Can you tell me, how to prove that every proüper and affine morphism between locally Noetherian schemes is finite? Every help will be appreciated.

There is an elementary proof of the result "universally closed + affine $\Rightarrow$ integral" that I learnt from Olivier's paper "Going up along absolutely flat morphisms." In fact, it's so simple, I can present it here. Observation 1: Say $\phi:A \to B$ is an injective ring map that is closed on $\mathrm{Spec}$. Then $\phi^{1}(B^\ast) = A^\ast$. (This proof was edited and corrected to reflect xuhan's comment.) Proof: Fix $a \in A$ with $\phi(a) \in B^\ast$. We must show $a \in A^\ast$ or, equivalently, $a$ is nonzero in the residue field $\kappa(\mathfrak{p})$ of $A$ at any prime $\mathfrak{p} \in \mathrm{Spec}(A)$. Note that the last statement is clearly true if $\mathfrak{p}$ lies in the image $Z$ of $\mathrm{Spec}(\phi)$. So it suffices to show $Z = \mathrm{Spec}(A)$. By closedness, $Z = V(I)$ for some ideal $I \subset A$ (settheoretically). Localizing at any prime $\mathfrak{p}$ shows $I \subset \mathfrak{p}$ by the injectivity hypothesis. Then $I$ is contained in all primes of $A$, so it contains only nilpotents, and hence $Z = \mathrm{Spec}(A)$. Observation 2: Say $\phi:A \to B$ is an injective ring map, and $\phi[T]:A[T] \to B[T]$ is closed on $\mathrm{Spec}$. Then $\phi$ is integral. Proof: Fix some $f \in B$, and consider the surjective map $B[T] \to B[\frac{1}{f}]$ given by $T \mapsto \frac{1}{f}$. If we write $C \subset B[\frac{1}{f}]$ for the image of the composite $A[T] \to B[T] \to B[\frac{1}{f}]$, then $C \to B[\frac{1}{f}]$ is an injective ring map that is closed on $\mathrm{Spec}$. The image of $T$ in $C$ becomes a unit in $B[\frac{1}{f}]$, and hence must be a unit on $C$ by Observation 1, so we can write $f = \sum_{i=0}^n a_i \big(\frac{1}{f}\big)^i$ in $B[\frac{1}{f}]$ for $a_i \in A$. Clearing denominators shows that $f \in B$ satisfies a monic polynomial over $A$. Observation 2 + killing the kernel shows: Theorem: If $\phi:A \to B$ is a ring map that is universally closed on $\mathrm{Spec}$, then it is integral. 


Have a look at Ravi Vakil's notes on Algebraic geometry, 18.1.8, 18.9.A (and possibly tracing through the results used in these sections) http://math.stanford.edu/~vakil/216blog/. The point is that $f:X\rightarrow Y$ is affine, then $X\cong \underline{\mathrm{Spec}}(f_*\mathcal{O}_X)$, and if $f$ is proper, then $f_*\mathcal{O}_X$ is coherent. Hence $X$ is the relative spectrum of a coherent $\mathcal{O}_Y$algebra, and hence finite. 


See also Liu, Lemma 3.3.17 (for a proof without using the coherence theorem). 

