Two formulations of the Nullstellensatz I'm reading Anand Pillay's lecture notes on model theory and encountered the following formulation of the Nullstellensatz:

Let $K$ be an algebraically closed field. Let $\bar{P}(\bar{x}) = 0$ be a finite system of polynomial equations in $(x_1, .., x_n)$ with coefficients in $K$. Suppose that this system has a common solution in
  some field extending $K$. Then it has a common solution in $K$.

The formulation I am familiar with is like the wikipedia version, i.e.,

Let $k$ be a field (such as the rational numbers) and $K$ be an algebraically closed field extension (such as the complex numbers), consider the polynomial ring $k[X_1,X_2,..., X_n]$ and let $I$ be an ideal in this ring. The algebraic set $V(I)$ defined by this ideal consists of all n-tuples $x = (x_1,...,x_n)$ in $K^n$ such that $f(x) = 0$ for all $f$ in $I$. Hilbert's Nullstellensatz states that if $p$ is some polynomial in $k[X_1,X_2,..., X_n]$ that vanishes on the algebraic set $V(I)$, i.e. $p(x) = 0$ for all $x$ in $V(I)$, then there exists a natural number $r$ such that $p^r$ is in $I$.

Can anyone explain in what way these two formulations express the same thing?
 A: I don't think they mean the same thing, but the first formulation you wrote is a direct corollary from the second one (from Wikipedia).
Corollary (Nullstellensatz, weak version). Let $K$ be an algebraically closed field, and let $I$ be an ideal of $K\left[x_1,\dots,x_n\right]$. If $I\neq K\left[x_1,\dots,x_n\right]$, then $V\left(I\right)\neq\varnothing$.
(this follows directly from the Nullstellnsatz.)
Now, consider our case. Suppose $\overline{P}\left(\overline{x}\right)=0$ has no solutions in $K$, where $K$ is algebraically closed. Let $I$ be the ideal of $K\left[x_1,\dots,x_n\right]$ generated by the polynomials in $\overline{P}$. Hence $I=K\left[x_1,\dots,x_n\right]$, which means that $1$ can be written as a combination of the polynomials in $\overline{P}$. Thus, there cannot be a solution in any field extension of $K$.
A: I'd like to add that weak Nullstellensatz can be deduced from statement in the lecture notes.
 Suppose that a system $\bar P$ of polynomial equations over algebraically closed field $K$ has no common solutions. That is, $V(I)=\emptyset$, where $I$ is an ideal of $K[ x_1, \dots, x_n]$ generated by $\bar P$.
Either $I=K[ x_1, \dots, x_n]$ or $I$ is contained in a maximal ideal $m$. In that case $K[ x_1, \dots, x_n] / m$ is a field extending K, and $x_1, \dots, x_n$ is a common solution of $\bar P$ which is impossible by the hypothesis. Therefore, $I=K[ x_1, \dots, x_n]$.
Now, full version of Nullstellensatz can be obtained from this by localization (also known as Rabinowitz trick). I personally like the explanation here at MathOverflow.
