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I'm reading the proof here.

I'm at the line where they say $$ \psi\pi(f)=\psi(f+FR)=\varphi(f)+PR.$$

Since $\psi\pi$ is surjective, it should follow that $\{\varphi(f)+PR:f\in F\}=P/PR$. I don't understand the notation $\mathrm{im}(\varphi)+PR=P$. I assume $\mathrm{im}(\varphi)+PR=\{\varphi(f)+PR:f\in F\}$? If that's a subset of $P/PR$, how can it equal $P$?

cdfd
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1 Answers1

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(This should just be a comment, but too many of these questions are left without posted answers...)

The notation $\text{im}(\varphi) + PR$ refers to the sum of the two submodules $\text{im}(\varphi)$ and $PR$ (these are subsets of $P$, not $P/PR$). Explicitly,

$$\text{im}(\varphi) + PR = \{\varphi(f) + \sum_{\text{finite}} pr \mid f \in F, p \in P, r \in R \}$$

The reason that $\text{im}(\varphi) + PR = P$, is the equality $\{\varphi(f)+PR:f\in F\}=P/PR$, which says that $\{\varphi(f)\}$ forms a complete set of coset representatives of $PR$ in $P$, so every element of $P$ is a sum of some $\varphi(f)$ with an element of $PR$.

zcn
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  • I see, thanks. I thought the $+$ was still being used for the coset notation in $P/PR$ like in the previous line, not just a normal sum of modules in $P$. – cdfd Mar 05 '14 at 07:45
  • Yes, overloading the operator $+$ can be confusing at times. Thanks for accepting the answer, and saving this question from the unanswered queue – zcn Mar 05 '14 at 07:48