Let $R$ be a finite commutative ring with identity. Considere the matrix ring $A=M_n(R)$. What is the order of a maximal commutative subring of $A$ that contains all scalar matrices?

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 $1+1$.
– Will SawinAug 7 '12 at 13:17

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@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?
– Florian EiseleAug 7 '12 at 13:35

Yes, I mean free as an $R$- module.
– zacariasAug 7 '12 at 13:39

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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.
– Tom GoodwillieAug 7 '12 at 14:04

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@Zacarias: you should then edit your question to make it meaningful, which is not the case at the moment.
– YCorAug 7 '12 at 18:02

`$\left\{\left(\begin{array}{cc}a&b\\ 0&a\end{array}\right)\mid a \in R, b \in I\right\}$`

is a subalgebra of $M_2(R)$ which is neither free nor finitely generated. What dimension would you assign to it? – Florian Eisele Aug 7 '12 at 13:35