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Trying to study isomorphism classes of certain commutative Artinian $\mathbb{C}$-algebras I was lead to the following problem about matrices.

Suppose you have a (non-zero) nilpotent matrix $A\in M_n(\mathbb{C})$. Think of matrix algebras over the Artin algebra $\mathbb{C}[A]$. To be more specific, if another (linearly independent with $A$) nilpotent matrix $B\in M_n(\mathbb{C})$ which commutes with $A$ is given, I want to understand $\mathbb{C}[A]$-isomorphism classes of the subalgebra $\mathbb{C}[A,B]\subset M_n(\mathbb{C})$ of polynomial expressions in $A,\,B$. Clearly if $M\in GL_n(\mathbb{C})$ commutes with $A$, the conjugated algebra $M(\mathbb{C}[A,B])M^{-1}=\mathbb{C}[A,MBM^{-1}]$ is $\mathbb{C}[A]$-isomorphic to the original one. Here's my question:

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If $\mathbb{C}[A,B]\simeq \mathbb{C}[A,B_1]$ as $\mathbb{C}[A]$-algebras, is it true that the isomorphism is induced by conjugation with a matrix $M\in GL_n(\mathbb{C})$ which commutes with $A$? (so $B_1=MBM^{-1}$)

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I've been trying to work out an example to convince myself in the small case $n=4$. Assume $$A=\begin{pmatrix}0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\end{pmatrix}.$$ In this case the general form of a - nilpotent - matrix $B$ which commutes with $A$ is $$B_{yxab}=\begin{pmatrix}0 & y & x & a\\ 0 & 0 & 0 & b\\ 0 & 0 & 0 & x\\ 0 & 0 & 0 & 0\end{pmatrix},$$ where $x,y,a,b$ are arbitrary complex numbers. As we are interested in the algebra $\mathbb{C}[A,B]$ we can simplify $B_{yxab}$ changing it with $B_{yxab}-xA-aA^2$ to get a new $$B_{yb}=\begin{pmatrix}0 & y & 0 & 0\\ 0 & 0 & 0 & b\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}.$$ Of course (almost all) the $B_{yb}$'s are conjugated as we can check looking at Jordan forms. However, the conjugation matrices $$\begin{pmatrix}by & 0 & 0 & 0\\ 0 & b & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\end{pmatrix}$$ do not commute with $A$ so the algebras are not $\mathbb{C}[A]$-isomorphic...

My question is related to Conjugation between commutative subalgebras of a matrix algebra?, where one of the answers give a counter-example. However it seems that here I have more hypotheses (and more hope...)

Anyway, even if the answer for my question is NO I would appreciate any hint/reference to understand the previously mentioned isomorphism classes. Thanks.

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    $\begingroup$ What about $A=0$? Say, take $B_1^2=B_2^2=0$ such that $B_1$ and $B_2$ are not zero and not conjugate, then both $\mathbb C[A,B_1]=\mathbb C[B_1]$ and $\mathbb C[A,B_2]=\mathbb C[B_2]$ are isomorphic to $\mathbb C[x]/(x^2)$. $\endgroup$ Jul 16, 2020 at 17:10
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    $\begingroup$ You ask about $\mathbb C[A, B] \simeq \mathbb C[A, B_1]$ as $\mathbb C[A]$-algebras, but (because you conclude that $B$ and $B_1$ are conjugate by the centraliser of $A$) you seem actually to want to impose the stronger question that there is an isomorphism carrying $B$ to $B_1$. Is that correct? $\endgroup$
    – LSpice
    Jul 16, 2020 at 17:36
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    $\begingroup$ For example, let $n=6$, take a basis $e_1$, ..., $e_6$ and let $Ae_1=e_2$ and $Ae_i=0$ for $i>1$. Next, take $B_1$, $B_2$ as follows: $B_1e_3=e_4$, $B_1e_5=e_6$, $B_2e_3=e_4$ and all other $B_ie_j$ zero. Then both $\mathbb C[A,B_1]$ and $\mathbb C[A,B_2]$ are isomorphic to $C[x,y]/(x^2,xy,y^2)$ via isomorphisms under which $A$ corresponds to $x$. But $B_1$ and $B_2$ are not conjugate. $\endgroup$ Jul 16, 2020 at 17:57
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    $\begingroup$ @LSpice You're right. The conclusion $B_1=MBM^{-1}$ is not part of the question and it was misleading. That was not what I meant. I appologize. $\endgroup$
    – amateur
    Jul 16, 2020 at 18:36
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    $\begingroup$ Your question can also be phrased as asking about the fiber over $(0,0)$ of Hilbert scheme of points $Hilb_n(\mathbb C^2)$. The Hilbert scheme is the moduli of colength $n$ ideals in $\mathbb C[x,y]$; when the underlying variety is the origin, this corresponds exactly to commutative nilpotent $\mathbb C$-algebras with two generators of dimension $n$. $\endgroup$ Jul 16, 2020 at 19:12

