Say $X\subset \mathbb{C}^{n}$ is an algebraic set. Is it true the irreducible components of $X$ in the algebraic sense coincide with irreducible components of $X$ in the analytic sense?
A way to rephrase this question is to ask if the irreducible components of $X$ in the algebraic sense coincide with the closure of the connected components (In the euclidian topology) of $X\backslash Sing(X)$?
Over $\mathbb{R}$ this is incorrect, for instance $y^2=x^3$ is algebraically irreducible but has 2 analytic components. Over $\mathbb{C}$ I suspect it is correct.
Thank you!
Edit: As the answer bellow shows, it is true. However I actually was interested in this question when $\mathbb{C}^{n}$ is switched by some domain like the unit disk. In this case it is false, since one algebraic component may intersect a domain along many analytic components.
Yes, this is true. The set of smooth points of an irreducible algebraic set over $\Bbb C$ are connected, and thus the connected components of $X\setminus X^{sing}$ correspond exactly to the irreducible components of $X$. This characterization works in both the algebraic and analytic worlds.
By the way, your example over $\Bbb R$ does not work - the cuspidal cubic is the image of $\Bbb R$ under the map $t\mapsto (t^2,t^3)$, and the continuous image of an irreducible set is again irreducible. $\Bbb R$ is irreducible, as the analytic closed sets are discrete - if we had an analytic function who's zero set had an accumulation point, it would define an analytic function in a small complex neighborhood of $\Bbb R\subset \Bbb C$, and this function would be identically zero by the properties of analytic functions over $\Bbb C$.