Let \[ X\subseteq\AA^n_k \] be an affine variety, embedded in projective space by \[ \AA^n_k\hookrightarrow\PP^n_k, \qquad (a_1,\ldots,a_n) \longmapsto [1:a_1:\ldots:a_n]. \] Let \(\overline X\subseteq\PP^n_k\) be its projective Zariski closure, and let \[ H_0=\{x_0=0\} \] be the hyperplane at infinity. Show that \[ \overline X=X\union\bd X, \qquad \bd X=H_0\intersect\overline X. \]
Solution
Put \[ U_0=\{x_0\ne0\}\subseteq\PP^n_k. \] Under the standard embedding, \(U_0\cong\AA^n_k\) and \(X\) is identified with a closed subset of \(U_0\).
1. The projective closure satisfies\[ \boxed{\overline X\intersect U_0=X.} \]
Let \[ j:U_0\hookrightarrow\PP^n_k \] be the open immersion. For any subset \(S\) of a topological space and any open subset \(U\), closure restricts by \[ \overline S^{\,T}\intersect U = \overline{S\intersect U}^{\,U}. \] Applying this with \[ T=\PP^n_k, \qquad S=X, \qquad U=U_0 \] gives \[ \overline X\intersect U_0 = \overline X^{\,U_0}. \]
But \(X\) is an affine variety in \[ U_0\cong\AA^n_k, \] so \(X\) is Zariski closed in \(U_0\). Therefore its closure inside \(U_0\) is itself: \[ \overline X^{\,U_0}=X. \] Hence \[ \overline X\intersect U_0=X. \]
2. The complement of \(U_0\) in \(\PP^n_k\) is exactly the hyperplane atinfinity:\[ \PP^n_k\sm U_0=H_0. \]
By definition, \[ U_0=\{[x_0:\ldots:x_n]:x_0\ne0\}. \] Its complement is therefore \[ \{[x_0:\ldots:x_n]:x_0=0\} = H_0. \]
3. The projective closure decomposes as\[ \boxed{ \overline X = X\union(H_0\intersect\overline X). } \]
By step 2, \[ \PP^n_k = U_0\union H_0. \] Intersecting with \(\overline X\) gives \[ \overline X = (\overline X\intersect U_0) \union (\overline X\intersect H_0). \] Step 1 identifies the first term with \(X\). Therefore \[ \overline X = X\union(H_0\intersect\overline X). \]
4. With\[ \bd X \coloneqq H_0\intersect\overline X, \]one has\[ \boxed{\overline X=X\union\bd X.} \]
This is the identity in step 3 with \(\bd X\) written for the boundary at infinity.
5. Q.E.D.
Step 1 identifies the affine part of the projective closure with \(X\), and steps <1>2–<1>4 identify the remaining part with the boundary at infinity.