Sign up ×

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.


share|cite|improve this question

1 Answer 1

up vote 8 down vote accepted

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 $X$.

share|cite|improve this answer
Dear Andrea, This is a nice answer, except that I wouldn't call the statement about local rings in your first sentence "hard". Maybe "non-trivial". Regards, – Matt E May 19 '12 at 12:25

Your Answer


