I found a paper 
"some applications of a model theoretic fact to 
(semi-) algebraic geometry" by Lou van den Dries.

In this paper, the author uses model theoretical method to prove the completness of projective variety.

So my question is "can I translate the proof into modern lanquage in algebrai geometry?".

The central part of the proof is this lemma.

Lemma

Let "$T$" be a first order theory and $\phi(C)$, an $L(T)$-formula. 

Suppose that for each model $K$ of $T$ and each homomorphism $f:A\rightarrow L$ where $A\subset K$ and $L$ is also model of $T$, we have

if $c\in A^m$ and $K$ satisfy $\phi(c)$,then $L$ also satisfy $\phi(f(c))$

Then there is positive quantifier free formula $\psi(C)$ such that $\phi(C)$ is equivalent to $\psi(C)$ in the theory $T$.

The completness of projective variety says; let $p_1(C,X),...,p_k(C,X)\in \mathbb{Z}[C,X]$ be homogeneous in $X=(X_0\cdots X_n)$. Then there are polynomials $q_1(C),\cdots,q_l(C)$ in $\mathbb{Z}[C]$ such that for each algebraically closed field $K$ and $c\in K^m$ the system $p_1(c,X)=\cdots=p_k(c,X)$ has a non trivial solution in $K$ if and only if
$q_1(c)=\cdots=q_l(c)=0$ 

The proof goes as follows.

Let $\phi(C)$ be the formula $\exists X (p_1(C,X)=...=p_k(C,X)=0)$.Let $T$ be the theory of algebraically closed fields. We have to show that $\phi(C)$ is equivalent to a positive quanfier free formula,relative to T. By the model theoretic lemma we are reduced to showing;

if $f:A\rightarrow L$  is a morphism of a subring $A$ of an algebraically closed field $K$ into another algebraically closed field $L$ and if the system $p_1(C,X)=...=p_k(C,X)=0$ has a non trivial in K, then the system $p_1(C,X)=...=p_k(C,X)=0$ has a non trivial solution in L.

Now by Chevalley's place extension theorem we may as well assume that $A$ is a valuation ring of $K$. Multiplying a non-trivial solution in $K$ by a suitble constant in $A$,we may as well assume that we have $x\in A^{n+1}$ with at least one coordinate invertible in $A$.
Then, applying $f$ to $p_1(C,x)=...=p_k(C,x)=0$,
we get a solution $f(x)$ of $p_1(C,f(x))=...=p_k(C,f(x))=0$,and $f$ maps the invertible coordinate of $x$ into a zero element of $L$.