Posted in Algebraic Geometry

## Weak Nullstellensatz

In algebraic geometry, one of  the fundamental theorems is Hilbert’s Nullstellensatz.

Weak version:

Weak Hilbert’s Nullstellensatz: If $k$ is algebraic closed field then the maximal ideals of $k[x_1, \dots, x_n]$ are exactly of the form $(x_1 - a_1, \dots, x_n -a_n)$ for some $a_i \in k$.

Some cases:

If $k = \mathbb{C}$, a system of polynomials equation have a root ’cause $V(I) \ne \emptyset$.

If $k = \mathbb{R}$, there exists an ideal $I \ne (x_1 - a_1, \dots, x_n -a_n)$ such that $V(I) = \emptyset$.

Another version:
\begin{theorem} Ideal $I \subseteq k[X_1, \dots, X_n]$ with $k$ is algebraic closed. Then $V(I) = \emptyset$ implies $I = k[X_1, \dots,X_n]$.
\end{theorem}

Moreover, $k = \mathbb{C}$, if we have a system of polynomials equation $(f_1 = 0, f_2 = 0, \dots, f_k = 0)$,  if this one have no root then there exist $g_1, \dots, g_k \in \mathbb{C}[X_1,\dots,X_n]$

s.t. $f_1(X)g_1(X) + f_2(X)g_2(X) + \dots + f_k(X)g_k(X) = 1.$

For example (Parrilo): Consider following polynomials over $\mathbb{C}$:

$f_1(x) = x^2 + y^2 - 1 =0$,

$f_2(x) = x + y = 0$,

$f_3(x) = 2x^3 + y^3 +1 = 0$.

There exist:

$g_1(x) = \frac{1}{7}(1 - 16x - 12y - 18xy - 6y^2)$,

$g_2(x) = \frac{1}{7}(-7y - x +4y^2 - 16 + 12xy + 2y^3 + 6y^2x)$,

$g_3(x) = \frac{1}{7}(8 + 4y)$

s.t. $f_1g_1 + f_2g_2 + f_3g_3 = 1$.