Let $k$ be a Noetherian commutative ring with unity (I am happy adding hypotheses, as I will ultimately want $k = \mathbb Z$) and $$\varphi\colon A = k[x_1,\ldots,x_n] \to k[y_1,\ldots,y_p] = B$$ a surjective $k$-algebra homomorphism. (I can assume $\varphi$ sends $(x_1,\ldots,x_n)$ to $(y_1,\ldots,y_p)$, but don't think this will help.)

I believe $\ker \varphi$ can be generated by exactly $n-p$ elements $f_{p+1},\ldots,f_{n} \in (x_1,\ldots,x_n)$. Certainly at least this many are needed.

Must $f_{p+1},\ldots,f_n$ be a regular sequence?

(It would be desirable if in fact $C = k[f_{p+1},\ldots,f_n]$ were a tensor factor of $A$, meaning there would be a subring $\tilde B$ of $A$ generated by lifts $\tilde y_j \in A$ of $y_j$ and such that $$A = \tilde B \otimes C = k[\tilde y_1,\ldots,\tilde y_p,f_{p+1},\ldots,f_n].$$ This becomes true after completing at $(x_1,\ldots,x_n)$, but is apparently false(!) for $k = \mathbb Z$, and open (the Abhyankar–Sathaye embedding problem) over $\mathbb C$ as soon as $p \geq 2$ and $n-p \geq 1$.)

One way forward might be the solution I saw to this MSE question. It would mean it is necesary only to show the Krull dimension of each $A/(f_1,\ldots,f_m)$ is $(\dim k) + n - m$ for $0 \leq m \leq p$. We know this is true for $m = 0,p$, and that the Krull dimension goes down as one quotients out the ideal generated by (the image of) each successive $f_j$, so to show it decreases by exactly $1$ each time, it suffices to show it never decreases by $0$. It seems clear this holds of the transcendence degree over $k$, at least; is this enough to show the same holds of Krull dimension?

(This is a repost of this MSE question, which got little traction.)