Let \(A = \CC[x,y]/(y^2-(x-1)^3 - (x-1)^2)\).
Show that \(A\) is an integral domain and sketch the \(\RR\)-points of \(\text{Spec} A\).
Find the integral closure of \(A\). Recall that for an integral domain \(A\) with fraction field \(K\), the integral closure of \(A\) in \(K\) is the set of all elements of \(K\) integral over \(A\).
Solution
1. Proof that \(A\) is an integral domain:
1.1. The defining relation is \(f(x, y) = y^2 - (x-1)^3 - (x-1)^2 = y^2 - x(x-1)^2\).View \(f(x, y)\) as a monic polynomial of degree 2 in \(y\) over the UFD \(R = \mathbb{C}[x]\):\[ f(x, y) = y^2 - x(x-1)^2 \in R[y]. \]
1.2. The polynomial \(f(x, y)\) is irreducible in \(\mathbb{C}(x)[y]\) if and only if \(x(x-1)^2\) is not a square in \(\mathbb{C}(x)\).Since \(x\) has odd multiplicity \(1\) in \(\mathbb{C}[x]\), \(\sqrt{x(x-1)^2} = (x-1)\sqrt{x} \notin \mathbb{C}(x)\).Because \(f\) is monic in \(y\), Gauss’s Lemma implies \(f(x, y)\) is irreducible in \(\mathbb{C}[x, y]\).
1.3. Since \(\mathbb{C}[x, y]\) is a commutative ring and \(f(x, y)\) is irreducible, the principal ideal \(\langle f(x, y) \rangle\) is prime.Therefore \(A = \mathbb{C}[x, y]/\langle f(x, y) \rangle\) is an integral domain.
2. Description of the real points of \(\operatorname{Spec}(A)\):
2.1. The real locus consists of all \((x, y) \in \mathbb{R}^2\) satisfying \(y^2 = x(x-1)^2\).Since \(y^2 \ge 0\) and \((x-1)^2 \ge 0\), real solutions exist if and only if \(x \ge 0\).
2.2. Geometric features of the curve \(y = \pm (x-1)\sqrt{x}\):- At \(x = 0\), \(y = 0\), forming a smooth endpoint with vertical tangent (\(x \sim y^2\)).- For \(0 < x < 1\), the curve splits into two branches \(y = \pm (1-x)\sqrt{x}\), forming a closed loop bounded between \(x = 0\) and \(x = 1\).- At \((1, 0)\), the loop crosses itself: the two branches have tangents \(y \approx \pm(x-1)\), creating an ordinary double point (node).- For \(x > 1\), the two branches \(y = \pm (x-1)\sqrt{x}\) extend to infinity.
3. Integral closure of \(A\):
3.1. Let \(K = \operatorname{Frac}(A)\) be the fraction field of \(A\).Define \(t = \frac{y}{x-1} \in K\).Then:\[ t^2 = \frac{y^2}{(x-1)^2} = \frac{x(x-1)^2}{(x-1)^2} = x \implies x = t^2. \]
3.2. Express \(y\) in terms of \(t\):\[ y = t(x - 1) = t(t^2 - 1) = t^3 - t. \]Thus \(A = \mathbb{C}[x, y] = \mathbb{C}[t^2, t^3 - t] \subseteq \mathbb{C}[t] \subset K\).
3.3. The element \(t\) is integral over \(A\) since \(t^2 - x = 0\) is a monic polynomial in \(A[T]\).Thus \(\mathbb{C}[t]\) is an integral extension of \(A\).
3.4. The polynomial ring \(\mathbb{C}[t]\) is a PID, hence integrally closed in its fraction field \(\operatorname{Frac}(\mathbb{C}[t]) = \mathbb{C}(t) = K\).Therefore \(\mathbb{C}[t]\) is the integral closure of \(A\) in \(K\), with normalization isomorphism \(\widetilde{A} \cong \mathbb{C}[t]\).
4. Conclusion:\(A\) is an integral domain, its real points form a nodal cubic curve with a loop on \([0, 1]\), and its integral closure is \(\mathbb{C}[t]\) where \(t = \frac{y}{x-1}\). Q.E.D.