# Is this plane curve irreducible?

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!

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 $$\mathbb{A}^2_{\mathbb{C}}$$.
• $y^2-x^2(x+1)=(y-p(x))(y-q(x))$ is impossible. Is this a proof? Nov 4, 2022 at 7:29
• 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. Nov 4, 2022 at 8:27