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Let $R$ be a ring and $S\subseteq R$. Then usually $R[S]$ denoted the subring generated by $S$ (especially in the context of field extensions I have seen this notation to be used).

My question is: How is this notation consistent with the notation $R[X]$ for the set of all polynomials with coefficients in $R$ ? Since if $R$ is the set of all polynomials, then for the element $X\in R$, the set $R[X]$ (in the sense of the definition above) is, since it is the smallest subring containing the element $X$, the set $\{ \sum_i X^i \mid i\in \mathbb{N}\}$.

So obviously $R\neq R[X]$ and therefore the notation $R[X]$ for the set of all polynomials doesn't seem to me to be consistent with the notation $R[S]$ for the smallest subring containing $S$.


Thus I wonder, why one doesn't use a different notation for the set of all polynomials. After all, when (rigorously) defining the set of all polynomials with coefficients in $R$, one usually defines $R[X]$ to be the set of all function $\mathbb{N} \rightarrow R$, that are are nonzero only for finitely many values (here $X$ is just a letter with not any mathematical meaning yet) and then one shows that if one defines $X$ as $X=(0,1,0,0,\ldots )$, that one can write every such function $f$ as $\sum_n f(n)X^n$, which yields, when making yet another definition $f(n)=a_n$, a mathematical object that comprises what we usually understand a polynomial to be.

But, as I showed, even if $X=(0,1,0,0,\ldots )$, one can't say that $R[X]$ can also be understood as the subring generated by $X$, so $R[X]$ has to remain just a string of symbol, if one wants to remain contradiction free (where this contradiction is on a syntactical/notational level and not a mathematical level).

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I interpret $R[S]$ as "the smallest $R$-subalgebra of some context-specified $R$-algebra generated by $S$". Under this reading $R[X]$ makes perfect sense, since $X$ generates $R[X]$ as an $R$-subalgebra of $R[X]$. – Zhen Lin Oct 27 '12 at 13:56
@ZhenLin And is it also custom to denote with "$R[S]$" the smallest $R$-algebra generated by $S$ ? – temo Oct 27 '12 at 14:35
No further comments/answer from anybody :( ? – temo Oct 27 '12 at 16:22
Ok, seems I have to prepare myself for a bounty...(bump!!) – temo Oct 28 '12 at 17:14
For $S\subseteq R$, you would have $R[S] = R$. Most likely you have seen this when you have two rings $k$ and $R$ with $k\subseteq S\subseteq R$. Then $k[S]$ is used for the smallest subring of $R$ containing $S$. If $k$ lies in the center of $R$ then both $R$ and $k[S]$ are $k$-algebras. – Marc Olschok Oct 29 '12 at 19:50

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