a very basic question about commutative algebra

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.

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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Why is it not maximal? – Mariano Suárez-Alvarez Jan 16 '12 at 22:09
"[$\ldots$] 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. – kahen Jan 16 '12 at 22:11

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 $\phi^{-1}(0)=\mathfrak{a}$.
Thanks Zev , now I have a question related. If I have a ring homomorphism $f: A \to B$ and J is a maximal ideal of B , then it´s not always true that $f^-1 (J)$ it´s a maximal ideal of A , example $A = {\Bbb Z},B = Q,J = \left( 0 \right)$ but I don´t like this example , and I don´t have more, there exist conditions that ensure that the preimage of a maximal ideal must also be maximal? ( for example in A and $A/a$ it´s an example , as we seen ) – August Jan 16 '12 at 22:20
@August If $f$ is surjective then this is the case, as then $B$ is isomorphic to $A/\mathfrak{a}$ for some ideal $\mathfrak{a}\subseteq A$. – Alex Becker Jan 16 '12 at 23:37
@August There are certain classes of ring homomorphisms $f:A\rightarrow B$ that have this property. Surjective ring homomorphisms, and more generally, finite ring homomorphisms (homomorphisms making $B$ into a finitely generated $A$-module), and more generally still, integral ring homomorphisms have the property. Another class of homomorphisms for which this holds consists of the homomorphisms of finite type with $A$ a Jacobson ring. Another non-example is the inclusion of $\mathbb{Z}$ into $\mathbb{Q}[[T]]$ (the power series over $\mathbb{Q}$). The inverse image of the maximal ideal $(T)$ is – Keenan Kidwell Jan 17 '12 at 0:45
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. – Keenan Kidwell Jan 17 '12 at 0:47
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 $\pi(\mathfrak{a}')$.