Lemma 16.3.3. Let $R \to A$ be a syntomic ring map. Then there exists a smooth $R$-algebra map $A \to C$ with a retraction such that $C$ is a global relative complete intersection over $R$, i.e.,

flat over $R$ and all fibres of dimension $n - c$.

Lemma 16.3.3. Let $R \to A$ be a syntomic ring map. Then there exists a smooth $R$-algebra map $A \to C$ with a retraction such that $C$ is a global relative complete intersection over $R$, i.e.,

\[ C \cong R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c) \]

flat over $R$ and all fibres of dimension $n - c$.

**Proof.**
Apply Lemma 16.3.1 to get $A \to C$. By Algebra, Lemma 10.136.6 we can write $C = R[x_1, \ldots , x_ n]/(f_1, \ldots , f_ c)$ with $f_ i$ mapping to a basis of $J/J^2$. The ring map $R \to C$ is syntomic (hence flat) as it is a composition of a syntomic and a smooth ring map. The dimension of the fibres is $n - c$ by Algebra, Lemma 10.135.4 (the fibres are local complete intersections, so the lemma applies).
$\square$

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