Difference between $R[c_1,c_2,\dots, c_n]$ and a finitely generated $R$-algebra. What is the difference between $R[c_1,c_2,\dots, c_n]$ ($c_1, c_2,\dots, c_n\notin R$), where $R$ is a ring, and a finitely generated $R$-algebra?  
Is the difference that if $c_1, c_2,\dots, c_n$ are the generators of the $R$-algebra, then their highest powers are bounded in the $R$-algebra, while they are not in $R[c_1,c_2,\dots, c_n]$?
Thanks in advance. 
 A: When you write the expression
$$R[c_1,\ldots,c_n],$$
what that signifies (to me at least) is a quotient of a polynomial ring $R[x_1,\ldots,x_n]$  by some ideal $I\subset R[x_1,\ldots,x_n]$ such that the composition of the maps
$$R\longrightarrow R[x_1,\ldots,x_n]\longrightarrow R[x_1,\ldots,x_n]/I$$
is injective; the $c_i$'s denote the equivalence classes of the $x_i$'s in the quotient ring. 
Any such ring certainly a finitely-generated $R$-algebra, but the converse is not true. For example, $\mathbb{Z}/p\mathbb{Z}$ is a finitely-generated $\mathbb{Z}$-algebra, but it cannot be expressed as $\mathbb{Z}[c_1,\ldots,c_n]$ for any $c_i$'s in the manner described above.
If, however, you do not require the above map to be injective, then any finitely-generated $R$-algebra can be expressed as $R[c_1,\ldots,c_n]$, and any $R$-algebra $R[c_1,\ldots,c_n]$ is finitely-generated (both directions essentially by definition).
By the way, you seem confused about what it means for an $R$-algebra to be finitely generated. The polynomial ring $R[x]$ is an $R$-algebra, the powers of $x$ occurring in the elements of $R[x]$ are not bounded, and $x$ is a generator of $R[x]$ as an $R$-algebra.
A: If you're talking about $R[c_1,\dots,c_n]$ for $c_1,\dots,c_n\not\in R$, then surely the $c_i$'s must live somewhere (don't we all?). So, perhaps you've got a ring extension $R\subseteq S$; here, $R$ is a subring of $S$. Also, $c_1,\dots,c_n\in S$ (but they aren't elements of $R$ by your assumption). 
In this case, $R[c_1,\dots,c_n]=\{p(c_1,\dots,c_n):p\in R[x_1,\dots,x_n]\}$ is a subring of $S$. So, we have a triple $R\subseteq R[c_1,\dots,c_n]\subseteq S$ of rings. What is the structure of the ring $R[c_1,\dots,c_n]$? Maybe you don't care as long as it's a subring of $S$. But if you do care, then thankfully Zev has explained this point very nicely above. We have a surjective homomorphism of $R$-algebras $R[x_1,\dots,x_n]\to R[c_1,\dots,c_n]$ obtained by evaluation at $(c_1,\dots,c_n)$, i.e., $p(x_1,\dots,x_n)\to p(c_1,\dots,c_n)$. If you check this is indeed a surjective homomorphism of $R$-algebras, then Zev explains the rest! The ring $R[c_1,\dots,c_n]$ is isomorphic to a quotient ring of $R[x_1,\dots,x_n]$! Here's an easy exercise in case you're interested:
Exercise 1: Prove that $R[c_1,\dots,c_n]$ is isomorphic to $R[x_1,\dots,x_n]$ as $R$-algebras if and only if the set $\{c_1,\dots,c_n\}$ is algebraically independent over $R$ (meaning: there is no polynomial in $n$ variables with coefficients in $R$ that vanishes at the tuple $(c_1,\dots,c_n)$). 
I hope this helps!
