The Stacks project

Proof. Let us use More on Morphisms, Lemma 37.12.1 to show that $U_ d \to \mathop{\mathrm{Spec}}(\mathbf{Z})$ is smooth. Namely, suppose that $\mathop{\mathrm{Spec}}(A) \to U_ d$ is a morphism and $A' \to A$ is a small extension. Then $B = A \otimes _{A_ d} B_ d$ is a finite free $A$-algebra which is syntomic over $A$ (by construction of $U_ d$). By Smoothing Ring Maps, Proposition 16.3.2 there exists a syntomic ring map $A' \to B'$ such that $B \cong B' \otimes _{A'} A$. Set $e'_1 = 1 \in B'$. For $1 < i \leq d$ choose lifts $e'_ i \in B'$ of the elements $1 \otimes e_ i \in A \otimes _{A_ d} B_ d = B$. Then $e'_1, \ldots , e'_ d$ is a basis for $B'$ over $A'$ (for example see Algebra, Lemma 10.101.1). Thus we can write $e'_ i e'_ j = \sum \alpha _{ij}^ l e'_ l$ for unique elements $\alpha _{ij}^ l \in A'$ which satisfy the relations $\sum _ l \alpha _{ij}^ l \alpha _{lk}^ m = \sum _ l \alpha _{il}^ m \alpha _{jk}^ l$ and $\alpha _{ij}^ k = \alpha _{ji}^ k$ and $\alpha _{i1}^ j - \delta _{ij}$ in $A'$. This determines a morphism $\mathop{\mathrm{Spec}}(A') \to X_ d$ by sending $a_{ij}^ l \in A_ d$ to $\alpha _{ij}^ l \in A'$. This morphism agrees with the given morphism $\mathop{\mathrm{Spec}}(A) \to U_ d$. Since $\mathop{\mathrm{Spec}}(A')$ and $\mathop{\mathrm{Spec}}(A)$ have the same underlying topological space, we see that we obtain the desired lift $\mathop{\mathrm{Spec}}(A') \to U_ d$ and we conclude that $U_ d$ is smooth over $\mathbf{Z}$. $\square$