1
$\begingroup$

I want to define a plane curve in $\mathbb{A}^2(\mathbb{C})$ by the polynomial $f(x,y)=x(x-1)^2-(y-1)^2=0$ where $(x,y)\in\mathbb{A}^2(\mathbb{C})$, but my goal is for the plane curve to be irreducible.

How do I tell/prove that this curve is irreducible?

Thanks so much!

$\endgroup$
0

1 Answer 1

6
$\begingroup$

I start from a change of coordinates: \begin{equation*} \begin{cases} x^{\prime}=x-1\\ y^{\prime}=y-1 \end{cases} \end{equation*} so the plane curve in the new coordinates has equation $\left(x^{\prime}+1\right)\left(x^{\prime}\right)^2-\left(y^{\prime}\right)^2=0$; for simplicity, I rewrite the new equation in $y^2=x^2(x+1)$.

The best and simpler easy reasoning to prove the irreducibility of $X=\{(x,y)\in\mathbb{A}^2_{\mathbb{C}}\mid y^2=x^2(x+1)\}$, in my knowldge, is the following: because $x^2(x+1)$ must be a square of a polynomial then $x+1=t^2$, so $x=t^2-1$ and $y^2=(t^2-1)^2t^2\Rightarrow y=\pm t(t^2-1)$. From all this, $X$ is the image of $\mathbb{A}^1_{\mathbb{C}}$ via the continuous map (with respect to Zariski topology) \begin{equation*} \varphi:t\in\mathbb{A}^1_{\mathbb{C}}\to\left(t^2-1,t\left(t^2-1\right)\right)\in\mathbb{A}^2_{\mathbb{C}}; \end{equation*} because the image of irreducible sets via continuous map is irreducible as well, $X$ is an irreducible subset of $\mathbb{A}^2_{\mathbb{C}}$.

$\endgroup$
2
  • $\begingroup$ $y^2-x^2(x+1)=(y-p(x))(y-q(x))$ is impossible. Is this a proof? $\endgroup$
    – Bob Dobbs
    Nov 4, 2022 at 7:29
  • $\begingroup$ No, you have to prove $\not\exists p(x,y),q(x,y)\in\mathbb{C}[x,y]$ such that $\deg p,\deg q\geq1$ and $p\cdot q=y^2-x^2(x+1)$. Maybe, you could reduce this statement to your claim claim. $\endgroup$ Nov 4, 2022 at 8:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .