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I am reading Eisenbud’s book Commutative Algebra, and have gotten stuck on the proof of Theorem 11.6c. Let $R$ be a ring, $K(R)$ its total quotient ring, and $I,J\subset K(R)$ invertible $R$-modules (i.e. locally free of rank one). Then the claim is that the map $\Psi:I^{-1}J\rightarrow\operatorname{Hom}_R(I,J)$ obtained by $I^{-1}J\ni t\mapsto\varphi_t:I\rightarrow J$, $\varphi_t(a)=ta$, is an isomorphism. I have no trouble believing it is a monomorphism. We know that every invertible fractional ideal of $R$ contains a nonzerodivisor in $R$, so in the proof that $\Psi$ is an epimorphism, Eisenbud selects a nonzerodivisor $v\in I\cap R$ and lets $\varphi\in\operatorname{Hom}_R(I,J)$ be arbitrary. He then puts $w:=\varphi(v)$ and claims $\varphi$ coincides with $\varphi_{w/v}$. I buy the subsequent argument that these two homomorphisms are the same, but I fail to convince myself that $w/v\in I^{-1}J$. Clearly $w\in J$ by hypothesis, but I do not see why $(1/v)I\subset R$; i.e. that $1/v\in I^{-1}$.

My question is, why is $w/v\in I^{-1}J$?

Tomo
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1 Answers1

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$\varphi(v)=w\implies a\varphi(v)=aw,\forall a\in I\implies v\varphi(a)=aw,\forall a\in I\implies a(w/v)\in J,\forall a\in I\implies$ $(w/v)I\subseteq J\implies w/v\in I^{-1}J$.

user26857
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  • Some more words would be helpful. The long code of TeX is also making reading more complicated. – Pedro Dec 20 '15 at 20:08
  • It's true. Though @user26857 has written true statements that show that $(w/v)I\subset J$, I remain a little bit confused. Does $I^{-1}J$ coincide with the $R$-submodule of $K(R)$ generated as a module by elements $ij$ for $i\in I^{-1}, j\in J$, where multiplication is in the ring $K(R)$, or is it the $R$-submodule of $K(R)$ consisting of elements $s\in K(R)$ which take $I$ into $J$; i.e. such that $Is\subset J$? And do these two notions in fact coincide? One inclusion is obvious to me, but the other isn't. – Tomo Dec 20 '15 at 20:44
  • $I$ and $J$ are locally free of rank one hence generated as $R$-modules by nonzerodivisors $s,t\in K(R)$ since we have assumed $R$ local. Given $m\in K(R)$ s.t. $Im\subset J$, we must have that $Im\subset Rt$, hence that $m=pt/s$ for $p\in R$, and now note that $pt\in J$ and $s^{-1}\in I^{-1}$. So I guess this shows that the two notions coincide and I've answered my own question. – Tomo Dec 20 '15 at 20:57