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I am referring to Theorem 6.3 p. 143 from Lang's Algebra (but the question description will be as self-contained as possible).

Let $A$ be a commutative ring and $M$, $W$, $V$, $U$ be $A$-modules. Let the sequence $$0 \longrightarrow W \stackrel{\lambda}{\longrightarrow} V \stackrel{\phi}{\longrightarrow} U \longrightarrow 0 $$ be exact.

By Proposition 2.1 p. 122, the induced sequence $$0\longrightarrow Hom_A(U,M) \stackrel{\phi'}\longrightarrow Hom_A(V,M) \stackrel{\lambda'}\longrightarrow Hom_A(W,M) $$ is exact. If we let $M$ be $A$ as a module over itself, then we obtain $$0 \longrightarrow Hom_A(U,A) \stackrel{\phi'}{\longrightarrow} Hom_A(V,A) \stackrel{\lambda'}{\longrightarrow} Hom_A(W,A)$$ exact or in Lang's notation $$0 \longrightarrow U^{\vee} \stackrel{\phi'}{\longrightarrow} V^{\vee} \stackrel{\lambda'}{\longrightarrow} W^{\vee},$$ where $V^{\vee}$ is the dual module of $V$.

Now, this is where i have a problem: it is mentioned in the proof of theorem 6.3 that since $A$ is projective (because it is free), then we also have exactness from the right, i.e. $$0 \longrightarrow U^{\vee} \stackrel{\phi'}{\longrightarrow} V^{\vee} \stackrel{\lambda'}{\longrightarrow} W^{\vee} \longrightarrow 0$$ is exact. I don't see where this exactness from the right comes from since $A$ being projective means that the functor $Hom_A(A,\cdot)$ is exact, while to obtain dual spaces we use the functor $Hom_A(\cdot,A)$.

Any insights? Thank you:-)

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You are missing that Lang assumes that the $A$-modules $W,V,U$ are finite free. The exactness of the dual sequence is then immediate.

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  • $\begingroup$ I am studying the same problem after several months, i got stuck exactly at the same point and now i can not see why the exactness is immediate. Could you please give me a hint? Thanks. $\endgroup$ – Manos Dec 10 '11 at 18:31

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