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Tag 07F1

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.

Situation 16.8.1. Here $R \to \Lambda$ is a regular ring map of Noetherian rings.

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 15.38.3. 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. $\square$

    The code snippet corresponding to this tag is a part of the file smoothing.tex and is located in lines 1892–2054 (see updates for more information).

    \section{Warmup: reduction to a base field}
    \label{section-reduction}
    
    \noindent
    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 \ref{lemma-reduce-to-field}.
    
    \begin{situation}
    \label{situation-global}
    Here $R \to \Lambda$ is a regular ring map of Noetherian rings.
    \end{situation}
    
    \noindent
    Let $R \to \Lambda$ be as in Situation \ref{situation-global}.
    We say {\it PT holds for $R \to \Lambda$} if $\Lambda$ is a
    filtered colimit of smooth $R$-algebras.
    
    \begin{lemma}
    \label{lemma-product}
    Let $R_i \to \Lambda_i$, $i = 1, 2$ be as in Situation \ref{situation-global}.
    If PT holds for $R_i \to \Lambda_i$, $i = 1, 2$, then PT holds for
    $R_1 \times R_2 \to \Lambda_1 \times \Lambda_2$.
    \end{lemma}
    
    \begin{proof}
    Omitted. Hint: A product of colimits is a colimit.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-delocalize-base}
    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 $R$.
    \end{lemma}
    
    \begin{proof}
    We first apply Lemma \ref{lemma-smooth-standard-smooth} 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 \ref{lemma-standard-smooth-include-generators}.
    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.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-reduce-to-field}
    If for every Situation \ref{situation-global} where $R$
    is a field PT holds, then PT holds in general.
    \end{lemma}
    
    \begin{proof}
    Assume PT holds for any Situation \ref{situation-global} where $R$ is a field.
    Let $R \to \Lambda$ be as in Situation \ref{situation-global} arbitrary.
    Note that $R/I \to \Lambda/I\Lambda$ is another regular ring map
    of Noetherian rings, see
    More on Algebra, Lemma \ref{more-algebra-lemma-regular-base-change}.
    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 \ref{proposition-lift}
    that $R$ is a reduced ring.
    
    \medskip\noindent
    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 \ref{algebra-lemma-when-colimit}.
    
    \medskip\noindent
    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 \ref{algebra-lemma-total-ring-fractions-no-embedded-points} and
    \ref{algebra-lemma-Noetherian-irreducible-components}.
    By Lemma \ref{lemma-product} 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 $S^{-1}R$.
    
    \medskip\noindent
    We apply Lemma \ref{lemma-delocalize-base}
    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 \ref{lemma-lifting}
    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 \ref{lemma-desingularize-strictly-standard}
    we can find $A \to B \to \Lambda$ with $B$ smooth over $R$
    which finishes the proof.
    \end{proof}

    Comments (2)

    Comment #2759 by Anonymous on August 4, 2017 a 8:47 am UTC

    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 Johan (site) on October 4, 2017 a 6:13 pm UTC

    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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