Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
add comment

1 Answer 1

up vote 7 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|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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.