Inconsistent system of simultaneous equations Let $F$ be an algebraically closed field, and $f_1,\ldots,f_n$ polynomials in $k$ variables over $F$. The system of simultaneous equations 
$$\mathcal{F}: f_1(x_1,\ldots,x_k)=0,\ldots,f_n(x_1,\ldots,x_k)=0$$
is said to be inconsistent if there exist polynomials $g_1,\ldots,g_n$ such that $\sum \limits_{i=1}^n f_i g_i = 1$.
I am trying to prove that the system of equations $\mathcal{F}$ has a solution if and only if it is not inconsistent. It seems very clear in a straightforward way: if $(x_1,\ldots,x_k)$ is a solution of this system, then $\sum \limits_{i=1}^{n} f_i(x_1,\ldots,x_k) g_i(x_1,\ldots,x_k) = 0 \neq 1$. But, can anyone suggest, how to prove the converse (if system is not inconsistent, then it has a solution)? And by the way, could you please explain the inconsistence intuitionally, like in the same way as for linear systems?  
 A: If $(x_1,...,x_k)$ is a solution of the system, then $\sum_{i=1}^n f_i g_i$ is zero at the point $(x_1,...,x_k)$. Therefore $\sum_{i=1}^n f_i g_i = 1$ implies there can't be any solutions, hence inconsistent.
The point is to prove if there are no solutions, this always happens: i.e. there are $g_i$ such that $\sum_{i=1}^n f_i g_i = 1$. 
The key is that $F$ is algebraically closed. By Hilbert's Nullstellensatz that implies the points $(x_1,...,x_k)$ of the affine space $\mathbb{A}^k_F$ are in one-to-one bijection with the maximal ideals of the polynomial ring $R=F[T_1,T_2,...,T_k]$ via the correspondence
$$ (x_1,x_2,...,x_k)\in\mathbb{A}^k_F \leftrightarrow (T_1 - x_1, T_2 - x_2,..., T_k - x_k) \subset R$$ 
Now $(x_1,x_2,...,x_k)$ is a solution of the system $\mathcal{F}$ if and only if the corresponding maximal ideal $(T_1 - x_1,...,T_k - x_k)$ contains the ideal generated by the $f_i$. That means $\mathcal{F}$ has a solution if and only if the ideal generated by $f_i$ is contained in some maximal ideal. Since every proper ideal is contained in some maximal ideal, if $\mathcal{F}$ has no solutions then $f_i$ must generate the entire ring.
The polynomials $f_i$ generate the entire ring $R$ if and only if they generate the unit element. That means there exist $g_i$ such that $\sum_{i=1}^n g_i f_i = 1$.
