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Let $(R,\mathfrak{m})$ be a Noetherian local ring of positive prime characteristic $p$ and let $F$ be the Frobenius functor. Write $d$ for dimension of $R$. Assume that for some $0\leq i< d $ the local cohomology module $\mathrm{H}^i_{\mathfrak{m}}(R)$ is of finite length. If we write $\ell(\cdot)$ for length of a module of finite length, I am interested to know whether it is true that $$\ell(F(\mathrm{H}^i_{\mathfrak{m}}(R)))=p^d\cdot\ell(\mathrm{H}^i_{\mathfrak{m}}(R))?$$ The trivial case is when $\mathrm{H}^i_{\mathfrak{m}}(R)=0$. Any results from the literature in special cases where this formula holds, or any example in which the formula does not hold would be interesting. I don't know of any special case or example, except for the trivial case.

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By the Frobenius functor. What functor do you mean? Do you mean the tensor functor, or the restriction of scalars functor, or a different functor? I assume you mean the former (which Long explains well below). –  Karl Schwede Nov 29 '11 at 3:18
    
Yes, I meant the tensor functor over Frobenius. –  Mahdi Majidi-Zolbanin Nov 29 '11 at 4:11
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up vote 5 down vote accepted

Let $p=2$, $R=k[[x,y]]/(x^2,xy)$. Then $H^0_m(R) \cong R/m= k$. $F(k) = R/m^{[2]}= k[[x,y]]/(x^2,xy,y^2)$ has length $3 \neq 2\times 1$.

In general for a finite length module $M$, the condition that $\ell(F(M))=p^d\ell(M)$ is pretty restrictive. For $M=k$ this forces $R$ to be regular (Kunz). Also, when $R$ is a complete intersection such condition forces $M$ to have finite projective dimension, see Claudia Miller's survey.

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Your example is good. So more generally every time $\mathrm{H}^i_{\mathfrak{m}}(R)\cong k$, the formula cannot hold by Kunz, as you pointed out. Over a complete intersection the local cohomology modules are zero in the specified range which does not give any information about my formula. It would be interesting to describe all modules of finite length for which $\ell(F(M))=p^d\cdot\ell(M)$ holds (eigenmodules of Froben). As you pointed out, when $R$ is a complete intersection, the answer is all modules of finite projective dimension. But the answer is not always finite projective dimension. –  Mahdi Majidi-Zolbanin Nov 29 '11 at 4:10
    
@ Hailong: I have a question as follows: Let $M$ be a finitely generated $R$-module of dimension $d$ such that $H^i_{\frak m} (M)$ has finite length for all $i < d$. In what coditions $H^i_{\frak m} (F(M))$ has finite length for all $i < d$? I think if $R$ is regular, it is true since Frobenius functor is flat. But is there any weaker condition? –  Pham Hung Quy Dec 2 '11 at 11:19
    
Moreover, we can ask a more general question about the finilety generated property of $H^i_{\frak a} (F(M))$ for all $i < t$, if $H^i_{\frak a} (M)$ is finitely generated for all $i < t$. –  Pham Hung Quy Dec 2 '11 at 11:25
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