Equivalent formulations of Hilbert's Nullstellensatz I'm trying to figure out this proof. Can anyone help out?
We know the following statement of the theorem:
For any idea $I \subset \mathbb{C} [X_1,\ldots,X_n],\mathbb{I}(\mathbb{V}(I)) = \sqrt{I}$
We want to show that the above statement implies this one:
Let $k$ be an algebraically closed field and $F_1 ,\ldots,F_m \in k[T_1,\ldots,T_n]$. If the ideal $(F_1,\ldots,F_m) \neq (1)$ Then the system of equations $F_1 = \cdots = F_m = 0$ has a solution in $k.$

Note that for an ideal $I$ in a commutative ring, $\sqrt{I}=(1) \Rightarrow I=(1)$ (because if some power of an element is a unit, it must be a unit itself), this implies that in your situation $\Bbb{I}(\Bbb{V}(F_1,\ldots,F_m)) \neq (1)$, so that $\Bbb{V}(F_1, \ldots, F_m) \neq \varnothing$, because $\Bbb{I}(\varnothing) = (1)$.
But $\Bbb{V}(F_1, \ldots, F_m) \neq \varnothing$ means precisely that there is some solution to $F_1 = \ldots = F_m = 0$.
