TITLE: Intuition for Model Theoretic Proof of the Nullstellensatz
QUESTION [13 upvotes]: I recently read the model-theoretic proof of the Nullstellensatz using quantifier elimination (see www.msri.org/publications/books/Book39/files/marker.pdf). I'm convinced that the Nullstellensatz is true, i.e. by showing that $\exists y_1 \cdots \exists y_n (\bigwedge_{i} (f_i(y) = 0)$ is equivalent to a quantifier free formula, hence if it is true in an algebraically closed extension, it is true in the original algebraically closed field. However, I don't have any intuition for how this formal sequence of steps leads to the truth of the Nullstellensatz. In particular, what I don't see is what the quantifier-free formulas are that these are equivalent to. For example, what is a quantifier-free formula in $a,b,c,d,e,f,g,h,i,j,k,l$ which is equivalent to the system $ax^2+bxy+cy^2+dx+ey+f=gx^2+hxy+iy^2+jx+ky+l=0$ having a non-trivial solution? And generalizations?
Can one simplify the model-theoretic proof by showing more directly that the existence of solutions to systems of equations (or non-equations) in algebraically closed fields are equivalent to polynomial conditions on the coefficients? I.e. apply essentially the same argument but restrict to the case of algebraically closed fields and avoid general results like Godel's Completeness Theorem to make the argument more clear. My expectation is that this might involve proving an algorithm for determining the quantifier-free formula from the formula with quantifiers.

REPLY [3 votes]: After some more searching, I found the notes written by Ford for an REU. These notes prove the Nullstellensatz using model theory, though the presentation is a little bit more concrete than that in Marker's notes.
On page 7, in Theorem 4.3, the notes prove quantifier elimination for ACF without reference to any preceding lemmas. The basic idea is this: First note that we only need consider formulas of the form $\exists x \phi(\bar{v},x)$. In other words, if our system of equations has $n$ variables, we need to first find quantifier-free conditions on $x_1,\cdots,x_{n-1}$ which will give the system of equations a solution in $x_n$. (In general, we will be dealing with a system of equations and inequations joined by and/or, but I ignore that for simplicity). Next, we find conditions which will make those conditions have a solution in $x_{n-1}$, etc. Repeating this process, we have a series of algebraic equations and inequations in the coefficients of the polynomials joined by and/or which gives a necessary and sufficient condition for the equation to have solutions.
Thus the problem is to find an algebraic condition (i.e. a sequence of expressions of the form $p(x)=0$ and $p(x) \neq 0$ joined by and/or operators) on the coefficients $\bar{v}$ of a system of $m$ equations in one variable $x$ which will ensure that the equations have a solution over an algebraically closed field.
To give a rough sketch, assume our system contains at least two non-constant polynomials in $x$, with degrees $1 \le d_1 \le d_2$ and leading coefficients $a_i(\bar{v})$. Then our new condition is $$(a_1(\bar{v})=0 \wedge p_1(x)-a_1(\bar{v})x^{d_1}=0 \wedge p_2(x)=0 \wedge\cdots \wedge p_m(x)=0) \vee (a_1(\bar{v}) \neq 0 \wedge p_1(x)=0 \wedge a_1(\bar{v})p_2(x)-a_2(\bar{v}) x^{d_2-d_1} p_1(x) = 0 \wedge p_3(x)=0 \wedge \cdots \wedge p_m(x)=0)$$ The sums of the degrees of the polynomials is smaller than before, and we can repeat this process until we don't have at least two non-constant polynomials. Then we use the Euclidean algorithm to deal with the $\neq$, and we are done.
What's nice is that this actually gives an elementary and algorithmic proof of the Nullstellensatz: first reduce the existence of a solution over an algebraically closed field to a system of equations and inequations in the coefficients. Next, note that there is a solution over some algebraically closed extension. Therefore, the coefficients satisfy the necessary conditions, so there is a solution over our original algebraically closed field, and we are done. I believe this is correct and much simpler than the one which Terry Tao provides.
Now we tackle the example which I give (again, a relatively rough sketch). We want to find conditions on $a,b,c,d,e,f,g,h,i,j,k,l,x$ which ensure that the equations have a solution in $y$.
Let the first equation be $p_1$, the second $p_2$. We thus reduce it to either $c=0$ and $(e+bx)y+ax^2+dx+f=iy^2+(hx+k)y+gx^2+jx+l=0$ (Case 1) or $c \neq 0$ and $cy^2+(e+bx)y+ax^2+dx+f= (chx+ck-ie-ibx)y+cgx^2+cjx+cl-iax^2-idx-if=0$ (Case 2).
In Case 1, the first polynomial is $p_1$, the second is $p_2$, and if $e+bx=0$, the equation having a solution is equivalent to $ax^2+dx+f=0$ and either $i \neq 0$ or $hx+k \neq 0$ or $gx^2+jx+l=0$. If $e+bx \neq 0$, we subtract $iy$ times the first polynomial from $e+bx$ times the first to get $
(e+bx)(iy^2+(hx+k)y+gx^2+jx+l)-iy((e+bx)y+ax^2+dx+f)=$
$bx gx^2 + e gx^2 + bx hx + e hx - ax^2 iy - bx iy - dx iy - e iy - 
 f iy + bx iy^2 + e iy^2 + bx jx + e jx + bx k + e k + bx l + e l$ We now have a system of simultaneous linear equations, so to check whether there's a solution, we only need to check whether each has a solution and, degenerate cases aside, whether a certain determinant is $0$, i.e. an algebraic condition.
In Case 2, the first polynomial is $p_2$, the second is $p_1$. If $chx+ck-ie-ibx=0$, then the existence of a solution is equivalent to $cgx^2+cjx+cl-iax^2-idx-if=0$ and either $c \neq 0$, $e+bx \neq 0$, or $ax^2+dx+f=0$. If $chx+ck-ie-ibx \neq 0$, then we compute $(chx+ck-ie-ibx)(cy^2+(e+bx)y+ax^2+dx+f)-cy((chx+ck-ie-ibx)y+cgx^2+cjx+cl-iax^2-idx-if)=$
$ax^2 chx + ax^2 ck - cgx^2 cy - cjx cy - cl cy + chx cy^2 + ck cy^2 + 
 chx dx + ck dx + chx f + ck f + cy iax^2 - ax^2 ibx - cy^2 ibx - 
 dx ibx - f ibx + cy idx - ax^2 ie - cy^2 ie - dx ie - f ie + cy if$. We then proceed as in Case 1.