Lemma 15.33.1. Let $A \to B$ be a finite type ring map. If for some presentation $\alpha : A[x_1, \ldots , x_ n] \to B$ the kernel $I$ is a Koszul-regular ideal then for any presentation $\beta : A[y_1, \ldots , y_ m] \to B$ the kernel $J$ is a Koszul-regular ideal.

**Proof.**
Choose $f_ j \in A[x_1, \ldots , x_ n]$ with $\alpha (f_ j) = \beta (y_ j)$ and $g_ i \in A[y_1, \ldots , y_ m]$ with $\beta (g_ i) = \alpha (x_ i)$. Then we get a commutative diagram

Note that the kernel $K$ of $A[x_ i, y_ j] \to B$ is equal to $K = (I, y_ j - f_ j) = (J, x_ i - f_ i)$. In particular, as $I$ is finitely generated by Lemma 15.32.2 we see that $J = K/(x_ i - f_ i)$ is finitely generated too.

Pick a prime $\mathfrak q \subset B$. Since $I/I^2 \oplus B^{\oplus m} = J/J^2 \oplus B^{\oplus n}$ (Algebra, Lemma 10.134.15) we see that

Pick $p_1, \ldots , p_ t \in I$ which map to a basis of $I/I^2 \otimes \kappa (\mathfrak q) = I \otimes _{A[x_ i]} \kappa (\mathfrak q)$. Pick $q_1, \ldots , q_ s \in J$ which map to a basis of $J/J^2 \otimes \kappa (\mathfrak q) = J \otimes _{A[y_ j]} \kappa (\mathfrak q)$. So $s + n = t + m$. By Nakayama's lemma there exist $h \in A[x_ i]$ and $h' \in A[y_ j]$ both mapping to a nonzero element of $\kappa (\mathfrak q)$ such that $I_ h = (p_1, \ldots , p_ t)$ in $A[x_ i, 1/h]$ and $J_{h'} = (q_1, \ldots , q_ s)$ in $A[y_ j, 1/h']$. As $I$ is Koszul-regular we may also assume that $I_ h$ is generated by a Koszul regular sequence. This sequence must necessarily have length $t = \dim I/I^2 \otimes _ B \kappa (\mathfrak q)$, hence we see that $p_1, \ldots , p_ t$ is a Koszul-regular sequence by Lemma 15.30.15. As also $y_1 - f_1, \ldots , y_ m - f_ m$ is a regular sequence we conclude

is a Koszul-regular sequence in $A[x_ i, y_ j, 1/h]$ (see Lemma 15.30.13). This sequence generates the ideal $K_ h$. Hence the ideal $K_{hh'}$ is generated by a Koszul-regular sequence of length $m + t = n + s$. But it is also generated by the sequence

of the same length which is thus a Koszul-regular sequence by Lemma 15.30.15. Finally, by Lemma 15.30.14 we conclude that the images of $q_1, \ldots , q_ s$ in

form a Koszul-regular sequence generating $J_{h''}$. Since $h''$ is the image of $hh'$ it doesn't map to zero in $\kappa (\mathfrak q)$ and we win. $\square$

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