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Show that if $Q \notin m$, $Q$ ideal,$m$ maximal ideal then $\phi(Q)R_m=R_m$.

$R_m$ is the localisation. $\phi$ is the homomorphism from $R$ to $R_m$ sending $r$ to $r/1$

This is the actual question: Inducing homomorphisms on localizations of rings/modules

I basically need to know:

If $M$ does not contain $I_k$, we have $(I_k)_M=R_M$ because $I_k∩(R∖M)≠∅$. Why does this reason imply $(I_k)M=R_M$(this is what I'm asking above I think)

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It's probably $Q\not\subset m$. In this case, take $q\in Q$, $q\notin m$; then $q/1$ belongs to $\phi(Q)R_m$ and is invertible.

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    Why does the last bit show $\phi(Q) R_m=R_m$? – 2017-02-14
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    @Bob If an ideal contains an invertible element, it is the whole ring. – 2017-02-14