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Let $M$ be a finitely presented module. Show that $M$ is flat if and only if $M$ is projective. I think I am fine with the part "projective" implies flatness but I need help on showing that flatness + finitely presented modules implies projective.

Any help on this part will be great.

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    $\begingroup$ This is Theorem 3.56 in Rotman's Homological Algebra. It uses several preceding technical lemmas and the Bourbaki-Lambek theorem, so I don't know if there is necessarily a quick proof. $\endgroup$
    – Ben West
    Apr 11, 2017 at 2:40

1 Answer 1

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This proof maybe is too long, but anyways I decided to post it. We begin with the following definition: if $B$ is a right $R$-module, the its character module is defined by $B^*=\text{Hom}_{\Bbb{Z}}(B,\Bbb{Q}/\Bbb{Z})$. This is a left $R$-module by defining $(rf)(m)\mapsto f(mr)$ for $r\in R$ and $f\in \text{Hom}_{\Bbb Z}(B,\Bbb{Q}/\Bbb{Z})$.

We have the following lemmas:

Lemma 1: A sequence of right $R$-modules $$A_1\longrightarrow A\longrightarrow A_2$$ is exact if only if the sequence of character modules $$A_2^*\longrightarrow A^*\longrightarrow A_1^*$$ is exact.

Proof: I let you to prove this nice result.

Lemma 2: Let $R, S$ be rings. If $M$ is a finitely presented (f.p.) left $R$-module and $N$ is a $(R,S)$-bimodule, then $$\sigma:N^*\otimes_{R}M\longrightarrow \text{Hom}_{R}(M,N)^*$$ is an isomorphism.

Proof: As $M$ is f.p. then there are $m,n\in \Bbb{N}$ such that $$R^m\longrightarrow R^n\longrightarrow M\longrightarrow 0$$ is an exact sequence of left $R$-modules.

If $M=R$, then as $N^*\otimes R\cong N^*$ and $\text{Hom}_R(R,N)\cong N$ it follows that $\text{Hom}_{R}(R,N)^*\cong N^*\otimes R$. Therefore $$N^*\otimes_{R} R^m=N^*\otimes \bigoplus_{i=1}^m R \cong \bigoplus_{i=1}^m (N^*\otimes_{R} R)\cong \bigoplus_{i=1}^m \text{Hom}_R(R,N)^*=\bigoplus_{i=1}^m\text{Hom}_{\Bbb Z}\Bigl(\text{Hom}_R(R,N),\Bbb Q/\Bbb Z\Bigr)\cong \text{Hom}_{\Bbb Z}\Bigl(\bigoplus_{i=1}^m \text{Hom}_R(R,N),\Bbb Q/\Bbb Z\Bigr)\cong \text{Hom}_{\Bbb Z}(\text{Hom}_R(R^m,N),\Bbb Q/\Bbb Z)=\text{Hom}_R(R^m,N)^*.$$

Now, if we apply $N^*\otimes_R$ to the exact sequence given lines above we get the following commutative diagram $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} N^*\otimes_R R^m & \longrightarrow & N^*\otimes_R R^n & \longrightarrow & N^*\otimes_R M & \longrightarrow & 0 \\ \da{\cong} & & \da{\cong} & & \da{\sigma} \\ \text{Hom}_R(R^m,N)^* & \longrightarrow & \text{Hom}_R(R^n,N)^* & \longrightarrow & \text{Hom}_R(M,N)^* & \longrightarrow & 0 \\ \end{array} $$

Where the exactness of the top row follows by applying $\text{Hom}_R(\_\,,N)$ and $\text{Hom}_R(\_\,,\Bbb Q/\Bbb Z)$, noting that $\Bbb Q/\Bbb Z$ is an injective $R$-module. Finally, by the three lemma we deduce that $\sigma$ is an isomorphism.


Now we'll prove the theorem. Let $$\begin{array}{c} N & \ra{\phi} & N_0\longrightarrow 0 \end{array} $$

be an exact sequence. It's enough to prove that
$$\begin{array}{c} \text{Hom}_R(M,N) & \ra{\phi_*} & \text{Hom}_R(M,N_0)\longrightarrow 0 \end{array} $$ is an exact sequence. Remember that $\phi_*\colon \text{Hom}_R(M,N)\rightarrow \text{Hom}_R(M,N_0)$ is defined by $\phi_*(f)=\phi\circ f$.

By lemma 1 we have that $$0\longrightarrow N_0^*\longrightarrow N^*$$ is exact. Applying $\otimes_R M$ we find the commutative diagram $$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\ras}[1]{\kern-1.5ex\xrightarrow{\ \ \smash{#1}\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} 0 & \longrightarrow & N_0^*\otimes_R M & \longrightarrow & N^*\otimes_R M\\ & & \da{\cong} & & \da{\cong}\\ 0 & \longrightarrow & \text{Hom}_R(M,N_0)^* & \longrightarrow & \text{Hom}_R(M,N)^*\\ \end{array} $$

As $M$ is flat then the top row is exact, and the vertical maps are isomorphisms by lemma 2, so the bottom row is also exact. Thus by lemma 1 we deduce that $$\text{Hom}_R(M,N)\longrightarrow \text{Hom}_R(M,N_0)\longrightarrow 0$$ is an exact sequence. Hence, $M$ is projective.

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  • $\begingroup$ You haven't specified what you mean by the map $\sigma$ in Lemma 2. I suppose $\sigma$ ought to be a map of functors, so that in your later argument "...we get the following commutative diagram..." you actually get a commutative diagram. $\endgroup$
    – Stephen
    Oct 9, 2020 at 12:36
  • $\begingroup$ Presumably $\sigma$ is given by $\phi \otimes m \mapsto [\psi \mapsto \phi(\psi(m))]$ for $\phi \in N^*, m \in M$ and $\psi \in \mathrm{Hom}_R(M,N)$? $\endgroup$
    – Stephen
    Oct 9, 2020 at 12:50

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