Lemma 93.17.3. Let $k$ be a field. Let $A$ be a global complete interesection over $k$. There exists a flat finite type ring map $k[[t]] \to B$ with $B/tB \cong A$ such that $B[1/t]$ is smooth over $k((t))$.
Proof. Write $A = k[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ as in Algebra, Definition 10.135.1. We are going to choose $a_{ij} \in (t) \subset k[[t]]$ and set
\[ g_ j = f_ j + a_{0j} + a_{1j}x_1 + \ldots + a_{nj}x_ n \in k[[t]][x_1, \ldots , x_ n] \]
After doing this we take $B = k[[t]][x_1, \ldots , x_ n]/(g_1, \ldots , g_ c)$. We claim that $k[[t]] \to B$ is flat at every prime ideal lying over $(t)$. Namely, the elements $f_1, \ldots , f_ c$ form a regular sequence in the local ring at any prime ideal $\mathfrak p$ of $k[x_1, \ldots , x_ n]$ containing $f_1, \ldots , f_ c$ (Algebra, Lemma 10.135.4). Thus $g_1, \ldots , g_ c$ is locally a lift of a regular sequence and we can apply Algebra, Lemma 10.99.3. Flatness at primes lying over $(0) \subset k[[t]]$ is automatic because $k((t)) = k[[t]]_{(0)}$ is a field. Thus $B$ is flat over $k[[t]]$.
All that remains is to show that for suitable choices of $a_{ij}$ the generic fibre $B_{(0)}$ is smooth over $k((t))$. For this we have to show that we can choose our $a_{ij}$ so that the induced morphism
\[ (a_{ij}) : \mathop{\mathrm{Spec}}(k[[t]]) \longrightarrow \mathbf{A}^{c(n + 1)}_ k \]
maps into the nonempty Zariski open of Lemma 93.17.2. This is clear because there is no nonzero polynomial in the $a_{ij}$ which vanishes on $(t)^{\oplus c(n + 1)}$. (We leave this as an exercise to the reader.) $\square$
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Your email address will not be published.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)