# Intersection of irreducible components of a smooth curve

Say $V$ is a smooth projective curve. Is it true that the irreducible components of $V$ don't meet? I've heard something along the lines of "a point contained in two components can't be smooth", but I'm failing to see why this is true.

Thanks!

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However for plane curves things are much simpler. Suppose you have an affine curve $C \subseteq \mathbb{C}^2$ which is defined by the polynomial $F = F_1 \cdot F_2$, where $F_1, F_2 \in \mathbb{C}[x,y]$ are two polynomials. Suppose that there exists a point $p \in \mathbb{C}^2$ such that $F_1(p) = F_2(p) = 0$, i.e. the curves $\{ F_1 = 0 \}$ and $\{ F_2 = 0 \}$ intersect in the point $p$. You should be able to prove that the partial derivatives of $F$ in $p$ vanish, hence $p$ is a singular point of $C$.
Notice that the theorem cited at the beginning is true in all dimensions and the reasoning with partial derivatives works for hypersurfaces in $\mathbb{C}^n$ (or $\mathbb{P}^n$). Hence, a smooth point of a (reduced) algebraic variety $X$ over $\mathbb{C}$ belongs to a unique irreducible component of $X$.