Separability of $\mathbb{C}[x,y_1,\ldots,y_r]$ over $\mathbb{C} + (h,y_1,\ldots,y_r)$

Question

The answer to this MO question says the following:

Lemma 1. Let $h \in \mathbf C[x]$ be a polynomial of degree $n \geq 2$. Then $\mathbf C+(h) \subseteq \mathbf C[x]$ is unramified if and only if $h$ is squarefree.

I wonder what happens in higher dimensions, namely:

Question: Is the following claim true: Let $h \in \mathbb{C}[x]$ be a polynomial of degree $n \geq 2$. Then $\mathbb{C}[x,y_1,\ldots,y_r]$ is unramified (equivalently: separable) over $\mathbb{C}+(h,y_1,\ldots,y_r)$ if and only if $h$ is squarefree.

What I think is that if Lemma 1 can be generalized to the following Lemma 2, then the answer to my current question is positive.

Lemma 2. Let $A$ be a commutative $\mathbb{C}$-algebra of characteristic zero. Let $h \in A[x]$ be a polynomial of degree $n \geq 2$. Then $A +(h) \subseteq A[x]$ is unramified if and only if $h$ is squarefree.

Indeed: If Lemma 2 is true, then $\mathbb{C}[y_1,\ldots,y_r]+(h) \subset \mathbb{C}[y_1,\ldots,y_r][x]= \mathbb{C}[x,y_1,\ldots,y_r]$ is unramified if and only if $h$ is aquarefree.

Recall a known result about separability (can be found here) which says that given $R_1 \subseteq R_2 \subseteq R_3$, if $R_1 \subseteq R_3$ is separable, then $R_2 \subseteq R_3$ is separable.

Therefore here, if $h$ is squarefree, then $\mathbb{C}[y_1,\ldots,y_r]+(h) \subset \mathbb{C}[x,y_1,\ldots,y_r]$ is unramified, and then $\mathbb{C}+ (h,y_1,\ldots,y_r) \subset \mathbb{C}[x,y_1,\ldots,y_r]$ is unramified.

Any hints and comments are welcome! Thank you.

Answer 1

In Lemma 2, I'm not sure what "squarefree" means. However, the meaning is clear if $h\in\mathbb{C}[x]$, and then Lemma 2 is true. Indeed, the inclusion $\mathbb{C}+(h)\subset \mathbb{C}[x]$ induces an injection $A\otimes_\mathbb{C}(\mathbb{C}+(h))\subset A[x]$. The image is easily checked to be $A+(h)$. The lemma follows because being unramified is preserved by base change.