Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The following is a corollary of the Briançon-Skoda theorem:

If $R$ is a regular Noetherian ring of Krull dimension $d$ and $f_1,f_2,...,f_{d+1}\in R$. Then, $f_1^df_2^d...f_{d+1}^d \in (f_1^{d+1},f_2^{d+1},...,f_{d+1}^{d+1})R$

Now, given $R=k[[x_1,...,x_d]]$ where $k$ is a field, then $R$ is a regular local ring, so the corollary applies. I was wondering if there is direct proof of the above corollary for the case of the power series ring. I remember reading that there is a direct proof for the case when $f_1,...,f_{d+1}$ are polynomials. Can anyone point to a reference for this.

share|improve this question
    
I took the liberty of adding the missing cedillas. (I also erased the corresponding apology.) –  José Figueroa-O'Farrill Nov 16 '10 at 4:35
1  
Thanks José (I just copy pasted the accent on that one) –  Timothy Wagner Nov 16 '10 at 5:17
add comment

1 Answer

up vote 2 down vote accepted

It depends on what do you mean by "direct". There is no elementary proof, as far as I known, even for polynomials when $d=2$, see page 8 of this note. When $d=1$, the statement is an easy exercise: one can write $f_1= da, f_2=db$ with $(a,b) = R$ since $R$ is an UFD.

There are a few proofs of this very interesting theorem, analytic (the original one, over complex numbers), using duality theory (Lipman-Sathaye, for all regular rings) and reduction to characteristic $p$ + tight closure (Hochster-Huneke, when $R$ contains a field, see the note in the first paragraph for some exposition of this method) but I am not sure any of them can be called direct. Any new proof will be very exciting!

share|improve this answer
    
Thanks a lot Hailong for all the information. I shall look into these. –  Timothy Wagner Nov 16 '10 at 5:16
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.