# The smallest subalgebra $R[M]$ that contains a subset $M$ of a $R$ algebra $A$

$$R$$ is a commutative ring with $$1$$

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

• In the end it amounts to taking finite sums yes, but it's just an index. In principle, there are infinitely many $\alpha$. – Levi Nov 8 at 15:40
• Specifically, this is multi-index notation. Though instead of the indices being used for powers, the entries are regarded themselves as indices into $M$. – jgon Nov 8 at 15:43
• One then could also write instead of just $\alpha$, $\alpha \in \mathbb{N}$? – New2Math Nov 8 at 15:45
• @New2Math The length of $\alpha$ may vary, also $M$ might be uncountable. – jgon Nov 8 at 15:47
• 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. – jgon Nov 8 at 15:48