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!
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!
The most mature answer to this question would be mentioning the theorem "a regular local ring is a domain", which is a hard fact of commutative algebra. I don't know how to explain this in an elementary way.
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 $X$.