Let $V=V(y^2-x(x^2-1)$. It's easy to know $y^2-x(x^2-1)$ is irreducible in $\mathbb{C}[x,y]$, then $V$ is an irreducible curve in $\mathbb{A}^2(\mathbb{C})$.
If we consider it in $\mathbb{A}^2(\mathbb{R})$, $y^2-x(x^2-1)$ is also irreducible and infinite, hence $V$ is also irreducible. But in this case, its picture shows $V$ may be reducible with two components.
I'm confused with this example. Can you give me an explaination? Thank you!
The picture shows that the curve has two connected components in the Euclidean topology. That does not mean it has two irreducible components in the Zariski topology. In fact, the curve is irreducible in the Zariski topology, and it is a nice exercise to show that if a Zariski closed set contains one of the two connected components (in the Euclidean topology) then it also contains the other.