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 $M_n(k)$.
answered Dec 30, 2011 at 17:06
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$\begingroup$ Dead link is dead. (Also, exorbis.pdf doesn't work either.) $\endgroup$ Commented Dec 30, 2011 at 17:45
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$\begingroup$ thank you Denis Serre. Your example responds to my question. Your answer made me think of another question: two unital subalgebras (that is, containing the scalar matrices) are conjugated? $\endgroup$– MiguelCommented Dec 30, 2011 at 18:05
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3$\begingroup$ Miguel, you could simply add the subspace $k\cdot I_n$ to each of the algebras - they would still be isomorphic but not conjugated. Also see my last comment for a different (diagonalizable) counterexample (which I hope to be correct). $\endgroup$ Commented Dec 30, 2011 at 18:10
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$\begingroup$ @darij. Sorry, I completed the link. $\endgroup$ Commented Dec 30, 2011 at 21:19