For commutative rings $R \subseteq S$,
recall that $S$ is [separable](https://en.wikipedia.org/wiki/Separable_algebra) 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](https://math.stackexchange.com/questions/3997681/properties-of-mathbbch-subseteq-mathbbcx?noredirect=1&lq=1) question may be relevant. Also asked in [MSE](https://math.stackexchange.com/questions/3997197/separability-of-mathbbch-subseteq-mathbbcx)).


**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](https://pdf.sciencedirectassets.com/272332/1-s2.0-S0021869380X80025/1-s2.0-0021869380902331/main.pdf?X-Amz-Security-Token=IQoJb3JpZ2luX2VjEAUaCXVzLWVhc3QtMSJHMEUCIQDjm2Fqy725SiofuzGVJP4E3G%2FXJjmL5QZYnjMqZZQ%2BngIgHvRHaCCWv2Ui%2F756H8sS%2BcspRvknkqHwuFpEzPXSn4wqvQMI3v%2F%2F%2F%2F%2F%2F%2F%2F%2F%2FARADGgwwNTkwMDM1NDY4NjUiDFFXR7yWAyjQFZFKZCqRA4CdnJpaE8ptww%2Be4b1zH4vmjQTihmg2a5fC5lwbEUv40uccQgS3ZX%2FPOfiKCHnCTz0LwNBl5zv3kG23a8UhQYmcA3ZptKgzdM8eMhg7KldR84XKyvRXgdNgbwIr6I6aZ7DBxP99h7BVRFZmvREhNchfjdHJ8uV0kvE9mB9knm6R8ugIunekOx9v2tH%2BAWpUzwtmbx0fsGHtzoY2tlAncO2T3schB1pFx5GRVq9e20Cv9hDW71fGhUztVZIb8AemGp5IbrmwIKVnWhGOOGTLdm0aSlV%2Bsh1je%2FqKYruVf3OXyYJD8QXsfOjA8aWdWa3%2FWT%2BOFFmHcskv3%2Fzm80SVzbr08pWVYT02yJR6GQxXceBX9zlejip31BbNvHWB8oFVKIat821Tp3epZZ1aErWhcuM20W0wKMCp74m8vXFMnIOgUP3GsPoOkRcAokAZbF7K288nLXR6kaU7a00Qm7VPxtz5qATPiOsO76lFLSO5%2BqfbSGI0Vnsu%2FsyPRTZ31xplininVoEtWZgLTRlvVvoK%2FlOXMOSVsoAGOusBdx2c%2F9oHzJYrKKKXQ8uFXI%2FZ%2FSTgrQJMtwW%2FhL9nCFhsqhFnVmRAs2BwywAiUx%2BCqdjJWZWmTkRbCFT%2BHHSoNxjacBDPuZ6Bnx9Pc1%2Fm%2FmkLEP9neiaMKXPWFiLtfeWG8QApGokJM1eIB7WiyX7RSBz%2FeSbON1nmgU7rqgpoxAWWoFt5R1BtmSdLiaTRl%2Bkr%2F%2BcVJS2sZQ85FgvsdRTnrRMIJpKPdwCsVlhm7D7AA797ASt8bz2QkC8MsX149RprtoELP8iBrTNhcQALdmhQIO4tnWCCFYFLzmIlm7vlRwJ9Jc99bLr09S9vGA%3D%3D&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Date=20210123T212144Z&X-Amz-SignedHeaders=host&X-Amz-Expires=300&X-Amz-Credential=ASIAQ3PHCVTYQ2CR5ZGR%2F20210123%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Signature=4497c8a33679d2b5cf46402ede6678840cc1506780ffff79f2854294ae02e1cd&hash=5ba89707ce0dcb87f4676f0f3411344167be8eebdc09980f6e9314345d8dc846&host=68042c943591013ac2b2430a89b270f6af2c76d8dfd086a07176afe7c76c2c61&pii=0021869380902331&tid=spdf-95359de1-2f03-4d31-abc4-e4e841532060&sid=7a58ddde1d338143876ab9766bc68efd0ffdgxrqb&type=client) 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](https://math.stackexchange.com/questions/106856/subalgebras-of-a-polynomial-ring) question we have: $B=\mathbb{C}+(h)=\mathbb{C}[h,xh,\ldots,x^{n-1}h]$.


[Lemma 4.1](https://www.ams.org/journals/tran/1966-122-02/S0002-9947-1966-0210699-5/S0002-9947-1966-0210699-5.pdf) 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!



Separability of $\mathbb{C}[x]$ over its $\mathbb{C}$-subalgebras
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 $\deg h \geq 2$.


