For commutative rings $R \subseteq S$, recall that $S$ is separable over $R$, if $S$ is a projective $S \otimes_R S$-module, via $f: S \otimes_R S \to S$ given by: $f(s_1 \otimes_R s_2)=s_1s_2$.

Question 1: Is $\mathbb{C}[x]$ separable over $\mathbb{C}[x^2,x^3]$?

More generally,

Question 2: Is it possible to characterize all $\mathbb{C}$-subalgebras $\mathbb{C} \subset R \subset \mathbb{C}[x]$ such that $\mathbb{C}[x]$ is separable over $R$?

According to wikipedia: "Moreover, an algebra $S$ is separable if and only if it is flat when considered as a right module of $S \otimes_R S$ in the usual way". Here $\mathbb{C}[x^2,x^3] \subset \mathbb{C}[x]$ is not flat; I am not sure if there is a connection between flatness or non-flatness of $\mathbb{C}[x^2,x^3] \subset \mathbb{C}[x]$ and $\mathbb{C}[x] \otimes_{\mathbb{C}[x^2,x^3]} \mathbb{C}[x] \subset \mathbb{C}[x]$.

If, for example, $\mathbb{C}[x^2,x^3] \subset \mathbb{C}[x] \otimes_{\mathbb{C}[x^2,x^3]} \mathbb{C}[x]$ is flat (I do not know if this is true or false), then flatness of $\mathbb{C}[x] \otimes_{\mathbb{C}[x^2,x^3]} \mathbb{C}[x] \subset \mathbb{C}[x]$ would imply flatness of $\mathbb{C}[x^2,x^3] \subset \mathbb{C}[x]$, which is false.

(This question may be relevant. Also asked in MSE).

Edit: After receiving a comment that "it's unlikely that you can characterise all $R$ for which $R \subseteq \mathbb{C}[x]$ is separable, I would like to change Question 2 to the following question:

Question 3: Is it possible to characterize all $h \in \mathbb{C}[x]$, such that $\mathbb{C}[x]$ is separable over:

(i) $A=\mathbb{C}[h]$.

(ii) $B=\mathbb{C}+(h)$, where $(h)$ denotes the ideal of $\mathbb{C}[x]$ generated by $h$.

Example: If $h=x^2$, then $B=\mathbb{C}+(x^2)=\mathbb{C}[x^2,x^3] \subseteq \mathbb{C}[x]$ is not separable (first comment below).

Remark: Denote $h(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0$.

According to Corollary 8 with $h(Z)=c_nZ^n+c_{n-1}Z^{n-1}+\cdots+c_1Z+c_0-h$, we obtain an answer to Question 3(i): $\mathbb{C}[x]$ is separable over $\mathbb{C}[h]$ iff $\deg(h)=1$ (namely, $B=\mathbb{C}[x]$).

Still, I am not sure what can be said about Question 3(ii); could it be that for any $h$ of degree $\geq 2$, $B= \mathbb{C}+(h) \subseteq \mathbb{C}[x]$ is inseparable? If not, could one present a counterexample of minimal $\deg(h)$?

Notice that by this question we have: $B=\mathbb{C}+(h)=\mathbb{C}[h,xh,\ldots,x^{n-1}h]$.

Lemma 4.1 seems relevant; however, here $B$ is integrally closed in its field of fractions $\mathbb{C}(x)$ iff $B=\mathbb{C}[x]$, so we are not able to apply Lemma 4.1.

Any help is welcome! Thank you very much!

