Let $X \subset \mathbb A_k^n$ be an affine variety, and let $Z_f = X \setminus Z(f)$. I'm looking at a proof of the fact that $\mathbb Z_f$ is isomorphic to an affine variety. The proof proceeds as follows:
Let $J = I(X) \subset k[x_1,\ldots , x_n]$ be the ideal of the variety $X$, and let $F \in k[x_1, \ldots , x_n]$ be a polynomial with $F \big|_X =f $. We set $J_F = \langle J,tF-1 \rangle \subset k[x_1, \ldots , x_n, t]$. Then we claim that $Z_f$ is isomorphic to the affine variety $W = Z(J_F) \subset \mathbb A^{n+1}$.
My question is this: Why do we need $J_F$ to be of this form? Wouldn't it work if we just set $J_f = \langle tF-1 \rangle$?
Thanks