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Let $k$ be an algebraically closed field. We know that $k[x_1,\cdots,x_n]$, with $x_i$ algebraically independent over $k$, is a graded commutative $k$-algebra. Is it true that any graded commutative $k$-algebra can be identified as $k[x_1,\cdots,x_n]/ \mathcal{I}$, where $\mathcal{I}$ is some ideal of $k[x_1,\cdots,x_n]$?

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Every commutative algebra can be made into a trivial graded algebra by defining the grade of everything to be $0$. So your conjecture is a rather strong claim; essentially it says that all commutative algebras are finitely generated. –  Henning Makholm Oct 5 '12 at 18:35
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For example $k[x_1,x_2,\ldots]$ has a natural grading by degree. –  Andrew Oct 5 '12 at 18:39
    
Nice point, if you make it an answer i'll accept it. –  Manos Oct 5 '12 at 18:39
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up vote 2 down vote accepted

The claim is not true. For example, $k[x_1,x_2,\ldots]$ has a natural grading by degree, however there is no $n$ such that $k[x_1,x_2,\ldots]\cong k[x_1,\ldots,x_n]/I$ for an ideal $I.$

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You are almost correct though : Any finitely generated commutative $k$-algebra is isomorphic to a quotient of the form $k[x_1,\ldots, x_n] / I$.

If $a_1, \ldots, a_n$ are generators for your $k$-algebra $A$, use the 1st isomorphism theorem on $k[x_1, \ldots, x_n] \to A$ which sends $1 \mapsto 1$ and $x_i \mapsto a_i$. Moreover, the grading of the quotient $k[x_1, \ldots, x_n]$ will induce a grading on $A$, which however, may not coincide with the initial grading of $A$, if there was one on $A$.

It would be interesting to see what happens, if someone can tell more about this, i.e., if every fin.gen. graded $k$-algebra is isomorphic to such a quotient.

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