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P. M. Cohn proved the following: A subalgebra of $\mathbb{C}[t]$ is free if and only if it is integrally closed. (Actually, Cohn's result deals with $k[t]$, $k$ any field, but I do not mind to take $k=\mathbb{C}$).

Let $f=f(t),g=g(t) \in \mathbb{C}[t]$, with $\deg(f)>\deg(g)> 1$, and consider $\mathbb{C}[f,g]$.

For example: $f=t^3$, $g=t^2$, and then $\mathbb{C}[t^3,t^2] \subsetneq \mathbb{C}[t]$ is not integrally closed, since $t \in \mathbb{C}(t)=\mathbb{C}(t^3,t^2)$ is of course integral over $\mathbb{C}[t^3,t^2]$ ($t$ is a root of $T^2-t^2$, and $t$ is also a root of $T^3-t^3$) but $t \notin \mathbb{C}[t^3,t^2]$.

Is it possible to characterize all (non-)integrally closed subrings of $\mathbb{C}[t]$ of the form $\mathbb{C}[f,g]$ as above?

It would be nice if it is possible to find a general form of $f$ and $g$ in terms of their coefficients and degrees, so denote: $f=a_nt^n+\cdots+a_1t+a_0$ and $g=b_mt^m+\cdots+b_1t+b_0$, where $n > m > 1$ and $a_i,b_i \in \mathbb{C}$.

What about characterizing all free subrings $F$ of $\mathbb{C}[t]$ generated by two elements? (see remark (2) below; I wish to exclude $\mathbb{C}[t^4,t^2]$). If the field of fractions of a free $2$-generated subring $F$ equals $\mathbb{C}(t)$ is necessarily $F=\mathbb{C}[t]$?

Remarks:

(1) The assumption $m>1$ excludes the case $g=b_1t+b_0$ and then $\mathbb{C}[f,g]=\mathbb{C}[t]$, so it is integrally closed.

(2) I think that I better further assume that the field of fractions of $\mathbb{C}[f,g]$ equals $\mathbb{C}(t)$, in order to exclude cases such as $f=t^4, g=t^2$, which I am not interested in.

(3) Abhyankar-Moh-Suzuki theorem says that if $\mathbb{C}[f,g]=\mathbb{C}[t]$ then $m$ divides $n$. Therefore, if $m$ does not divide $n$, then $\mathbb{C}[f,g] \subsetneq \mathbb{C}[t]$; in this case, is it still possible for $\mathbb{C}[f,g]$ to be integrally closed, under the further assumption that $\mathbb{C}(f,g)=\mathbb{C}(t)$?

(4) If $\mathbb{C}[f,g] \subseteq \mathbb{C}[t]$ is flat and $\mathbb{C}(f,g)=\mathbb{C}(t)$, then $\mathbb{C}[f,g]=\mathbb{C}[t]$ and we are done (= $\mathbb{C}[f,g]$ is integrally closed). This follows from the second answer to this question (trivially, $\mathbb{C}[f,g] \subseteq \mathbb{C}[t]$ is integral). Is there a criterion for flatness in terms of the coefficients of $f$ and $g$ and their degrees?

(5) Additional relevant questions and papers are: Necessary and sufficient condition that a localization of an integral domain is integrally closed 1; Generalized quotient rings (Richman) 2; (Some remarks on Richman simple extensions of an integral domain 3).

Thank you very much for any help!

user237522
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2 Answers2

1

If $A\subset \mathbb{C}[t]$ is an integrally closed $\mathbb{C}$-subalgebra with $A\neq \mathbb{C}$, then $A=\mathbb{C}[u]$ for some $u=p(t)$. Thus, if $A=\mathbb{C}[f,g]\subset\mathbb{C}[t]$ and $\deg f,\deg g$ neither dividing the other, then $A$ is not integrally closed by Abhyankar-Moh-Suzuki. Of course, this is not a characterization, since even if $\deg f$ diivided $\deg g$, $A$ still might be not integrally closed, like $f=t^3, g=t^2+t^6$.

