The Stacks project

[Corollary 1, Warfield-Purity]

Lemma 15.124.1. Let $P$ be a module over a ring $R$. The following are equivalent

  1. $P$ is a direct summand of a direct sum of modules of the form $R/fR$, for $f \in R$ varying.

  2. for every short exact sequence $0 \to A \to B \to C \to 0$ of $R$-modules such that $fA = A \cap fB$ for all $f \in R$ the map $\mathop{\mathrm{Hom}}\nolimits _ R(P, B) \to \mathop{\mathrm{Hom}}\nolimits _ R(P, C)$ is surjective.

Proof. Let $0 \to A \to B \to C \to 0$ be an exact sequence as in (2). To prove that (1) implies (2) it suffices to prove that $\mathop{\mathrm{Hom}}\nolimits _ R(R/fR, B) \to \mathop{\mathrm{Hom}}\nolimits _ R(R/fR, C)$ is surjective for every $f \in R$. Let $\psi : R/fR \to C$ be a map. Say $\psi (1)$ is the image of $b \in B$. Then $fb \in A$. Hence there exists an $a \in A$ such that $fa = fb$. Then $f(b - a) = 0$ hence we get a morphism $\varphi : R/fR \to B$ mapping $1$ to $b - a$ which lifts $\psi $.

Conversely, assume that (2) holds. Let $I$ be the set of pairs $(f, \varphi )$ where $f \in R$ and $\varphi : R/fR \to P$. For $i \in I$ denote $(f_ i, \varphi _ i)$ the corresponding pair. Consider the map

\[ B = \bigoplus \nolimits _{i \in I} R/f_ iR \longrightarrow P \]

which sends the element $r$ in the summand $R/f_ iR$ to $\varphi _ i(r)$ in $P$. Let $A = \mathop{\mathrm{Ker}}(B \to P)$. Then we see that (1) is true if the sequence

\[ 0 \to A \to B \to P \to 0 \]

is an exact sequence as in (2). To see this suppose $f \in R$ and $a \in A$ maps to $f b$ in $B$. Write $b = (r_ i)_{i \in I}$ with almost all $r_ i = 0$. Then we see that

\[ f\sum \varphi _ i(r_ i) = 0 \]

in $P$. Hence there is an $i_0 \in I$ such that $f_{i_0} = f$ and $\varphi _{i_0}(1) = \sum \varphi _ i(r_ i)$. Let $x_{i_0} \in R/f_{i_0}R$ be the class of $1$. Then we see that

\[ a' = (r_ i)_{i \in I} - (0, \ldots , 0, x_{i_0}, 0, \ldots ) \]

is an element of $A$ and $fa' = a$ as desired. $\square$

Comments (4)

Comment #7853 by Shota Inoue on

Typo in the proof of the converse implication: should be replaced by .

Comment #7854 by Shota Inoue on

The alphabet already appears in the sentence "To see this suppose ...". So the in the last displayed equation should be replaced by another symbol (say ), and should be (as we want to show the implication ).

Comment #7855 by Shota Inoue on

For the comment #7854, is a typo: it should be .

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