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.

  • $\begingroup$ By "free R-algebra" do you mean "free algebra over R" or "algebra over R that is free as an R-module"? From your example $S = M_n(R)$ it seems that you mean the second, but this isn't clear (to me) from the phrasing of the first sentence. $\endgroup$ Sep 19, 2010 at 16:40
  • $\begingroup$ Yes that is what i meant. Thanks for pointing that out. $\endgroup$
    – TonyS
    Sep 20, 2010 at 13:28
  • $\begingroup$ Dear Tony, you probably knew this already, but just in case: if $S$ is a maximal order in some $M_n(K)$ where $K$ is the quotient field of $R$, and $S$ is free as $R$-module, then $S=M_a(R)$ for some $a$. This can be found in Auslander-Goldman article "Maximal orders". $\endgroup$ Sep 20, 2010 at 15:11
  • $\begingroup$ Yes, i have this article on my desk. I think if $R$ is a complete d.v.r., $D$ a skew field over K=Quot(R) and $S$ is the unique maximal $R$-order in $D$, then $M=S/rad(S)$ is the unique simple $S$ module. We can compute $\ell_R(M)$ after an etale base change $R \rightarrow R'$, so we can assume $S$ is a hereditary order in $M_n(K)$, where $n^2=[D:K]$.This should give $\ell_R(M)=n$. $\endgroup$
    – TonyS
    Sep 20, 2010 at 15:39

1 Answer 1


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.)

  • 1
    $\begingroup$ Thanks for your answer. So basically i have to compute the $R$-length of simple $S$-modules. Just one question is left to your answer to my first question: Why should $R$ be artinian, i.e. of finite length over itself? I just assume its a regular local ring. $\endgroup$
    – TonyS
    Sep 20, 2010 at 13:34
  • $\begingroup$ Oops, thanks for the correction. I read "artinian local" rather than "regular local." I've edited my answer accordingly; hopefully there aren't too many more mistakes. I'm assuming that, rather than simply any finitely generated $R$-algebra, you want a module-finite $R$-algebra. Please correct me if I'm wrong. $\endgroup$ Sep 20, 2010 at 17:45
  • $\begingroup$ Thank you very much. That looks very good. I think i will take a deeper look in the book you mentioned. $\endgroup$
    – TonyS
    Sep 20, 2010 at 18:56

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