# A principal open set of an affine algebraic set is an affine variety

Notations

• $k$ is an algebraic closed field and $\mathbb A^n(k)$ is the topological space $k^n$ with the Zariski topology
• If $X\subseteq\mathbb A^n(k)$ is an affine algebraic set and $f\in\Gamma(X)$, then $D(f)=\{x\in X\,:\, f(x)\neq0\}$
• An affine variety for me is a ringed space $(X,\mathcal O_X)$, where $\mathcal O_X$ is a sheaf of $k$-valued function, that is isomorphic (as ringed space) to $(V,\mathcal O_V)$, where $V$ is an IRREDUCIBLE affine algebraic set and $\mathcal O_V$ is the sheaf of regular functions on $V$.

I want to find an affine variety that is not an irreducible affine algebraic set. If $X\subseteq\mathbb A^n(k)$ is an irreducible affine algebraic set, and $f\in\Gamma(X)$, I would prove that $(D(f),\mathcal O_{X|D(f)})$ is an affine variety:

Let's consider $\mathfrak a=I(X)\subseteq k[T_1,\ldots, T_n]$ as a subring of $k[T_1,\ldots, T_n, T_{n+1}]$ and let $F\in k[T_1,\ldots, T_n]$ be a representative of $f$. If $Y=V(\big<\mathfrak a, FT_{n+1}-1\big>)$, the function $$j: Y\longrightarrow D(f)$$ that is the restriction of the projection of $\mathbb A^{n+1}(k)$ on $\mathbb A^n(k)$ is clearly bijective and continuous with inverse $j^{-1}:(x_1,\ldots, x_n)\longmapsto (x_1,\ldots, x_n,\frac{1}{f(x_1,\ldots,x_n)})$. Now

1. How can I prove that $Y$ is irreducible?
2. Why is $j^{-1}$ continuous? (so $j$ is a homeomorphism)
3. Why is $j$ an isomorphism of ringed spaces? If $U\subset D(f)$ is open, should be proved that for all $g\in \mathcal O_X(U)$, then $g\circ f_{|f^{-1}(U)}\in\mathcal O_X(f^{-1}(U))$.
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1) $X$ is irreducible and $D(f)$ is open in $X$ implies that $D(f)$ is also irreducble. Now, since $j: Y \rightarrow D(f)$ is continuous and bijective, it is a homeomorphism. This implies that $Y$ is also irreducible (as it is topologically homeomorphic to $D(f)$).
3) To show that $j$ is an isomorphism of ringed spaces, you need to construct $j^{\#}: \mathcal{O}_{D(f)} \rightarrow j_{\ast} \mathcal{O}_Y$ and verify that it gives you an isomorphism of sheaves for each open subset $U \subset D(f)$. This can be done as follows. For any section $g \in \mathcal{O}_{D(f)}(U)$ that is represented by $G \in k[T_1, ...., T_n]$, we define its image to be the section represented by $H(T_1,...,T_n, T_{n+1}) = G(T_1,...,T_n)$. There is a little bit detailed and routine checking required here that $H$ does represent validly a unique section in $j_{\ast \mathcal{O}_Y(U)} = \mathcal{O}_Y(j^{-1}(U))$ and vice versa, i.e. a section in the $\mathcal{O}_Y(j^{-1}(U))$ has a preimage in $\mathcal{O}_{D(f)}(U)$. It probably uses some fact in step 1 but all in all, it is not too bad.
I don't understand why if $j$ Is bijective and continuous, then it is a homeomorphism. –  Dubious Apr 28 '13 at 18:10
You're right, I was missing a detail, say $j^{-1}$ is also continuous. –  mr.bigproblem Apr 28 '13 at 18:32
To do this, take a distinguished open subset $D(h) \subset k^{n+1}$ that gives a distinguished open subset $D(h) \cap Y$ in $Y$, and compute the preimage via $j^{-1}$, which is $j(D(h) \cap Y) = j(D(h)) \cap D(f)$ and verify that this is open in $X$. –  mr.bigproblem Apr 28 '13 at 18:37