Lemma 16.2.8. Let $R \to A \to \Lambda$ be ring maps with $A$ of finite presentation over $R$. Assume that $H_{A/R} \Lambda = \Lambda$. Then there exists a factorization $A \to B \to \Lambda$ with $B$ smooth over $R$.

Proof. Choose $f_1, \ldots , f_ r \in H_{A/R}$ and $\lambda _1, \ldots , \lambda _ r \in \Lambda$ such that $\sum f_ i\lambda _ i = 1$ in $\Lambda$. Set $B = A[x_1, \ldots , x_ r]/(f_1x_1 + \ldots + f_ rx_ r - 1)$ and define $B \to \Lambda$ by mapping $x_ i$ to $\lambda _ i$. To check that $B$ is smooth over $R$ use that $A_{f_ i}$ is smooth over $R$ by definition of $H_{A/R}$ and that $B_{f_ i}$ is smooth over $A_{f_ i}$. Details omitted. $\square$

Comment #6045 by Harry Gindi on

I found this last one a bit tricky. I wrote up a proof if you want to add it:

Observe that the ideal generated by the coefficients (f_1,...,f_r) c B is the unit ideal, so the family of maps {B→B_{f_i}}{i=1}^r is a Zariski open cover. It suffices therefore to show that each of the B{f_i} is smooth over R. But observe that B_{f_i}≅A_{f_i}?91?x1,...,x_{i-1},x_{i+1},...,x_r] (omitting the ith indeterminate) for each 1≤i≤r, which is witnessed by the map sending x_i to the polynomial

1- ((f_1/f_i) x_1 + … + (f_{i-1}/f_i) x_{i-1} + (f_{i+1}/f_i) x_{i+1} + … + (f_r/f_i) x_r)

in A?91?x_1,...,x_{i-1},x_{i+1},…,x_r].

But since f_i belongs to H_{A/R}, we know that R→A_{f_i} is smooth, and polynomial rings in finitely many indeterminates are smooth so we see A_{f_i}→B_{f_i} is smooth and therefore R→A_{f_i} → B_{f_i} is smooth, as desired.

Comment #6184 by on

Thanks. I added a bit more text to the hint. I was too lazy to make a full proof with the relevant results on smooth ring maps in the algebra chapter. See changes here.

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