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Tag 0BC2

15.3. Stably free modules

Here is what seems to be the generally accepted definition.

Definition 15.3.1. Let $R$ be a ring.

  1. Two modules $M$, $N$ over $R$ are said to be stably isomorphic if there exist $n, m \geq 0$ such that $M \oplus R^{\oplus m} \cong N \oplus R^{\oplus n}$ as $R$-modules.
  2. A module $M$ is stably free if it is stably isomorphic to a free module.

Observe that a stably free module is projective.

Lemma 15.3.2. Let $R$ be a ring. Let $0 \to P' \to P \to P'' \to 0$ be a short exact sequence of finite projective $R$-modules. If $2$ out of $3$ of these modules are stably free, then so is the third.

Proof. Since the modules are projective, the sequence is split. Thus we can choose an isomorphism $P = P' \oplus P''$. If $P' \oplus R^{\oplus n}$ and $P'' \oplus R^{\oplus m}$ are free, then we see that $P \oplus R^{\oplus n + m}$ is free. Suppose that $P'$ and $P$ are stably free, say $P \oplus R^{\oplus n}$ is free and $P' \oplus R^{\oplus m}$ is free. Then $$ P'' \oplus (P' \oplus R^{\oplus m}) \oplus R^{\oplus n} = (P'' \oplus P') \oplus R^{\oplus m} \oplus R^{\oplus n} = (P \oplus R^{\oplus n}) \oplus R^{\oplus m} $$ is free. Thus $P''$ is stably free. By symmetry we get the last of the three cases. $\square$

Lemma 15.3.3. Let $R$ be a ring. Let $I \subset R$ be an ideal. Assume that every element of $1 + I$ is a unit (in other words $I$ is contained in the radical of $R$). For every finite stably free $R/I$-module $E$ there exists a finite stably free $R$-module $M$ such that $M/IM \cong E$.

Proof. Choose a $n$ and $m$ and an isomorphism $E \oplus (R/I)^{\oplus n} \cong (R/I)^{\oplus m}$. Choose $R$-linear maps $\varphi : R^{\oplus m} \to R^{\oplus n}$ and $\psi : R^{\oplus n} \to R^{\oplus m}$ lifting the projection $(R/I)^{\oplus m} \to (R/I)^{\oplus n}$ and injection $(R/I)^{\oplus n} \to (R/I)^{\oplus m}$. Then $\varphi \circ \psi : R^{\oplus n} \to R^{\oplus n}$ reduces to the identity modulo $I$. Thus the determinant of this map is invertible by our assumption on $I$. Hence $P = \mathop{\rm Ker}(\varphi)$ is stably free and lifts $E$. $\square$

Lemma 15.3.4. Let $R$ be a ring. Let $I \subset R$ be an ideal. Assume that every element of $1 + I$ is a unit (in other words $I$ is contained in the radical of $R$). Let $M$ be a finite flat $R$-module such that $M/IM$ is a projective $R/I$-module. Then $M$ is a finite projective $R$-module.

Proof. By Algebra, Lemma 10.77.4 we see that $M_\mathfrak p$ is finite free for all prime ideals $\mathfrak p \subset R$. By Algebra, Lemma 10.77.2 it suffices to show that the function $\rho_M : \mathfrak p \mapsto \dim_{\kappa(\mathfrak p)} M \otimes_R \kappa(\mathfrak p)$ is locally constant on $\mathop{\rm Spec}(R)$. Because $M/IM$ is finite projective, this is true on $V(I) \subset \mathop{\rm Spec}(R)$. Since every closed point of $\mathop{\rm Spec}(R)$ is in $V(I)$ and since $\rho_M(\mathfrak p) = \rho_M(\mathfrak q)$ whenever $\mathfrak p \subset \mathfrak q \subset R$ are prime ideals, we conclude by an elementary argument on topological spaces which we omit. $\square$

The lift of Lemma 15.3.3 is unique up to isomorphism by the following lemma.

