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Let $R$ be a $k$-subalgebra of $\mathbb{C}[x_1,\ldots,x_n]$, $n \geq 1$.

Assume that $R$ is of Krull dimension $1$. Is $R$ isomorphic to a polynomial ring in one variable?

More generally,

Assume that $R$ is of Krull dimension $m \leq n$. Is $R$ isomorphic to a polynomial ring in $m$ variables?

A known result says that $\mathbb{C}[x_1,\ldots,x_n]$, $n\geq 1$, is of Krull dimension $n$; I am asking about 'the converse' result (in a given polynomial ring).

Remark: What about $n=1$, $R=\mathbb{C}[x^2,x^3] \subset \mathbb{C}[x]$? Is it a counterexample to my first question? (probably yes); I guess that the Krull dimension of $R$ should be $1$, as a subalgebra of $\mathbb{C}[x]$ which is of Krull dimension $1$. It is not a polynomial ring in any number of variables, since it is not a UFD (indeed, $x^2x^2x^2=x^3x^3$).

So I ask: Is there a nice condition which guarantees a positive answer to my questions?

Any hints and comments are welcome!

user237522
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    Yes to the remark. Another way of seeing it is that $\mathbf{C}[x^2,x^3]$ is isomorphic to $\mathbf{C}[x,y]/(y^2-x^3)$. This is the coordinate ring of a elliptic curve with a singularity at the origin, so it can't be isomorphic to $\mathbf{C}[t]$. – Ashwin Iyengar Sep 13 '18 at 23:30
  • Thank you; I like your comment. Please, do you think that notions from algebraic geometry will help to answer my third question? (= "So I ask...")? – user237522 Sep 13 '18 at 23:49
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    Considerations of fields of transcendence degree $d\ge1$, especially when $d>1$, involve algebraic geometry in a fundamental way. It’s just unavoidable. As to @AshwinIyengar’s comment, when a cubic has a singularity, it is not, strictly speaking, an elliptic curve anymore. Loosely speaking, though, your remark is good enough. – Lubin Sep 13 '18 at 23:58
  • Thank you for your comment; please, do you have any specific ideas concerning my third question? – user237522 Sep 14 '18 at 00:03
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    If $R$ is Noetherian, integrally closed and dimension one $\mathbb{C}$-subalgebra of the polynomial ring, it is isomorphic to $\mathbb{C}[t]$. – Mohan Sep 14 '18 at 13:29
  • @Mohan, thank you very much for your comment. (I remember I have seen this result a few years ago). Actually, your comment can serve as one answer to my (third) question with $m=1$. What about $m \geq 2$? (Perhaps I will later ask about a generalization for the result you mentioned in a separate question, namely, for $m \geq 2$; I guess that there exists a counterexample). – user237522 Sep 14 '18 at 13:38
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    Take a look at the paper "A note on finite dimensional subrings of polynomial rings" by Paul Eakin. – user26857 Sep 15 '18 at 09:10
  • @user26857, Oh, I apologize. Thank you for the paper. It seems that Theorem 1 on page 77 of the paper (which was proved by Zaks; see page 80) shows that: If $R$ is integrally closed and dimension one $\mathbb{C}$-subalgebra of a polynomial ring, then it is isomorphic to $\mathbb{C}[t]$. Notice that $R$ is not assumed to be Noetherian; am I missing something? – user237522 Sep 15 '18 at 22:11
  • BTW, I guess that in his answer to https://math.stackexchange.com/questions/2803632/a-sufficient-and-necessary-condition-for-mathbbcfx-gx-mathbbcx, Mohan implicitly referred to the above paper that user26857 mentioned (Lemma C, page 77). – user237522 Sep 15 '18 at 23:23
  • The following paper by P. M. Cohn https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/plms/s3-14.4.618 shows that $R \subseteq k[t]$ is integrally closed iff $R$ is isomorphic to $k[h]$, $h \in k[t]$ (=$R$ is free). – user237522 Sep 20 '18 at 16:13

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