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Let $h=h(X) \in \mathbb{C}[x]$, $\deg(h) \geq 2$, and denote $R_{h}:=\frac{\mathbb{C}[x]}{\langle h(x)} \rangle $.

Question 1: Is it possible to characterize all $h \in \mathbb{C}[x]$ such that $\bar{h'}$ (= the class of the formal derivative of $h$ in $R_{h}$) is invertible in $R_{h}$?

Examples:

(1) $h=x^2$, $h'=2x$, $\bar{h'}$ is not invertible in $R_{h}$, since it is a zero divisor: $\bar{h'}\bar{x}=\bar{2x}\bar{x}=\bar{2x^2}=\bar{0}$.

(2) $h=x^2+1$, $h'=2x$ is invertible in $R_{h}$, since $\bar{h'}\bar{x}=\bar{2x^2}=-\bar{2}$.

(3) $h=x^2+x$, $h'=2x+1$. Is $\bar{h'}$ invertible? (yes?).

I excluded the cases $\deg(h) \leq 1$ since they are easier:

(i) If $\deg(h)=0$, then $h= \lambda \in \mathbb{C}-\{0\}$, so $\langle \lambda \rangle = \mathbb{C}[x]$, hence $R_h=0$ (the zero ring).

(ii) If $\deg(h)=1$, then $h= \lambda x + \mu$, so $\langle h \rangle$ is a maximal ideal of $R_h$, hence $R_h=\mathbb{C}$.

Question 2: Same question with $\mathbb{C}$ replaced by $\mathbb{R}$ or $\mathbb{Q}$ or $\mathbb{Z}$.

A relevant question is this (and perhaps also this).

Thank you very much!

user237522
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  • Hint: a polynomial if coprime to its derivative iff it is squarefree. You can compute the inverse via the extended Euclidean algorithm (to get the Bezout equation for $\gcd(f,f'))\ \ $ – Bill Dubuque Sep 29 '20 at 23:31
  • @BillDubuque, thank you for your relevant comment. – user237522 Sep 29 '20 at 23:35
  • Please, are there interesting generalizations for $k[x_1,\ldots,x_n]$, $n \geq 2$? For example, $\frac{k[x,y]}{\langle h,y \rangle}$, $h \in k[x]$, but the answer is the same. Perhaps taking $h \in k[x,y]$ and considering the partial derivatives $h_x,h_y$ may yield something interesting? – user237522 Sep 29 '20 at 23:45
  • Let me say that over $\mathbb Z$ the condition $\gcd(h,h')=1$ is necessary but not sufficient. Consider $h=X^2+X+1$. – user26857 Oct 02 '20 at 19:28
  • I reopened this question in case someone has a necessary and sufficient characterization for $\mathbb Z$. – user26857 Oct 02 '20 at 19:28

1 Answers1

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The derivative $h'$ should be coprime with $h$. In other words, $h$ should have no multiple complex roots - this is the answer for $\Bbb C,\Bbb Q$.

markvs
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  • Yes, because if $h,h'$ are coprime, then $ah+bh'=1$ for some $a,b \in k[x]$, from which one sees that $h'$ is invertible, with inverse $b$. Thank you very much. What about $\mathbb{R}$? – user237522 Sep 29 '20 at 23:33
  • The same for $\Bbb R$. – markvs Sep 29 '20 at 23:36
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    If by "characterize" the OP means an iff condition, then let me say that over $\mathbb Z$ the condition $\gcd(h,h')=1$ is necessary but not sufficient. Consider $h=X^2+X+1$. – user26857 Sep 30 '20 at 08:39