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Let $h=h(x) \in \mathbb{C}[x]-\mathbb{C}$. Write: $h=c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0$, where $c_j \in \mathbb{C}$, $1 \leq j \leq n$.

Denote by $R$ the following $\mathbb{C}$-subalgebra of $\mathbb{C}[x]$: $R= \mathbb{C}+\langle h \rangle$, where $\langle h \rangle$ is the ideal of $\mathbb{C}[x]$ generated by $h$.

Question 1: Is it possible to characterize, in terms of the $c_j$'s or in another way, all $h$'s such that :

(1) $R$ is a UFD or at least $R$ is integrally closed in its field of fractions (=normal)?

(2) $R \subseteq \mathbb{C}[x]$ is flat?

Remarks:

(1) if $\deg(h)=1$, then $h=ax+b$, for some $a \in \mathbb{C}^{\times}, b \in \mathbb{C}$, so $R=\mathbb{C}[x]$ which is of course a UFD and $R=\mathbb{C}[x] \subseteq \mathbb{C}[x]$ is flat.

(2) Actually, the field of fractions of $R$ is $\mathbb{C}(x)$, since $x=\frac{xh}{h}$. Therefore, $R$ is normal if and only if $R=\mathbb{C}[x]$... since $x$ is obviously integral over $R$. So the answer to my first question is trivial: $R$ is normal iff $R=\mathbb{C}[x]$ iff $h=ax+b$, $a,b \in \mathbb{C}$.

(3) It is not difficult to see that $h=x^2$ yields: $R=\mathbb{C}+\langle x^2 \rangle=\mathbb{C}[x^2,x^3]$ which is not a UFD, since $x^2x^2x^2=x^3x^3$ are two different factorizations of $x^6$. It is also not integrally closed in its field of fractions $\mathbb{C}(x)$, since $x$ belongs to $\mathbb{C}(x)$ but $x$ does not belong to $R$. It is well-known that $\mathbb{C}[x^2,x^3] \subseteq \mathbb{C}[x]$ is not flat.

What about $h=x^2+x+1$? $h=x^2+1$? According to the above remark (2), $R$ of those $h$'s is not normal. Still what about flatness?

The following questions are perhaps relevant: i and ii.

Question 2: Assuming we have a complete answer to Question (1) or (2); is it true that the same answer holds for $\mathbb{C}+ \langle h,y_1\ldots,y_m \rangle \subseteq \mathbb{C}[x,y_1,\ldots,y_m]$? I think the answer is positive.

Any hints and comments are welcome!

user237522
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  • For both 1 and 2, you will get that $\deg h=1$. – Mohan Jan 24 '21 at 21:32
  • Thank you. Yes, I have answered question 1 in the remarks ($h=ax+b$). I have tried to answer question 2 in the remarks ($R=\mathbb{C}[x]$ is equivalent to $h$ is linear). If you have a simple argument for 2, I will accept it as an answer. However, I am still curious about separability https://math.stackexchange.com/questions/3997197/separability-of-mathbbch-subseteq-mathbbcx?noredirect=1&lq=1 – user237522 Jan 24 '21 at 22:35
  • We cannot apply Corollary 8 https://www.sciencedirect.com/science/article/pii/0021869380902331, since the ideal in the denominator of the quotient ring is probably not generated by one polynomial, but by several polynomials ($R$ is not a UFD if $\deg(h) \geq 2$, since $R$ is not integrally closed if $\deg(h) \geq 2$). – user237522 Jan 24 '21 at 22:39
  • We can apply Theorem 5 (not Corollary 9) with (1) iff (3), but the problem is that it seems difficult (hopefully I am wrong) to explicitly find all the other $h$'s (= the generators of the ideal in the quotient), and I do not see how this would yield an exact form of the original $h$. Any ideas are welcome! – user237522 Jan 24 '21 at 22:56
  • 2 implies 1. More generally, if you have $R\subset S$ finite and flat with $S$ regular, then so is $R$. So, in your case (there may be easier arguments), $R$ will be integrally closed, assuming 2. – Mohan Jan 24 '21 at 23:50
  • Thank you, nice idea! Please, any ideas concerning separability? – user237522 Jan 25 '21 at 07:03

1 Answers1

1

Perhaps (please correct me if I am wrong! Truly, I am not sure) according to the anwer to this question (first comment): $R \subseteq \mathbb{C}[x]$ is flat iff $R=\mathbb{C}[x]$. (Without assuming separability of $R \subseteq \mathbb{C}[x]$?).

user237522
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