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My question is pertaining to a definition: can we "define" a commutative ring $R$ as $R=F[x^q \mid q\in \mathbb Q]$, where $F$ is a ring? I mean, can we freely replace "any things" in the unknowns in the ring $F[x_1,x_2,x_3,...]$ with countably many unknowns $x_1,x_2,...$? One sees rings such as $F[x^2,x^3]$ in the literature without any other explanations about $F[x,y]$ (such as algebras and relations) when introducing the rings.

Any help is appreciated!

karparvar
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    Here is the `standard' definition of polynomial rings. If $F$ is a commutative ring, $S$ is any set, there exists a commutative ring which we call $F[S]$, the polynomial ring over $F$ with variables $S$ containing $S$, such that given any commutative $F$-algebra $A$ and a set map $S\to A$, there exists a unique $F$-algebra map $F[S]\to A$ which respects the given set map on $S$. – Mohan Sep 30 '16 at 16:45
  • @Mohan: Thanks for the comment. Now, how could one define the ring $R=F[x^q \mid q\in \mathbb Q]$? – karparvar Oct 02 '16 at 07:11
  • Are you going to think of $\mathbb{Q}$ as just a set and $x^q$ a variable representing $q$? In which case there is no problem from my definition. But, if you want to incorporate the additional structures of rational numbers, then, you have a completely different problem. – Mohan Oct 02 '16 at 14:38
  • @Mohan: I require the usual addition and multiplication of two elements of this $R$, to the effect that $x^s.x^t=x^{s+t}$ for any $s,t\in \mathbb Q^{+}$. – karparvar Oct 02 '16 at 17:39
  • Then you have relations and it is no longer a polynomial ring as far as I can see. Why $\mathbb{Q}^+$? If you take all of rationals, may be you are thinking in terms of group ring, with just sddition, but even then multiplication can not be accounted for. – Mohan Oct 02 '16 at 18:42
  • $F[x^2,x^3]$ denotes the ring-adjunction of $,x^2,x^3$ to $F$, i.e. the subring of $F[x]$ generated by $R,x^2,x^3.\ \ $ – Bill Dubuque Jul 21 '17 at 02:48

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