Prove that the cuspidal cubic \(Y \subseteq \AA^2_\CC\) with equation \(x^3 - y^2 = 0\) is irreducible.
Hint
Express \(Y\) as the image of \(\AA^1\) under a continuous map.
Solution
1. The polynomial map\[ \phi:\AA^1_\CC\longrightarrow\AA^2_\CC, \qquad t\longmapsto(t^2,t^3) \]has image contained in \(Y\).
For every \(t\in\CC\), \[ (t^2)^3-(t^3)^2=t^6-t^6=0. \] Hence \(\phi(t)\in Y\).
2. The map \(\phi:\AA^1_\CC\to Y\) is surjective.
Let \[ (x,y)\in Y, \qquad x^3=y^2. \] If \(x=0\), then \(y=0\), and \[ (x,y)=\phi(0). \]
If \(x\ne0\), put \[ t=\frac yx. \] Then \[ t^2 = \frac{y^2}{x^2} = \frac{x^3}{x^2} =x \] and \[ t^3 = t\,t^2 = \frac yx\,x =y. \] Thus \((x,y)=\phi(t)\).
3. The variety \(Y\) is irreducible.
The affine line \(\AA^1_\CC\) is irreducible because its coordinate ring \(\CC[t]\) is an integral domain. The polynomial map \(\phi\) is a morphism, hence continuous in the Zariski topology. A continuous image of an irreducible topological space is irreducible, and step 2 gives \[ \phi(\AA^1_\CC)=Y. \] Therefore \[ \boxed{Y\text{ is irreducible}.} \]
4. Q.E.D.
Steps <1>1–<1>2 identify \(Y\) as the image of the irreducible affine line, and step 3 applies the irreducible-image criterion.