  • 2
    $\begingroup$ Because $\mathbf C[x]$ is finitely generated over $\mathbf C$ (in particular over $R$), separable is just another word for unramified. This can be checked either using differentials, or fibrewise (it is equivalent to each fibre being a finite product of separable field extensions). So $\mathbf C[x^2,x^3] \subseteq \mathbf C[x]$ is inseparable since the fibre over $(x^2,x^3)$ is $\mathbf C[x]/x^2$, but it's unlikely that you can characterise all $R$ for which $R \subseteq \mathbf C[x]$ is separable. $\endgroup$ Sep 17, 2020 at 19:40
  • $\begingroup$ @R.vanDobbendeBruyn, thank you very much for your helpful comment! Just to make sure, please what is the definition of being unramified that you use? Is it being separable and flat? (I apologize for my ignorance). Thank you. $\endgroup$
    – user237522
    Sep 17, 2020 at 19:57
  • 2
    $\begingroup$ No, unramified means finite type and $\Omega_{S/R} = 0$. If $R \to S$ is moreover flat, then it is called étale. See for example Tags 00UT and 08WD. The equivalence with separability is explained in Thm. 8.3.6 of Ford. $\endgroup$ Sep 17, 2020 at 20:03
  • $\begingroup$ Thank you very much again! (ok, I recall now. I was confused and forgot about the kernel). $\endgroup$
    – user237522
    Sep 17, 2020 at 20:06
  • $\begingroup$ @R.vanDobbendeBruyn, please, what if we restrict Question 2 to $R=\mathbb{C}[h]$ or $R=\mathbb{C}+(h)$, where $h \in \mathbb{C}[x]$ with $\deg(h) \geq 2$? By $(h)$ I mean the ideal of $\mathbb{C}[x]$ generated by $h$. Thank you very much! (Should I ask this as a new question?). $\endgroup$
    – user237522
    Jan 23, 2021 at 19:52

1 Answer 1


As I wrote in my comment, these extensions are known in algebraic geometry as unramified, which can be tested computationally by the vanishing of $\Omega_{S/R}$. (In differential geometry, this means $Y \to X$ is an immersion, i.e. injective on tangent spaces.)

Lemma. Let $h \in \mathbf C[x]$ be a polynomial of degree $n \geq 2$. Then $\mathbf C+(h) \subseteq \mathbf C[x]$ is unramified is and only if $h$ is squarefree.

Proof. The subalgebra $\mathbf C + (h)$ is generated as $\mathbf C$-algebra by $h, xh, x^2h, \ldots, x^{n-1}h$, since $x^n h$ is a linear combination of $h^2$ and $x^ih$ for $i < n$. So the question is whether the map \begin{align*} \mathbf A^1 &\to \mathbf A^n\\ x &\mapsto \big(h(x),xh(x),\cdots,x^{n-1}h(x)\big) \end{align*} is unramified. The Jacobian is $$\Big(\begin{matrix}h'(x) & xh'(x)+h(x) & x^2h'(x)+2xh(x) & \cdots & x^{n-1}h'(x) + (n-1)x^{n-2}h(x)\end{matrix}\Big),$$ and the ideal $I$ generated by the entries is just $(h,h')$. So $\Omega_{\mathbf A^1/\mathbf A^n} = k[x]/I$ is trivial if and only if $h$ and $h'$ have no factors in common, i.e. if and only if $h$ is squarefree. $\square$

A similar argument easily computes that the map \begin{align*} \mathbf A^1 &\to \mathbf A^1\\ x &\mapsto h(x) \end{align*} is never unramified if $n \geq 2$, since $h'(x)$ has at least one root. Thus, $\mathbf C[h] \subseteq \mathbf C[x]$ is never unramified if $\deg h \geq 2$.

  • $\begingroup$ Thank you very much! $\endgroup$
    – user237522
    Jan 25, 2021 at 16:02
  • $\begingroup$ I really apologize for interrupting you again, but what about separability of $\mathbb{C}[x,y_1,\ldots,y_r]$ over $\mathbb{C}+\langle h,y_1,\ldots,y_r \rangle$, $r \geq 1$. $\endgroup$
    – user237522
    Jan 27, 2021 at 19:02
  • $\begingroup$ If you have more questions, it's probably best to post it as a new question, so that everyone can see it. $\endgroup$ Jan 28, 2021 at 0:51
  • $\begingroup$ Thank you for your response. I agree with you and will soon post it. $\endgroup$
    – user237522
    Jan 28, 2021 at 8:27
  • $\begingroup$ (I have posted it here: mathoverflow.net/questions/382422/… Thank you!). $\endgroup$
    – user237522
    Jan 28, 2021 at 16:13

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