Let $X$ be an affine algebraic variety of $k^{n}$, $k$ algebraically closed. $I(X)$ the ideal of the polynomials that vanish in $X$, prove that if $I(X)\not = (1)$, then $X$ is not empty.
What is the problem with taking $f\not \in I(X)$ and saying that by definition exists $x\in X$ with $f(x)\not = 0$, in particular exists $x\in X$?