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
\[ \xymatrix{ A[x_1, \ldots , x_ n, y_1, \ldots , y_ m] \ar[d]^{x_ i \mapsto g_ i} \ar[rr]_-{y_ j \mapsto f_ j} & & A[x_1, \ldots , x_ n] \ar[d] \\ A[y_1, \ldots , y_ m] \ar[rr] & & B } \]
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
\[ \dim J/J^2 \otimes _ B \kappa (\mathfrak q) + n = \dim I/I^2 \otimes _ B \kappa (\mathfrak q) + m. \]
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
\[ y_1 - f_1, \ldots , y_ m - f_ m, p_1, \ldots , p_ t \]
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
\[ x_1 - g_1, \ldots , x_ n - g_ n, q_1, \ldots , q_ s \]
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
\[ A[x_ i, y_ j, 1/hh']/(x_1 - g_1, \ldots , x_ n - g_ n) \cong A[y_ j, 1/h''] \]
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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)