Timeline for Commutative Subrings of Finite Matrix Rings
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 7, 2012 at 18:47 | history | edited | zacarias | CC BY-SA 3.0 |
deleted 129 characters in body; edited title
|
| Aug 7, 2012 at 18:02 | comment | added | YCor | @Zacarias: you should then edit your question to make it meaningful, which is not the case at the moment. | |
| Aug 7, 2012 at 15:56 | comment | added | zacarias | Yes, the interesting problem is for finite rings. | |
| Aug 7, 2012 at 15:37 | comment | added | Florian Eisele | @zacarias: If $R$ is a PID, then you can just tensor your maximal subalgebra of $M_n(R)$ with the quotient field $K$ of $R$ to get a subalgbera of $M_n(K)$. This shows that in this case the same bounds as for fields apply. The case of a finite $R$ could be interesting, though. | |
| Aug 7, 2012 at 15:30 | history | edited | zacarias | CC BY-SA 3.0 |
added 3 characters in body
|
| Aug 7, 2012 at 15:30 | comment | added | zacarias | well, we must impose restrictions on the ring $R$. If $R$ is finite maximal subrings mean subring of maximal order. If $R$ is infinite suppose that $R$ is a principal ideal domain. | |
| Aug 7, 2012 at 14:04 | comment | added | Tom Goodwillie | No, if $R=\mathbb Z/4$ then in the algebra of $2\times 2$ matrices there is a maximal commutative algebra consisting of the matrices that are scalar plus even. | |
| Aug 7, 2012 at 13:44 | comment | added | zacarias | I think that a non-free subalgebra is never maximal, because is contained in a free algebra. For example in the example of Florian Eisele if we take $b\in R$ (and not in the ideal I) the subring is free. | |
| Aug 7, 2012 at 13:39 | comment | added | zacarias | Yes, I mean free as an $R$- module. | |
| Aug 7, 2012 at 13:35 | comment | added | Florian Eisele |
@OP: What is the $R$-dimension of an $R$-algebra if $R$ is no field? This can be a serious problem for arbitrary commutative rings, for instance: Take $R=\mathbb C[x_i:i\in \mathbb N]$ (polynomial ring in infinitely many variables), and take $I$ to be the ideal generated by all $x_i$ ($I$ isn't finitely generated). Then $\left\{\left(\begin{array}{cc}a&b\\ 0&a\end{array}\right)\mid a \in R, b \in I\right\}$ is a subalgebra of $M_2(R)$ which is neither free nor finitely generated. What dimension would you assign to it?
|
|
| Aug 7, 2012 at 13:17 | comment | added | Will Sawin | That's not true over a field with $n=2$, because there are matrix algebras of rank two over a field, like the ring of diagonal matricies, but $2$ is not less than $1+1$. | |
| Aug 7, 2012 at 13:05 | history | asked | zacarias | CC BY-SA 3.0 |