2 Answers 2

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Decided to turn my comment into an answer since it seems to be independent of the ambiguity noticed by @LSpice.

Take $$ A= \begin{pmatrix} 0&1&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0 \end{pmatrix}, $$ $$ B_1= \begin{pmatrix} 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0 \end{pmatrix} $$ and $$ B_2= \begin{pmatrix} 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&1\\ 0&0&0&0&0&0 \end{pmatrix} $$ Then there are isomorphisms $\mathbb C[A,B_i]\cong\mathbb C[x,y]/(x^2,xy,y^2)$ carrying $A$ to $x$ and $B_i$ to $y$, $i=1,2$. But $B_1$ is not conjugate to $B_2$ (they both are in Jordan normal forms, which have different number of nonzero blocks).

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  • $\begingroup$ My guess is that you could find that because the length of $\mathbb{C}[A,B_i]$ is too small with respect to the dimension of the space. Anyway this is not part of the question and I accept your answer, but I'd appreciate any hint/comments on my 'guess'. $\endgroup$
    – amateur
    Jul 16, 2020 at 18:41
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    $\begingroup$ @amateur Sorry I cannot quite get your point. All I can say is that of course any isomorphism between two finite-dimensional algebras can be made into conjugation with some matrix on a large enough space - namely the space the size of these algebras. And maybe in some circumstances you can do it more economically if you can find a common module of smaller dimension on which both algebras act. But I do not see any systematic method of finding such module in your situation. $\endgroup$ Jul 16, 2020 at 20:58
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Assuming that you are asking for an isomorphism taking $B$ to $B$':

No. Take $A$ to be the $n\times n$ Jordan block with eigenvalue zero. The centralizer of $A$ is exactly $\mathbb C[A]$. Hence for any $B,B' \in \mathbb C[A]$, we have $\mathbb C[A,B] = \mathbb C[A] = \mathbb C[A,B']$. However, $B$ and $B'$ need not be conjugate: take $B = A^2, B' = A^3$. As long as $n \geq 4$, $\{1,A,A^2,A^3\}$ is linearly independent, and none are conjugate (since their kernels have different dimensions).

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    $\begingroup$ It's not quite clear what the question is, so this may not be an answer. The question says: "Are $\mathbb C[A, B]$ and $\mathbb C[A, B']$ isomorphic by conjugation by a matrix $M$ in $\mathrm C(A)$?", and in your case they are; but then the statement (no longer part of the question) goes on "so $M B M^{-1} = B'$", which, as you point out, is false. I have asked what the actual question is. $\endgroup$
    – LSpice
    Jul 16, 2020 at 18:01
  • $\begingroup$ Note that the example in the post itself involves changing the choice of $B$ without affecting $\mathbb C[A, B]$, so it seems that the specific choice of $B$ is not important (and so it may, for example, be taken to be $0$ on "both sides" of your example). $\endgroup$
    – LSpice
    Jul 16, 2020 at 18:02
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    $\begingroup$ @LSpice Indeed, I'll edit to reflect this ambiguity. $\endgroup$ Jul 16, 2020 at 18:07
  • $\begingroup$ @JoshuaMundinger I wanted a truly 2-generated algebra but that was not clear. As above, I appologize for the misleading comment (not part of the question). $\endgroup$
    – amateur
    Jul 16, 2020 at 18:44

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