Lemma 93.17.3. Let $k$ be a field. Let $A$ be a global complete intersection 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.) $\square$
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