16.8 Warmup: reduction to a base field

In this section we apply the lemmas in the previous sections to prove that it suffices to prove the main result when the base ring is a field, see Lemma 16.8.4.

Let $R \to \Lambda$ be as in Situation 16.8.1. We say PT holds for $R \to \Lambda$ if $\Lambda$ is a filtered colimit of smooth $R$-algebras.

Lemma 16.8.2. Let $R_ i \to \Lambda _ i$, $i = 1, 2$ be as in Situation 16.8.1. If PT holds for $R_ i \to \Lambda _ i$, $i = 1, 2$, then PT holds for $R_1 \times R_2 \to \Lambda _1 \times \Lambda _2$.

Proof. Omitted. Hint: A product of colimits is a colimit. $\square$

Lemma 16.8.3. Let $R \to A \to \Lambda$ be ring maps with $A$ of finite presentation over $R$. Let $S \subset R$ be a multiplicative set. Let $S^{-1}A \to B' \to S^{-1}\Lambda$ be a factorization with $B'$ smooth over $S^{-1}R$. Then we can find a factorization $A \to B \to \Lambda$ such that some $s \in S$ maps to an elementary standard element in $B$ over $R$.

Proof. We first apply Lemma 16.3.4 to $S^{-1}R \to B'$. Thus we may assume $B'$ is standard smooth over $S^{-1}R$. Write $A = R[x_1, \ldots , x_ n]/(g_1, \ldots , g_ t)$ and say $x_ i \mapsto \lambda _ i$ in $\Lambda$. We may write $B' = S^{-1}R[x_1, \ldots , x_{n + m}]/(f_1, \ldots , f_ c)$ for some $c \geq n$ where $\det (\partial f_ j/\partial x_ i)_{i, j = 1, \ldots , c}$ is invertible in $B'$ and such that $A \to B'$ is given by $x_ i \mapsto x_ i$, see Lemma 16.3.6. After multiplying $x_ i$, $i > n$ by an element of $S$ and correspondingly modifying the equations $f_ j$ we may assume $B' \to S^{-1}\Lambda$ maps $x_ i$ to $\lambda _ i/1$ for some $\lambda _ i \in \Lambda$ for $i > n$. Choose a relation

$1 = a_0 \det (\partial f_ j/\partial x_ i)_{i, j = 1, \ldots , c} + \sum \nolimits _{j = 1, \ldots , c} a_ jf_ j$

for some $a_ j \in S^{-1}R[x_1, \ldots , x_{n + m}]$. Since each element of $S$ is invertible in $B'$ we may (by clearing denominators) assume that $f_ j, a_ j \in R[x_1, \ldots , x_{n + m}]$ and that

$s_0 = a_0 \det (\partial f_ j/\partial x_ i)_{i, j = 1, \ldots , c} + \sum \nolimits _{j = 1, \ldots , c} a_ jf_ j$

for some $s_0 \in S$. Since $g_ j$ maps to zero in $S^{-1}R[x_1, \ldots , x_{n + m}]/(f_1, \ldots , x_ c)$ we can find elements $s_ j \in S$ such that $s_ j g_ j = 0$ in $R[x_1, \ldots , x_{n + m}]/(f_1, \ldots , f_ c)$. Since $f_ j$ maps to zero in $S^{-1}\Lambda$ we can find $s'_ j \in S$ such that $s'_ j f_ j(\lambda _1, \ldots , \lambda _{n + m}) = 0$ in $\Lambda$. Consider the ring

$B = R[x_1, \ldots , x_{n + m}]/ (s'_1f_1, \ldots , s'_ cf_ c, g_1, \ldots , g_ t)$

and the factorization $A \to B \to \Lambda$ with $B \to \Lambda$ given by $x_ i \mapsto \lambda _ i$. We claim that $s = s_0s_1 \ldots s_ ts'_1 \ldots s'_ c$ is elementary standard in $B$ over $R$ which finishes the proof. Namely, $s_ j g_ j \in (f_1, \ldots , f_ c)$ and hence $sg_ j \in (s'_1f_1, \ldots , s'_ cf_ c)$. Finally, we have

$a_0\det (\partial s'_ jf_ j/\partial x_ i)_{i, j = 1, \ldots , c} + \sum \nolimits _{j = 1, \ldots , c} (s'_1 \ldots \hat{s'_ j} \ldots s'_ c) a_ j s'_ jf_ j = s_0s'_1\ldots s'_ c$

which divides $s$ as desired. $\square$

Lemma 16.8.4. If for every Situation 16.8.1 where $R$ is a field PT holds, then PT holds in general.

Proof. Assume PT holds for any Situation 16.8.1 where $R$ is a field. Let $R \to \Lambda$ be as in Situation 16.8.1 arbitrary. Note that $R/I \to \Lambda /I\Lambda$ is another regular ring map of Noetherian rings, see More on Algebra, Lemma 15.40.3. Consider the set of ideals

$\mathcal{I} = \{ I \subset R \mid R/I \to \Lambda /I\Lambda \text{ does not have PT}\}$

We have to show that $\mathcal{I}$ is empty. If this set is nonempty, then it contains a maximal element because $R$ is Noetherian. Replacing $R$ by $R/I$ and $\Lambda$ by $\Lambda /I$ we obtain a situation where PT holds for $R/I \to \Lambda /I\Lambda$ for any nonzero ideal of $R$. In particular, we see by applying Proposition 16.5.3 that $R$ is a reduced ring.

Let $A \to \Lambda$ be an $R$-algebra homomorphism with $A$ of finite presentation. We have to find a factorization $A \to B \to \Lambda$ with $B$ smooth over $R$, see Algebra, Lemma 10.126.4.

Let $S \subset R$ be the set of nonzerodivisors and consider the total ring of fractions $Q = S^{-1}R$ of $R$. We know that $Q = K_1 \times \ldots \times K_ n$ is a product of fields, see Algebra, Lemmas 10.24.4 and 10.30.6. By Lemma 16.8.2 and our assumption PT holds for the ring map $S^{-1}R \to S^{-1}\Lambda$. Hence we can find a factorization $S^{-1}A \to B' \to S^{-1}\Lambda$ with $B'$ smooth over $S^{-1}R$.

We apply Lemma 16.8.3 and find a factorization $A \to B \to \Lambda$ such that some $\pi \in S$ is elementary standard in $B$ over $R$. After replacing $A$ by $B$ we may assume that $\pi$ is elementary standard, hence strictly standard in $A$. We know that $R/\pi ^8R \to \Lambda /\pi ^8\Lambda$ satisfies PT. Hence we can find a factorization $R/\pi ^8 R \to A/\pi ^8A \to \bar C \to \Lambda /\pi ^8\Lambda$ with $R/\pi ^8 R \to \bar C$ smooth. By Lemma 16.6.1 we can find an $R$-algebra map $D \to \Lambda$ with $D$ smooth over $R$ and a factorization $R/\pi ^4 R \to A/\pi ^4A \to D/\pi ^4D \to \Lambda /\pi ^4\Lambda$. By Lemma 16.7.2 we can find $A \to B \to \Lambda$ with $B$ smooth over $R$ which finishes the proof. $\square$

Comment #2759 by Anonymous on

In 16.8.3 first paragraph, it should read "... $A_S \rightarrow B'$ is given by ..."

In 16.8.4 second to last paragraph it should say $A_S \rightarrow B' \rightarrow \Lambda_S$

Comment #2867 by on

Did not understand your first comment (I think what it says is correct). The second comment I do understand and I agree and I fixed it here. Thanks!

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