Find the irreducible components of the following algebraic sets over \(\CC\).
\(V\big(y^4 - x^2,\ y^4 - x^2 y^2 + x y^2 - x^3\big) \subseteq \AA^2\).
\(V\big(y^2 - xz,\ z^2 - y^3\big) \subseteq \AA^3\).
Solution
1. In part (a), the first equation factors as\[ y^4-x^2 = (y^2-x)(y^2+x). \]Hence every point lies on one of the two branches\[ x=y^2 \qquad\text{or}\qquad x=-y^2. \]
This is the displayed factorization over \(\CC\).
2. On the branch \(x=-y^2\), the second equation vanishes identically.
Substituting \(x=-y^2\) gives \[ \begin{aligned} y^4-x^2y^2+xy^2-x^3 &= y^4-y^6-y^4+y^6\\ &=0. \end{aligned} \] Thus the entire parabola \[ C=V(x+y^2) \] lies in the algebraic set.
3. On the branch \(x=y^2\), the second equation vanishes exactly at\[ (0,0),\qquad(1,1),\qquad(1,-1). \]
Substituting \(x=y^2\) gives \[ \begin{aligned} y^4-x^2y^2+xy^2-x^3 &= y^4-y^6+y^4-y^6\\ &= 2y^4(1-y^2). \end{aligned} \] Over \(\CC\) this vanishes exactly when \[ y=0,\quad y=1,\quad\text{or}\quad y=-1. \] Since \(x=y^2\), these give the three displayed points.
4. The irreducible components in part (a) are\[ \boxed{ V(x+y^2),\qquad \{(1,1)\},\qquad \{(1,-1)\}. } \]
By steps <1>1–<1>3, the algebraic set is \[ C\cup\{(1,1),(1,-1)\}. \] The point \((0,0)\) from step 3 already lies on \(C\).
The curve \(C\) is irreducible because \[ \CC[x,y]/(x+y^2)\cong\CC[y] \] is an integral domain. Each singleton is irreducible and neither \((1,1)\) nor \((1,-1)\) lies on \(C\), since \[ 1+(\pm1)^2=2\ne0. \] Therefore none of the three listed irreducible closed subsets is contained in another, so they are precisely the irreducible components.
5. In part (b), every point satisfying\[ z^2=y^3 \]either has \(y=z=0\), or can be written uniquely in the form\[ y=t^2,\qquad z=t^3 \]with \(t=z/y\).
If \(y=0\), then \(z^2=0\), so \(z=0\).
If \(y\ne0\), put \[ t=\frac zy. \] Then \[ t^2 = \frac{z^2}{y^2} = \frac{y^3}{y^2} =y, \] and therefore \[ t^3=ty=\frac zy\,y=z. \]
6. On the locus \(y=z=0\), the first equation imposes no condition on\(x\), giving the line\[ L=V(y,z). \]
If \(y=z=0\), then \[ y^2-xz=0 \] for every \(x\in\CC\). Hence this locus is exactly the \(x\)-axis.
7. Away from \(y=z=0\), the first equation forces \(x=t\), so theremaining points lie on\[ C'=V(y-x^2,\ z-x^3). \]
Using step 5, \[ y=t^2,\qquad z=t^3 \] with \(t\ne0\). The equation \[ y^2=xz \] becomes \[ t^4=xt^3. \] Since \(t\ne0\), this gives \(x=t\). Thus such a point is \[ (t,t^2,t^3). \] Conversely every point of this form satisfies both original equations, including \(t=0\). Its image is exactly \[ V(y-x^2,z-x^3). \]
8. The irreducible components in part (b) are\[ \boxed{ V(y,z),\qquad V(y-x^2,\ z-x^3). } \]
Steps <1>5–<1>7 show that the algebraic set is \(L\cup C'\). Moreover, \[ \CC[x,y,z]/(y,z)\cong\CC[x] \] and \[ \CC[x,y,z]/(y-x^2,z-x^3)\cong\CC[x], \] so both \(L\) and \(C'\) are irreducible.
They are distinct and neither contains the other: for example \[ (1,0,0)\in L\sm C', \qquad (1,1,1)\in C'\sm L. \] Thus they are exactly the irreducible components.
9. Q.E.D.
Step 4 gives the components in part (a), and step 8 gives the components in part (b).