$R$ is a commutative ring with $1$

I face difficulties to understand a theorem that the above described set is equal to the set below:

smallest subalgebra

The translation for the German part is

Let $A$ be a $R$ algebra and $M\subseteq A$ a subset

Only finitely many $r_{\alpha}\neq 0$

What I don't understand is what the parameter $\alpha$ means. Do they mean a finite subset of $M$ ? Is a multiplication of same elements allowed? I.e. if $x\in M $ is also $1xxxx$ in the set for example?

  • 3
    $\begingroup$ In the end it amounts to taking finite sums yes, but it's just an index. In principle, there are infinitely many $\alpha$. $\endgroup$ – Levi Nov 8 at 15:40
  • $\begingroup$ Specifically, this is multi-index notation. Though instead of the indices being used for powers, the entries are regarded themselves as indices into $M$. $\endgroup$ – jgon Nov 8 at 15:43
  • $\begingroup$ One then could also write instead of just $\alpha$, $\alpha \in \mathbb{N}$? $\endgroup$ – New2Math Nov 8 at 15:45
  • $\begingroup$ @New2Math The length of $\alpha$ may vary, also $M$ might be uncountable. $\endgroup$ – jgon Nov 8 at 15:47
  • 1
    $\begingroup$ Essentially what it is saying is take all finite sums of products of some element of $R$ with a finite product of things in $M$ (allowing repeats). Also you shouldn't think of $\alpha$ as a set, not just because we allow repeats, but also because $A$ might not be commutative, so the order of multiplication might matter. $\endgroup$ – jgon Nov 8 at 15:48

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