Let \(Y\) be the algebraic set in \(\AA^3\) defined by the two polynomials \(x^2 - yz\) and \(xz - x\). Show that \(Y\) is a union of three irreducible components. Describe them and find their prime ideals.
Solution
Factor the second equation: \[ Y = V(x^2 - yz) \intersect V(xz - x) = V(x^2-yz) \intersect \left( V(x) \union V(z-1) \right) . \] So a point of \(Y\) has \(x = 0\) or \(z = 1\).
Case \(x = 0\). Then \(yz = 0\), so \(y = 0\) or \(z = 0\), giving the two lines \[ V(x, y) \quad \text{and} \quad V(x, z) . \]
Case \(z = 1\). Then \(x^2 = y\), giving the parabola \(V(z - 1,\ x^2 - y)\).
Hence \[ Y = V(x,y) \union V(x,z) \union V(z-1,\ x^2-y) , \] two lines and a parabola. Each is irreducible because its ideal is prime: the quotients are \(k[z]\), \(k[y]\) and \(k[x]\) respectively, all domains. None contains another, so these are the irreducible components.
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PR-TNVSIIrreducible decomposition in a Noetherian spaceuses