Lemma 15.3.5. Let $R$ be a ring. Let $I \subset R$ be an ideal. Assume that every element of $1 + I$ is a unit (in other words $I$ is contained in the radical of $R$). If $P$ and $P'$ are finite projective $R$-modules, then

  1. if $\varphi : P \to P'$ is an $R$-module map inducing an isomorphism $\overline{\varphi} : P/IP \to P'/IP'$, then $\varphi$ is an isomorphism,
  2. if $P/IP \cong P'/IP'$, then $P \cong P'$.

Proof. Proof of (1). As $P'$ is projective as an $R$-module we may choose a lift $\psi : P' \to P$ of the map $P' \to P'/IP' \xrightarrow{\overline{\varphi}^{-1}} P/IP$. By Nakayama's lemma (Algebra, Lemma 10.19.1) $\psi \circ \varphi$ and $\varphi \circ \psi$ are surjective. Hence these maps are isomorphisms (Algebra, Lemma 10.15.4). Thus $\varphi$ is an isomorphism.

Proof of (2). Choose an isomorphism $P/IP \cong P'/IP'$. Since $P$ is projective we can choose a lift $\varphi : P \to P'$ of the map $P \to P/IP \to P'/IP'$. Then $\varphi$ is an isomorphism by (1). $\square$

    The code snippet corresponding to this tag is a part of the file more-algebra.tex and is located in lines 47–175 (see updates for more information).

    \section{Stably free modules}
    \label{section-stably-free}
    
    \noindent
    Here is what seems to be the generally accepted definition.
    
    \begin{definition}
    \label{definition-stably-free}
    Let $R$ be a ring. 
    \begin{enumerate}
    \item Two modules $M$, $N$ over $R$ are said to be
    {\it stably isomorphic} if there exist $n, m \geq 0$ such
    that $M \oplus R^{\oplus m} \cong N \oplus R^{\oplus n}$
    as $R$-modules.
    \item A module $M$ is {\it stably free} if it is stably isomorphic
    to a free module.
    \end{enumerate}
    \end{definition}
    
    \noindent
    Observe that a stably free module is projective.
    
    \begin{lemma}
    \label{lemma-exact-category-stably-free}
    Let $R$ be a ring. Let $0 \to P' \to P \to P'' \to 0$ be a short
    exact sequence of finite projective $R$-modules. If $2$ out of $3$
    of these modules are stably free, then so is the third.
    \end{lemma}
    
    \begin{proof}
    Since the modules are projective, the sequence is split. Thus we can
    choose an isomorphism $P = P' \oplus P''$. If $P' \oplus R^{\oplus n}$
    and $P'' \oplus R^{\oplus m}$ are free, then we see that
    $P \oplus R^{\oplus n + m}$ is free. Suppose that $P'$ and $P$ are
    stably free, say $P \oplus R^{\oplus n}$ is free and $P' \oplus R^{\oplus m}$
    is free. Then
    $$
    P'' \oplus (P' \oplus R^{\oplus m}) \oplus R^{\oplus n} =
    (P'' \oplus P') \oplus R^{\oplus m} \oplus R^{\oplus n} =
    (P \oplus R^{\oplus n}) \oplus R^{\oplus m}
    $$
    is free. Thus $P''$ is stably free. By symmetry we get the last of the
    three cases.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-lift-stably-free}
    Let $R$ be a ring. Let $I \subset R$ be an ideal. Assume that
    every element of $1 + I$ is a unit (in other words $I$ is contained
    in the radical of $R$). For every finite stably free $R/I$-module $E$
    there exists a finite stably free $R$-module $M$ such that $M/IM \cong E$.
    \end{lemma}
    
    \begin{proof}
    Choose a $n$ and $m$ and an isomorphism
    $E \oplus (R/I)^{\oplus n} \cong (R/I)^{\oplus m}$.
    Choose $R$-linear maps $\varphi : R^{\oplus m} \to R^{\oplus n}$
    and $\psi : R^{\oplus n} \to R^{\oplus m}$ lifting the
    projection $(R/I)^{\oplus m} \to (R/I)^{\oplus n}$
    and injection $(R/I)^{\oplus n} \to (R/I)^{\oplus m}$.
    Then $\varphi \circ \psi : R^{\oplus n} \to R^{\oplus n}$
    reduces to the identity modulo $I$. Thus the determinant of
    this map is invertible by our assumption on $I$. Hence
    $P = \Ker(\varphi)$ is stably free and lifts $E$.
    \end{proof}
    
