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!
 A: I start from a change of cordinates:
\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, the new equation can be rewrite as $y^2=x^2(x+1)$.
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(t^2-1,t(t^2-1))\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 $\mathbb{A}^2_{\mathbb{C}}$.
