# Jordan-Hölder series of $k$-subalgebras?

I am reading Shurygin's survey "Smooth Manifolds over Local Algebras and Weil Bundles" (Journal of Math. Sciences, Vol. 108, No. 2, 2002) and it mentions the following basic fact which I don't quite understand:

Let $A$ be an $n$-dimensional commutative associative unital local $\mathbb{R}$-algebra with max. ideal $\mathfrak{m}$ such that $A/\mathfrak{m}=\mathbb{R}$. Then $A$ has a Jordan-Hölder series $A\supset\mathfrak{m}=:A_1\supset A_2\supset\dots\supset A_{n−1}\supset A_n:=0$ such that $A_i/A_{i+1}$ are $1$-dimensional algebras with zero multiplication.

Certainly, a composition series of length $n$ of $\mathbb{R}$-vector subspaces exists, but it is not immediately obvious to me how one arranges for the additional condition to hold. The only reference he gives for this fact is Pierce's book "Associative Algebras", where the author only treats Jordan-Hölder for (sub)modules without any special additional multiplicative structure from what I can tell. Hence my question is:

How do we obtain the existence of such Jordan-Hölder series with the above additional property?

In particular, it seems that implicitly $A_i$ are meant to be $\mathbb{R}$-subalgebras (without unit) and $A_{i+1}$ an ideal of $A_i$. On the other hand, not all $A_i$ are necessarily ideals of $A$ since in general the length of $A$ as an $A$-module need not equal $\dim_{\mathbb{R}}A$. Or perhaps $A_i$ are only vector spaces such that the quotient vector space $A_i/A_{i+1}$ still "magically" inherits multiplication from $A$? Or am I missing something entirely obvious? Thanks!

• $A=\mathbf{C}$ (viewed as 2-dimensional $\mathbf{R}$-algebra) doesn't seem to match your claim.
– YCor
Mar 17 '18 at 22:34
• You will need to assume $A/\mathfrak{m}=\mathbb{R}$ as the above comment suggests. Mar 17 '18 at 22:52
• Sorry, yes, I forgot to mention this, it is part of the definition of a local Weil algebra.
– M.G.
Mar 17 '18 at 23:42
• Assume you have found $A\supset A_1\supset\cdots \supset A_k$, where all these are ideals. If $A_k\neq 0$, then nor is $A_k/\mathfrak{m}A_k$ and this is a $A/\mathfrak{m}=\mathbb{R}$ vector space. So, take any vector space surjection $A_k\to A_k/\mathfrak{m}A_k\to A/\mathfrak{m}$ and let $A_{k+1}$ be the kernel. Rest you should be able to check. Mar 18 '18 at 2:23

Since $A$ is an $n$-dimensional commutative associative unital $\Bbb R$-albebra, it has finite length as a module over itself.

Claim: The length $\ell_A(A)$ as a module over itself equals $n$.

$\because$) It's immediate that $\ell_A(A)\leq n$. To see that $\ell_A(A)=n$, let $F$ be an $A$-module composition factor $F$. It suffices to show that $\dim_\Bbb R F= 1$. But $A$ is a local commutative associative unital $\Bbb R$-algebra with maximal ideal $\mathfrak{m}$, so $A/\mathfrak{m}=\Bbb R$ is the only simple $A$-module. Hence, $F\cong \Bbb R$. $\square$

(This next part is essentially Mohan's comment.) By the Claim, we can find an $A$-module composition series $$A=A_0\supsetneq \cdots \supsetneq A_n = 0.$$ Then each $A_i$ is an $A$-ideal and therefore a (non-unital) subalgebra.

• Yes, the essence is that the only simple $A$-module is $\mathbb{R}$, which, when applied to a composition series of ideals, immediately gives the result. Also, you meant to say $\ell_A(A)\leq n$.
– M.G.
Mar 19 '18 at 19:31
• Since the characterisation of the simple $A$-modules is the only not entirely obvious part, you may wish to include a short argument for it for completeness sake: let $0\neq x\in M$ and define $f:A\to M$, $a\mapsto ax$, then $f\neq 0$ as $f(1)=x$, hence $f(A)\subseteq M$ non-trivial submodule; therefore, $f$ is surjective and $M\cong A/\ker f$, which is thus simple, whence $\ker f = \mathfrak{m}$ by the correspondence for ideals of the factor ring.
– M.G.
Mar 21 '18 at 0:09