Given a regular local ring $(R,m)$ and a finitely generated $R$-algebra $S$, which is free as an $R$-module. Let $M$ be a left $S$-module of finite length, $\ell\_S(M)=r<\infty$.


Under what conditions is $\ell\_R(M)<\infty$? If this is the case, can we compute $\ell\_R(M)$ in terms of $\ell\_S(M)$?


For example if $S=M\_n(R)$, then i think we have $\ell\_R(M)=n\ell\_S(M)$.


If $S$ is commutative and local, with maximal ideal n, then according to Liu one has the following: $\ell\_R(M)=[S/n:R/m]l\_S(M)$.


Are there general formulas for length and "restriction of scalars"? I'm especially interested in the case when $S$ is a maximal $R$-order in a division algebra. Literature tips are also appreciated.



Length of a module over different rings
Let $\{V\_i\}$ be representatives from each of the isomorphism classes of simple left $S$-modules. For any finite length module ${}\_S M$, let $\ell\_S(M; V\_i)$ denote the number of times that $V\_i$ occurs in a composition series for $M$. Then the following formula holds (where almost all $\ell(M;V\_i)$ are zero because $M$ has finite length, but any single $\ell\_R(V\_i)$ could be infinite, and $\infty \cdot 0 = 0$):$$\ell\_R(M) = \sum \ell\_R(V\_i) \cdot \ell\_S(M; V\_i).$$


Thus, a finite length $S$-module $M$ has finite $R$-length if and only if every simple $S$-module that occurs in $M$ has finite $R$-length. This makes it clear that *every* finite length left $S$-module has finite $R$-length if and only if all simple left $S$-modules have finite $R$-length. The above discussion is true with no requirements on the ring $S$. 


Now let's assume that $S$ is finitely generated as an $R$-module. (I'm not sure if this is what you meant by "finitely generated as an $R$-algebra," but it's probably true in the cases that you're studying if you're looking at maximal orders.) I'll prove that *$S$ has finitely many simple modules, each of which has finite $R$-length.* Let $J(A)$ denote the Jacobson radical of any ring $A$. Then $\mathfrak{m} = J(R) \subseteq J(S)$ because $S$ is a module-finite $R$-algebra; for a proof of this fact, see Lam's *A First Course in Noncommutative Rings*, Proposition 5.7. Now the simple $S$-modules are the same as the simple $S/J(S)$ modules (see Proposition 4.8 of the same text). But $S/\mathfrak{m} S \twoheadrightarrow S/J(S)$. Because $S$ is module-finite over $R$, $S/\mathfrak{m} S$ has finite $R$-length. Thus $S/J(S)$ has finite $R$-length. So $S/J(S)$ is artinian (the terminology here is that $S$ is a *semilocal* ring) and thus has finitely many simple modules, each of which has finite $R$-length. The same must be true for the simple $S$-modules


The formula above also verifies your formula in the case of a matrix ring. In case $S = \mathbb{M}\_n(R)$, $S$ has a unique simple left module $V = (R/\mathfrak{m} \ \cdots \ R/\mathfrak{m})^T$, the set of all column vectors of length $n$ with entries in $R/\mathfrak{m}$. (At least one way to see that this is the only simple $S$-module is through Morita theory: the Morita equivalence between $R$-$\operatorname{Mod}$ and $S$-$\operatorname{Mod}$ sends the unique simple $R$-module $R/\mathfrak{m}$ to $V$, so that $V$ must be the unique simple $S$-module.) Since $\ell\_R(V) = n$, the formula above reduces to $\ell\_R(M) = n \cdot \ell\_S(M)$.


(If something above doesn't make sense, you may want to review [Jordan-Hoelder theory](http://en.wikipedia.org/wiki/Composition_series#For_modules).)


