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Let $k$ be a field and $R$ a finitely generated domain over $k$. Then there are $y_1, \ldots, y_n$ such that $$R=k[y_1, \ldots, y_n]/P,$$ where $P$ is some prime ideal of $k[y_1, \ldots, y_n]$. My question is the following: is $R$ a finitely generated $k$-algebra? It seems that every finitely generated $k$-algebra has the form $k[x_1, \ldots, x_m]$. But $R$ does not have this form. Thank you very much.

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It is generated by the classes of $y_1,\dotsc,y_n$. Not every finitely generated $k$-algebra is of the form $k[x_1,\dotsc,x_m]$. Since there can be relations between the generators you have to mod them out (this is the $P$ in your definition of $R$). I assume you mean "is isomorphic to" when writing "have this form". –  Hans Giebenrath Apr 10 '13 at 7:32
    
@HansGiebenrath, thank you very much. –  LJR Apr 10 '13 at 7:45

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up vote 1 down vote accepted

A ring $R$ is a $k$-algebra if there is a homomorphism $k \to R$ and it is finitely generated (as a $k$-algebra) if there are elements finitely many $\{x_i\} \subset R$ such that each $r \in R$ can be written as a polynomial of the $x_i$'s with coefficients in $k$.

This is certainly true for a polynomial ring $k[x_1,\cdots, x_n]$, and it is therefore also true for any quotient $R=k[x_1, \cdots,x_n]/I$, since the classes of the $x_i$ generate $R$ as a $k$-algebra.

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