Every open subset of an affine variety is itself an affine variety

Question

It seems that I have proved the title claim, but I'm not convinced by my own argument, so can you check where is the error, if one?

Let $X\subset\mathbb{A}^n$ be an affine variety and let $Y$ be an open subset of $X$. I want to prove that $Y$ is itself an affine variety. Since $Y$ is open by assumption, it must have the form $Y=X-Z$, where $Z$ is a closed subset of $X$, say $Z=V(f_1,\ldots,f_m)$ where $f_j\in k[X_1,\ldots,X_n]$. Putting $D(f_j):=X-V(f_j)$ we find that $Y=D(f_1)\cup\ldots\cup D(f_m)$. Suppose that $X$ is defined by equations: $g_1=\ldots=g_k=0$. For every $j=1,\ldots,m$ let $Z_j\subset\mathbb{A}^{n+1}$ be the variety defined by the equations: $g_1=\ldots g_k=0 ; f_jX_{n+1}-1=0$. Then the map $(X_1,\ldots,X_{n+1})\longmapsto (X_1,\ldots,X_n)$ is an isomorphism between $Z_j$ and $D(f_j)$, which proves that $D(f_j)$ is closed, hence also the finite union $\bigcup_j D(f_j)$ is closed.

Answer 1

As indicated in the comments above, the statement is false. Hoot points out that your error is that $D(f_j)$ is not in general closed. One has to remember that being closed is a relative property, it is not something preserved under isomorphism. Take any $U \subset X$ open but not closed, and consider the inclusion map $U \hookrightarrow X$. $U$ is closed in the domain (namely, $U$), and the map is an isomorphism onto its image, but $U$ is not closed in $X$.