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Sorry for asking this simple things... First I will put the results that are used. I have a stupid question about the proof of the corollary.enter image description here

I don´t know totally how to use the proposition 1.1, because I can guarantee that there exist a maximal ideal of $ A/a $ of the form $ a+J $ using the hint. And now? clearly it´s natural in consider the preimage of the projection $\pi^{-1}(a+J)$ since it is an ideal of A , but not necessarily a maximal ideal. So what can I do?

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    "[$\ldots$] there exist[s]$a$maximal ideal of $A/a$ of the form $a+J$ [$\ldots$]". Ehm... I think you're using "$a$" here two mean two different things. Notice how the book you're quoting is using `\mathfrak` for ideals.2012-01-16

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Suppose $J$ is a maximal ideal of $A/\mathfrak{a}$. Then $\phi^{-1}(J)$ is an ideal of $A$, and if $I$ is any ideal of $A$ with $\phi^{-1}(J)\subseteq I\subseteq A$, then applying $\phi$, we get $J\subseteq \phi(I)\subseteq A/\mathfrak{a}$ (this is because the correspondence is order-preserving, and $\phi(\phi^{-1}(J))=J$ because, as a quotient map, $\phi$ is surjective). Because $J$ is a maximal ideal of $A/\mathfrak{a}$, this implies that either $\phi(I)=J$ or $\phi(I)=A/\mathfrak{a}$. Thus, applying $\phi^{-1}$, we have that either $I=\phi^{-1}(J)$ or $I=A$. Thus, $\phi^{-1}(J)$ is a maximal ideal of $A$ containing $\phi^{-1}(0)=\mathfrak{a}$.

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    the zero ideal of $\mathbb{Z}$. I don't know if you'd like this better than your example of $\mathbb{Z}\rightarrow\mathbb{Q}$ though.2012-01-17
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There is some maximal ideal of $A/\mathfrak{a}$. By (1.1) it has a preimage \mathfrak{a}' in $A$ which contains $\mathfrak{a}$, which must be maximal as if I\supset \mathfrak{a}' then \pi(I)\supset \pi(\mathfrak{a}') (as by (1.1) these are distinct), which contradicts the maximality of \pi(\mathfrak{a}').