# For $X$ an affine variety, $X \setminus Z(f)$ is isomorphic to an affine variety

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

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Have you followed the rest of the proof? One direction of the morphism is given by the projection $p : W \to Z_f$ where $(x_1, \ldots , x_n , t) \mapsto (x_1, \ldots , x_n)$. This doesn't make sense unless $(x_1, \ldots , x_n)$ is actually in $Z_f$ (i.e. in $X$ and not $Z(f)$). The $tF-1= 0$ takes care of the "not $Z(f)$", and the fact that $Z(J) = X$ forces the first $n$ coordinates to correspond to a point in $X$. – Daniel Freedman Mar 12 '12 at 17:14
If $X$ is the whole of $\mathbb A^n$, then you can just take $J = \langle tF-1 \rangle = \langle tf -1 \rangle$ – Daniel Freedman Mar 12 '12 at 17:20