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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!

M Turgeon
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YZhou
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  • It seems that we should further assume that $G$ is non-vanishing on $\cap V_i$, otherwise there will be obvious counterexamples. – YZhou Mar 06 '12 at 12:28

1 Answers1

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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 $h_\alpha$.

brunoh
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    So sorry to bother you, as I realise this post is 12 years old, but I don't think I understand your correction. I agree that $f_\alpha k_\alpha g=f_\alpha h_\alpha$ on $U_f$, but I'm not seeing how to use this in the proof. Would you be able to spell it out a little more explicitly? – Joe Aug 14 '24 at 21:28
  • By the way, it is written in my copy that "we can assume that $U_\alpha=U_f\cap U_{f_\alpha}$ for some collection $f_\alpha$", so that part of the proof from Harris seems fine. – Joe Aug 14 '24 at 21:30
  • @Joe If I remember well, as the OP mentioned in his question, all the equations have to be defined over at least $U_f$. This is possible if you take into account the fact that as Harris suggested you can take the $U_\alpha$ to be distinguished. In that case the use of the product $f_\alpha.k_\alpha$ allows all the equalities in the proof to be valid on the whole $U_f$ right? – brunoh Aug 15 '24 at 10:41
  • I am also doubtful that it works (as the sum at the end is over all indices $\alpha$). But one may use the fact that the last equality in the book is satisfied on the nonempty intersection of the $U_\alpha$'s, which is an open set, and thus also on the whole space. – user111 Feb 26 '25 at 12:18
  • @user111 Why do you think my trick doesn't work? – brunoh Feb 27 '25 at 11:52
  • Indeed $f_\alpha k_\alpha g=f_\alpha h_\alpha$ on $U_f$ but in the final sum, one needs $f_1$ in the first term, $f_2$ in the second term,...So, I think it works if one multiply the final identity by the (finite) product of all the $f_\alpha$'s, and then one gets what is needed on the intersection of the $U_\alpha$ (as in my previous comment). – user111 Feb 27 '25 at 17:11
  • @user111 Thanks for your comment, but I explained that in the proof given by Prof. Harris, you can substitue $k_{\alpha}, h_{\alpha}$ by their products with $f_{\alpha}$, which are defined over the whole $U_f$. Then the final sum stay the same (no need for $f_1, f_2$ ...) – brunoh Feb 28 '25 at 15:14