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Tag 00QO

Chapter 10: Commutative Algebra > Section 10.126: Colimits and maps of finite presentation

Lemma 10.126.3. Let $\varphi : R \to S$ be a ring map. The following are equivalent

  1. $\varphi$ is of finite presentation,
  2. for every directed system $A_\lambda$ of $R$-algebras the map $$ \mathop{\rm colim}\nolimits_\lambda \mathop{\rm Hom}\nolimits_R(S, A_\lambda) \longrightarrow \mathop{\rm Hom}\nolimits_R(S, \mathop{\rm colim}\nolimits_\lambda A_\lambda) $$ is bijective, and
  3. for every directed system $A_\lambda$ of $R$-algebras the map $$ \mathop{\rm colim}\nolimits_\lambda \mathop{\rm Hom}\nolimits_R(S, A_\lambda) \longrightarrow \mathop{\rm Hom}\nolimits_R(S, \mathop{\rm colim}\nolimits_\lambda A_\lambda) $$ is surjective.

Proof. Assume (1) and write $S = R[x_1, \ldots, x_n] / (f_1, \ldots, f_m)$. Let $A = \mathop{\rm colim}\nolimits A_\lambda$. Observe that an $R$-algebra homomorphism $S \to A$ or $S \to A_\lambda$ is determined by the images of $x_1, \ldots, x_n$. Hence it is clear that $\mathop{\rm colim}\nolimits_\lambda \mathop{\rm Hom}\nolimits_R(S, A_\lambda) \to \mathop{\rm Hom}\nolimits_R(S, A)$ is injective. To see that it is surjective, let $\chi : S \to A$ be an $R$-algebra homomorphism. Then each $x_i$ maps to some element in the image of some $A_{\lambda_i}$. We may pick $\mu \geq \lambda_i$, $i = 1, \ldots, n$ and assume $\chi(x_i)$ is the image of $y_i \in A_\mu$ for $i = 1, \ldots, n$. Consider $z_j = f_j(y_1, \ldots, y_n) \in A_\mu$. Since $\chi$ is a homomorphism the image of $z_j$ in $A = \mathop{\rm colim}\nolimits_\lambda A_\lambda$ is zero. Hence there exists a $\mu_j \geq \mu$ such that $z_j$ maps to zero in $A_{\mu_j}$. Pick $\nu \geq \mu_j$, $j = 1, \ldots, m$. Then the images of $z_1, \ldots, z_m$ are zero in $A_\nu$. This exactly means that the $y_i$ map to elements $y'_i \in A_\nu$ which satisfy the relations $f_j(y'_1, \ldots, y'_n) = 0$. Thus we obtain a ring map $S \to A_\nu$. This shows that (1) implies (2).

It is clear that (2) implies (3). Assume (3). By Lemma 10.126.2 we may write $S = \mathop{\rm colim}\nolimits_\lambda S_\lambda$ with $S_\lambda$ of finite presentation over $R$. Then the identity map factors as $$ S \to S_\lambda \to S $$ for some $\lambda$. This implies that $S$ is finitely presented over $S_\lambda$ by Lemma 10.6.2 part (4) applied to $S \to S_\lambda \to S$. Applying part (2) of the same lemma to $R \to S_\lambda \to S$ we conclude that $S$ is of finite presentation over $R$. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 30531–30551 (see updates for more information).

    \begin{lemma}
    \label{lemma-characterize-finite-presentation}
    Let $\varphi : R \to S$ be a ring map. The following are equivalent
    \begin{enumerate}
    \item $\varphi$ is of finite presentation,
    \item for every directed system $A_\lambda$ of $R$-algebras
    the map
    $$
    \colim_\lambda \Hom_R(S, A_\lambda) \longrightarrow
    \Hom_R(S, \colim_\lambda A_\lambda)
    $$
    is bijective, and
    \item for every directed system $A_\lambda$ of $R$-algebras
    the map
    $$
    \colim_\lambda \Hom_R(S, A_\lambda) \longrightarrow
    \Hom_R(S, \colim_\lambda A_\lambda)
    $$
    is surjective.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Assume (1) and write $S = R[x_1, \ldots, x_n] / (f_1, \ldots, f_m)$.
    Let $A = \colim A_\lambda$. Observe that an $R$-algebra homomorphism
    $S \to A$ or $S \to A_\lambda$ is determined by the images of
    $x_1, \ldots, x_n$. Hence it is clear that
    $\colim_\lambda \Hom_R(S, A_\lambda) \to \Hom_R(S, A)$
    is injective. To see that it is surjective, let $\chi : S \to A$
    be an $R$-algebra homomorphism. Then each
    $x_i$ maps to some element in the image of some $A_{\lambda_i}$.
    We may pick $\mu \geq \lambda_i$, $i = 1, \ldots, n$ and
    assume $\chi(x_i)$ is the image of $y_i \in A_\mu$ for
    $i = 1, \ldots, n$. Consider $z_j = f_j(y_1, \ldots, y_n) \in A_\mu$.
    Since $\chi$ is a homomorphism the image of $z_j$ in
    $A = \colim_\lambda A_\lambda$ is zero. Hence there exists a
    $\mu_j \geq \mu$ such that $z_j$ maps to zero in $A_{\mu_j}$.
    Pick $\nu \geq \mu_j$, $j = 1, \ldots, m$. Then the
    images of $z_1, \ldots, z_m$ are zero in $A_\nu$. This
    exactly means that the $y_i$ map to elements
    $y'_i \in A_\nu$ which satisfy the relations $f_j(y'_1, \ldots, y'_n) = 0$.
    Thus we obtain a ring map $S \to A_\nu$. This shows that
    (1) implies (2).
    
    \medskip\noindent
    It is clear that (2) implies (3). Assume (3).
    By Lemma \ref{lemma-ring-colimit-fp} we may write
    $S = \colim_\lambda S_\lambda$ with $S_\lambda$
    of finite presentation over $R$. Then the identity map
    factors as
    $$
    S \to S_\lambda \to S
    $$
    for some $\lambda$. This implies that $S$
    is finitely presented over $S_\lambda$ by
    Lemma \ref{lemma-compose-finite-type} part (4)
    applied to $S \to S_\lambda \to S$. Applying part (2) of the same
    lemma to $R \to S_\lambda \to S$ we conclude that $S$ is of finite
    presentation over $R$.
    \end{proof}

    Comments (2)

    Comment #1096 by Antoine Chambert-Loir on October 18, 2014 a 1:21 pm UTC

    It is not true that a factorization $S\to S_\lambda\to S$ can only exist if $S=S_\lambda$. However, it follows from 00F4 (4) that this factorization implies that $S$ is finitely presented.

    Comment #1130 by Johan (site) on November 4, 2014 a 6:26 pm UTC

    Oops, yes, that is a mistake. Actually, this also simplifies the thing a bit (hopefully correct this time). Here is the fix.

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