TITLE: Can a the variety associated to a finitely generated $K$-subalgebra of $K[X]$ be embedded into $\mathbb{A}^3$?
QUESTION [5 upvotes]: 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 $\mathbb{R^3}$.

REPLY [3 votes]: 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 $A\not\simeq K[T_1,T_2,T_3]/I$.

REPLY [2 votes]: The subalgebra of $K[X]$ generated by $X^{10}$, $X^{11}$, $X^{12}$ and $X^{13}$ cannot be generated by three elements.