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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 methods to prove the completeness of projective varieties.

So my question is "can I translate the proof into modern language in algebraic 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 $K\models \phi(c)$, then $L\models \phi(f(c))$. Then there is positive quantifier free formula $\psi(C)$ such that $T\vdash \phi(C)\leftrightarrow \psi(C)$.

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

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 quantifier free formula relative to T. By the model theoretic lemma we are reduced to showing the following:

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 $p_1(C,X)=\dots=p_k(C,X)=0$ has a non-trivial solution in K, then the system $p_1(C,X)=\dots=p_k(C,X)=0$ 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 suitable 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 $p_1(C,x)=\dots=p_k(C,x)=0$, 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 non-zero element of $L$.

Improve this question
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  • $\begingroup$ I admit that I haven't had the time or energy to read your question carefully. But probably yes: elimination of quantifiers for algebraically closed fields on the model theory side is equivalent in algebraic geometry to the fact the class of constructible sets is stable under projections. $\endgroup$
    – Donu Arapura
    Commented Feb 21 at 14:48
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    $\begingroup$ @DonuArapura That's true, but the model-theoretic proof uses a refinement of the idea of quantifier elimination, namely that if a formula is preserved by all homomorphisms from subrings to algebraically closed fields (rather than all embeddings from subrings to algebraically closed) then it is equivalent to a positive quantifier-free formula, and as a consequence, it defines a Zariski-closed set, rather than just a constructible set. $\endgroup$
    – Alex Kruckman
    Commented Feb 26 at 16:16
  • $\begingroup$ @AlexKruckman OK, thanks for the comment. I hadn't realized that there was a strong statement. I assume "positive" means no negations. $\endgroup$
    – Donu Arapura
    Commented Feb 26 at 22:24
  • $\begingroup$ @DonuArapura That's right, no negations. $\endgroup$
    – Alex Kruckman
    Commented Feb 26 at 22:52

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