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I already asked this question some days ago on http://math.stackexchange.com/questions/158747/bloch-kato-conjecture-and-wiles-numerical-criterion but didn't receive any response.

In the introduction (p. 14) of their paper on Wiles' proof of FLT, Darmon, Diamond and Taylor state that a numerical criterion of Wiles "seems to be very close to a special case of the Bloch-Kato conjecture".

This numerical criterion is a statement in commutative algebra (Theorem 5.3 on p. 139 in the paper): A surjective morphism $\phi: R \to T$ of augmented complete intersections over the ring of integers $\mathcal{O}$ of a finite extension over $\mathbb{Q}_p$ is an isomorphism iff $$|\ker \pi_R/(\ker \pi_R)^2| \le |\mathcal{O}/\pi_T(\text{Ann}_T(\ker \pi_T))|,$$ where $\pi_R: R \to \mathcal{O}$ is the augmentation.

Question: What is the relation between Wiles' numerical criterion and the Bloch-Kato conjecture ?

In order to be more explicit, let $K$ be a number field and let $V$ be a geometric representation of the absolute Galois group $G_K$ (in essential $V$ is a finite dimensional vector space over $\mathbb{Q}_p$ with a continuous action of $G_K$). Then the Bloch-Kato conjecture can be formulated as $$\dim H_f^1(G_K,V^\ast(1))-\dim H^0(G_K,V^\ast(1))=\operatorname{ord}_{s=0}L(V,s)$$ where $H^1_f$ denotes the Selmer group, $V^\ast$ is the dual representation and $L$ the $L$-function.

Then the question is for which $K, V$ and in what way this formula is close to the numerical criterion described above ?

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Did you read Diamond-Flach-Guo? If so, does it answer your question? – Kevin Buzzard Jun 18 at 22:03
Do you mean archive.numdam.org/ARCHIVE/ASENS/… ? No, I didn't read it so far. But I'll have a look. Thanks. – Ralph Jun 18 at 22:20
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Not an answer, but a quick comment : the Bloch-Kato conjecture consists of two parts, the "rank" part predicting the order of vanishing, and the "Tamagawa number" part predicting the rational factor in the special value. This second part is probably more relevant to the numerical criterion you mention. – François Brunault Jun 19 at 19:48

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