    \begin{lemma}
    \label{lemma-lift-projective}
    Let $R$ be a ring. Let $I \subset R$ be an ideal.
    Assume that every element of $1 + I$ is a unit
    (in other words $I$ is contained in the radical of $R$).
    Let $M$ be a finite flat $R$-module such that
    $M/IM$ is a projective $R/I$-module.
    Then $M$ is a finite projective $R$-module.
    \end{lemma}
    
    \begin{proof}
    By Algebra, Lemma \ref{algebra-lemma-finite-flat-local}
    we see that $M_\mathfrak p$ is finite free for all prime ideals
    $\mathfrak p \subset R$.
    By
    Algebra, Lemma \ref{algebra-lemma-finite-projective}
    it suffices to show that the function $\rho_M : \mathfrak p \mapsto
    \dim_{\kappa(\mathfrak p)} M \otimes_R \kappa(\mathfrak p)$
    is locally constant on $\Spec(R)$. Because $M/IM$ is finite projective, this
    is true on $V(I) \subset \Spec(R)$. Since every closed point
    of $\Spec(R)$ is in $V(I)$ and since
    $\rho_M(\mathfrak p) = \rho_M(\mathfrak q)$
    whenever $\mathfrak p \subset \mathfrak q \subset R$
    are prime ideals, we conclude by
    an elementary argument on topological spaces which we omit.
    \end{proof}
    
    \noindent
    The lift of Lemma \ref{lemma-lift-stably-free}
    is unique up to isomorphism by the following lemma.
    
    \begin{lemma}
    \label{lemma-isomorphic-finite-projective-lifts}
    Let $R$ be a ring. Let $I \subset R$ be an ideal. Assume that
    every element of $1 + I$ is a unit (in other words $I$ is contained
    in the radical of $R$). If $P$ and $P'$ are finite
    projective $R$-modules, then
    \begin{enumerate}
    \item if $\varphi : P \to P'$ is an $R$-module map inducing an
    isomorphism $\overline{\varphi} : P/IP \to P'/IP'$, then $\varphi$
    is an isomorphism,
    \item if $P/IP \cong P'/IP'$, then $P \cong P'$.
    \end{enumerate}
    \end{lemma}
    
    \begin{proof}
    Proof of (1). As $P'$ is projective as an $R$-module we may
    choose a lift $\psi : P' \to P$ of the map
    $P' \to P'/IP' \xrightarrow{\overline{\varphi}^{-1}} P/IP$.
    By Nakayama's lemma (Algebra, Lemma \ref{algebra-lemma-NAK})
    $\psi \circ \varphi$ and $\varphi \circ \psi$ are surjective.
    Hence these maps are isomorphisms (Algebra, Lemma \ref{algebra-lemma-fun}).
    Thus $\varphi$ is an isomorphism.
    
    \medskip\noindent
    Proof of (2). Choose an isomorphism $P/IP \cong P'/IP'$.
    Since $P$ is projective we can choose a lift $\varphi : P \to P'$ of the map
    $P \to P/IP \to P'/IP'$. Then $\varphi$ is an isomorphism by (1).
    \end{proof}

    Comments (2)

    Comment #2442 by Peng Du on March 1, 2017 a 3:47 pm UTC

    In Definition 15.3.1. (1), "if there exist n,m≥0" should be "if there exist m≥0" as there is no 'n'.

    Comment #2485 by Johan (site) on April 13, 2017 a 10:41 pm UTC

    OK, the mistake was that one of the direct sums should be $R^{\oplus n}$ and the other should be $R^{\oplus m}$. Thanks for pointing this out. The fix is here.

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