Let \(k\) be an algebraically closed field of characteristic zero, and let \[ X=V(x(xy-1))\subseteq\AA^2_k. \] Show that \(X\) is reducible and has exactly two irreducible components.
Solution
Put \[ L=V(x), \qquad H=V(xy-1). \]
1. The cubic decomposes as\[ \boxed{X=L\cup H.} \]
For any point \((a,b)\in\AA^2_k\), \[ a(ab-1)=0 \] if and only if \[ a=0 \qquad\text{or}\qquad ab-1=0. \] Therefore \[ V(x(xy-1)) = V(x)\cup V(xy-1) = L\cup H. \]
2. The closed subset \(L\) is irreducible.
Its coordinate ring is \[ k[L] = k[x,y]/(x) \cong k[y], \] which is an integral domain. Hence \(L\) is irreducible.
3. The closed subset \(H\) is irreducible.
Its coordinate ring is \[ k[H] = k[x,y]/(xy-1) \cong k[x,x^{-1}], \] which is an integral domain. Hence \(H\) is irreducible.
4. Neither \(L\) nor \(H\) contains the other.
The point \[ (0,0) \] lies in \(L\) but not in \(H\), while \[ (1,1) \] lies in \(H\) but not in \(L\). Therefore \[ L\nsubseteq H \qquad\text{and}\qquad H\nsubseteq L. \]
5. The irreducible components of \(X\) are exactly\[ \boxed{ L=V(x) \quad\text{and}\quad H=V(xy-1). } \]
By steps <1>2–<1>3, both \(L\) and \(H\) are irreducible closed subsets of \(X\). Step 4 shows that neither is contained in the other.
Let \(Z\subseteq X\) be irreducible. By step 1, \[ Z = (Z\cap L)\cup(Z\cap H), \] where both intersections are closed in \(Z\). Since \(Z\) is irreducible, one of these intersections must equal \(Z\). Thus \[ Z\subseteq L \qquad\text{or}\qquad Z\subseteq H. \] Hence every irreducible closed subset of \(X\) lies in one of \(L\) or \(H\), so the maximal irreducible closed subsets are exactly \(L\) and \(H\).
6. The variety \(X\) is reducible.
Step 1 writes \(X\) as the union of the two proper closed subsets \(L\) and \(H\). Therefore \(X\) is reducible.
7. Q.E.D.
Step 5 identifies the two irreducible components, and step 6 proves reducibility.