Mohan
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  • Thank you very much! Please, why: "If $A\subset \mathbb{C}[t]$ is an integrally closed $\mathbb{C}$-subalgebra with $A\neq \mathbb{C}$, then $A=\mathbb{C}[u]$ for some $u=p(t)$". – user237522 May 03 '18 at 18:01
  • Did you apply Cohn's result + free subrings $F \neq k$ of $k[t]$ are (isomorphic to $k[t]$ hence are) necessarily of the form $k[p]$ for some polynomial $p$? – user237522 May 03 '18 at 18:39
  • @user237522 There are many ways of seeing that. Luroth's theorem would imply that the fraction field of $A$ is $\mathbb{C}(u)$ for some $u$ and then show that $u$ can be chosen to be in $\mathbb{C}[t]$ and since every element of $\mathbb{C}[t]$ is integral over $A$, one gets $u\in A$. – Mohan May 03 '18 at 20:34
  • Thank you. Please, do you think it is possible to find a criterion concerning the coefficients of $f$ and $g$? I have suggested several ways: flatness, LCM-stability, normality (and there are more ways such as Jacobian criterion for flatness/smoothness, separability, dimension and regularity https://mathoverflow.net/questions/204596/a-generalization-of-miracle-flatness-theorem?noredirect=1&lq=1). – user237522 May 03 '18 at 22:17
1

Question $(3)$ is easily answered . . .

Let $m,n$ be positive integers with $m \le n$.

Suppose $f,g\in \mathbb{C}[t]$ are such that

  • $\deg(f)=m$.$\\[4pt]$
  • $\deg(g)=n$.$\\[4pt]$
  • $\mathbb{C}(f,g)=\mathbb{C}(t)$.

Claim:$\;$If $\mathbb{C}[f,g]$ is integrally closed, then $m{\,\mid\,}n$.

Proof:

Assume $f,g,m,n$ satisfy the specified conditions, and suppose $\mathbb{C}[f,g]$ is integrally closed.

Since $f\in \mathbb{C}[t]$ is non-constant, it follows from the identity $f(t) - f=0$, that $t$ is integral over $\mathbb{C}[f]$, hence $t$ is integral over $\mathbb{C}[f,g]$.

From $t \in \mathbb{C}(t)=\mathbb{C}(f,g)$, we get that $t$ is in the field of fractions of $\mathbb{C}[f,g]$, hence, since $\mathbb{C}[f,g]$ is integrally closed, it follows that $t\in \mathbb{C}[f,g]$, so $\mathbb{C}[f,g]=C[t]$

By the results of Abhyankar-Moh-Suzuki, it follows that $m{\,\mid\,}n$, which proves the claim.

quasi
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  • Thank you very much! Any thoughts about my other questions? – user237522 May 03 '18 at 18:04
  • A regards your question: "Is there a criterion for flatness in terms of the coefficients of f and g and their degrees?", I suspect that an algorithm (rather than a closed-form criterion) is the best you can hope for. – quasi May 03 '18 at 18:13
  • Thank you. (1) Perhaps some algebraic geometry arguments may help? (2) Isn't it possible to apply https://math.stackexchange.com/questions/158488/necessary-and-sufficient-condition-that-a-localization-of-an-integral-domain-is?rq=1 ? (first we will have to know how prime ideals in $\mathbb{C}[f,g]$ look like https://math.stackexchange.com/questions/2764108/prime-ideals-in-mathbbcft-gt-where-ft-gt-in-mathbbct). – user237522 May 03 '18 at 18:21
  • Perhaps, but still, I doubt that you can get a closed-form criterion in terms of the degrees of $f,g$ and their coefficients. – quasi May 03 '18 at 18:28
  • ok, thank you (I still hope it is possible...). – user237522 May 03 '18 at 18:31