The Stacks project

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 filtered colimits is a filtered 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 (Definition 16.2.3) 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. $\square$

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 15.41.3. 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 10.127.4.

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 10.25.4 and 10.31.6. 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. $\square$

Comments (7)

Comment #2759 by Anonymous on

In 16.8.3 first paragraph, it should read "... is given by ..."

In 16.8.4 second to last paragraph it should say

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!

Comment #4928 by Shane on

It took me quite a while to track down the definition of an "elementary standard element". Perhaps add a hyperlink?

Comment #6042 by Harry Gindi on

Small nitpick in the hint for 07F3. This is only true for filtered colimits not all colimits (more or less by definition).

Comment #6043 by Harry Gindi on

(It's true more generally for sifted colimits (and since we're looking at products rather than finite limits) in the category of sets, but we're using here that filtered colimits in CRing are computed as the filtered colimit of the underlying sets, which I don't think holds for sifted colimits in CRing).

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