integral ring extension, maximal ideals 
Let $\varphi:A\rightarrow A'$ be an integral ring extension. 
1) Show that for every maximal ideal $m'\subset A'$ the ideal $\varphi^{-1}(m')\subset A$ is maximal. 
2) and that for every maximal ideal $m\subset A$ there is a maximal ideal $m'\subset A'$ with $\varphi^{-1}(m')=m$. 

Now there is a tip given: for $m\subset A$ look at $S=A-m$ and the localization $S^{-1}A$ and $S^{-1}A'$. Now I didn't get any further with this tip! Please help me!
1) What I did: $\varphi^{-1}(m')$ is a prime ideal (else $m'$ wouldn't be one in $A'$ as well, so also not maximal). Now I define the map $\psi:A/\varphi^{-1}(m')\rightarrow A'/m',a+\varphi^{-1}(m')\mapsto a+m$. Now $\psi$ is well defined as if $a-b\in \varphi^{-1}(m')$ then $\psi(a-b)=0\iff\psi(a)=\psi(b)$. So now if $x\in A'$ then there are $a_i\in A$ s.t. $x^n+a_1x^{n-1}+\cdots+a_n=0$. But this is inducing $(x+m)^n+(a_1+m)(x+m)^{n-1}+...+(a_n+m)=m$, so $\psi$ is an integral ring extension of integral domains (as $\varphi^{-1}(m')$ is prime) with $A'/m'$ a field, so then $A/\varphi^{-1}(m')$ is also a field and thus $\varphi^{-1}(m')$ is maximal in $A$.
For 2) I didn't use that the extension is integral, so I'm not sure if it's correct at all: $m\subset A\Rightarrow \exists m'\subset A'$ s.t. $m\subset m'$. Now $\varphi^{-1}(m')=m$, because if $x\in\varphi^{-1}(m')-m$ then $\varphi^{-1}(m')$ is still an ideal, not containing $1$, as $m'$ is maximal, so its a proper ideal which contains $m$ in $A$.
So now here the only detail I'm missing is in this step: $\exists m'\subset A'$ s.t. $m\subset m'$. But I figured I can look at the ideal $(m)$ in $A'$, which is the ideal generated by every element of $m$, so then there exists a maximal ideal which contains the ideal $(m)$ properly, which isn't possible. Now I would only need to show that $(m)$ is a proper ideal, but im not sure how, if its true at all in general, but I guess it should be...
So now my question is 
a) how can I use the tip? 
b) did I make any mistakes or not show something important?
 A: Hint: (1). Show that if $\phi: A \to B$ be an integral extension, then for any multiplicative closed sets $S \subset A, S^{-1}A \to S^{-1}B$ is an integral extension.
(2). Take $S:= A - \mathfrak m.$ Let $\alpha: A \to S^{-1}A, \beta: B \to S^{-1}B, \gamma:S^{-1}A \to S^{-1}B $ be natural maps. Then $\beta \phi = \gamma \alpha.$ Let $\eta$ be a maximal ideal of $S^{-1}B.$ Show that $\mathfrak m = \phi^{-1}\beta^{-1}(\eta).$ Here $\beta^{-1}(\eta)$ is a prime ideal in $B.$
A: The first part looks okay, modulo this fact about integral extensions of domains. I assume this is something you can just use. For the second part, we definitely need to use the integral hypothesis because $\mathfrak{m}A' \neq A'$ needn't be true otherwise: $\mathbf{Z} \subset \mathbf{Q}$ probably yields the simplest example.
The condition $\mathfrak{m}A' \neq A'$ ought to remind you of Nakayama's lemma; that's why we localize at $\mathfrak{m}$ and henceforth assume that $A$ is local. There is some work hiding here: After localization, why do I still have an integral extension? How does one transport the result for that extension back to the original problem? Having justified this, we would be done if we knew that $A'$ were a finitely generated $A$-module, but you can use the usual trick of an ideal being the unit ideal iff it contains $1$ in order to work with a ring that is a finitely generated $A$-module.
I'm happy to say more, but I wanted to give the main ideas and see how that went first.
