# Does a surjection between polynomial rings lose one Krull dimension per generator quotiented?

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.)

• If $k$ is not noetherian, then the statement about the kernel of $\varphi$ need not hold. – Fred Rohrer Mar 22 at 14:14
• I'll add "Noetherian." My target application is really going to be rings between the integers and the rationals. – jdc Mar 22 at 17:21
• @FredRohrer, what is the counterexample if k is not Noetherian? – jdc Mar 25 at 12:17
• 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 – Fred Rohrer Mar 25 at 12:23
• Perhaps the answers to my 2nd ever MO question might be of interest: mathoverflow.net/questions/68386/… – David White Mar 26 at 20:56