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.)

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    $\begingroup$ If $k$ is not noetherian, then the statement about the kernel of $\varphi$ need not hold. $\endgroup$ – Fred Rohrer Mar 22 at 14:14
  • $\begingroup$ I'll add "Noetherian." My target application is really going to be rings between the integers and the rationals. $\endgroup$ – jdc Mar 22 at 17:21
  • $\begingroup$ @FredRohrer, what is the counterexample if k is not Noetherian? $\endgroup$ – jdc Mar 25 at 12:17
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    $\begingroup$ Let $k$ be a polynomial ring in infinitely many indeterminates $y_0,y_1,y_2,\ldots$ over a nonzero ring, let $n=1$ and $p=0$. Choose a natural number $m$. Map $x_1$ wherever you like, map $y_0,\ldots,y_m$ to $0$, and map $y_i$ to $y_{i-m}$ for $i>m$. This yields a surjective morphism of rings $\varphi$ whose kernel is generated by at least $m$ elements $\endgroup$ – Fred Rohrer Mar 25 at 12:23
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    $\begingroup$ Perhaps the answers to my 2nd ever MO question might be of interest: mathoverflow.net/questions/68386/… $\endgroup$ – David White Mar 26 at 20:56

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