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.

I think the proof in for Lemma 2.1 in Joe Harris's book Algebraic Geometry, A First Course, does not work. (The statement is on Page 19, and the proof on Page 61.) The proof fails because that $gk_\alpha=h_\alpha$ is valid on $U_\alpha$ only, while the last but one equation on Page 61 must work on the whole domain of the regular function.

I have been trying to give a proof or find a counterexample. The proof for the case in which the variety is irreducible is easy. In the reducible case, I managed to reduce the question to the following one:


Suppose we are working with a algebraically closed field K and a regular function on an open subset of an affine variety is by definition a function locally representable as $F/G$, with $F,G\in K[x_1,...,x_n]$, and $G$ non-vanishing in a neighbourhood. Suppose we have a variety $V\subset \mathbb A^n$, $V=\cup V_i$ is the decomposition into irreducible components. Suppose we have $F,G\in K[x_1,...,x_n]$, such that $G$ divides $F$ in each $K[x_1,...,x_n]/I(V_i)$. Does it follow that $G$ divides $F$ in $K[x_1,...,x_n]/I(V)$?


Intuitively, this means that $I(F)~$ contains $I(G)~$ in each component implies the same thing on the whole variety. It seems quite reasonable. But I am not sure whether there will be any abnormality when considering "multiplicities".

P.S.You may assume that $\cap V_i$ is nonempty, this is adequate for my purpose. Though I don't think this further assumption will help.

Thank you!

share|improve this question
    
It seems that we should further assume that $G$ is non-vanishing on $\cap V_i$, otherwise there will be obvious counterexamples. –  Andrew Mar 6 '12 at 12:28

1 Answer 1

Well, I noticed the same problem than you. But in my opinion, the proof on page 61 can be made correct with just the slight following precision : you get $k_\alpha g=h_\alpha$ only on $U_\alpha$. But then you can choose the $U_\alpha$ to be distinguished open sets $U_{f_\alpha}$. Then you have $f_\alpha k_\alpha g=f_\alpha h_\alpha$ on the whole $U_f$ (or even $\mathbb{A}^n)$ since $f_\alpha$ is zero outside of $U_{f_\alpha}$. Then you finish the proof as indicated using $f_\alpha k_\alpha$ and $f_\alpha h_\alpha$ instead of $k_\alpha$ and $h_\alpha$.

share|improve this answer

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.