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:



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> 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*.
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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:



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



Jordan-Hölder series of $k$-subalgebras?
Since $A$ is an $n$-dimensional commutative associative unital $\Bbb R$-albebra, it has finite length as a module over itself.



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> **Claim:** The length $\ell\_A(A)$ as a module over itself equals $n$.
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> *$\because$)* 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, $F\cong \Bbb R$. $\square$
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(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.


