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Let $K$ be a field.

Is there an example of a finitely generated $K$-subalgebra $$ A\subseteq K[X] $$ which is not isomorphic to $K[T_1,T_2,T_3]/I$ for some ideal $I$?

As $A$ is finitely generated, we may write $A\cong K[X_1,X_2,\ldots, X_n]/I$ for some $n$. Geometrically this means, that the variety associated to $A$ can be embedded into $\mathbb{A^n}$. This variety is $0$-dimensional or $1$-dimensional. I have the vague topological intuition, that such an object sould be embeddable into $\mathbb{A^3}$ just like a graph into $\mathbb{R^3}$.

user8463524
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2 Answers2

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Set $A=K[X^4,X^5,X^6,X^7]$. The maximal ideal $(X^4,X^5,X^6,X^7)$ of $A$ is generated by four elements and not less (why?). This shows that $A\not\simeq K[T_1,T_2,T_3]/I$.

user26857
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  • Thank you for your answer, user26857. Do you have a geometric (''hand-waving'') argument for why your example works? There is a surjective $K$-algebra map from $K[X_1,X_2,X_3,X_4]$ onto your $A$. I guess, calculating the kernel should give some information what's going on geometrically, right? Unfortunately I have no idea... – user8463524 Jun 25 '15 at 08:35
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The subalgebra of $K[X]$ generated by $X^{10}$, $X^{11}$, $X^{12}$ and $X^{13}$ cannot be generated by three elements.

Mariano Suárez-Álvarez
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