Let $G$ be a Galois group and $H$ be its normal subgroup. Let $M$ be a $G$-module.

Consider the restriction map $res: H^1(G,M) \to H^1(H,M)$. Its kernel is given by $H^1(G/H,M^H)$.

On the other hand, there exists the corestriction map $cores: H^1(H,M) \to H^1(G,M)$.

What is known about the kernel of $cores$? Are there any known results that explicitly describe the kernel or embed it into certain cohomology?

I am particularly interested in the following case: Let $K$ be a quadratic number field. Consider $G=Gal(\overline{K}/\Bbb{Q})$ and $H= Gal(\overline{K}/K)$. Let $E$ be an elliptic curve defined over $\Bbb{Q}$. Let $M = E(\overline{K})$. For the restriction map, we can express $ker(res)$ as $H^1(Gal(K/\Bbb{Q}), E(K))$ as mentioned above. However, I am facing difficulty in controlling $ker(cores)$.

  • $\begingroup$ What is $E$? An elliptic curve? Over what field? $\endgroup$ Commented Jun 17, 2023 at 18:42

1 Answer 1


(Not sure why this question comes up naturally. The more interesting question, and the one analogue to the kernel of restriction, is to ask what is the cokernel of corestriction. That turns up a lot. Like in class field theory where the cokernel of the norm map is a central question, while the kernel of the norm map is large and not of the same level of interest.)

The Tate spectral squence (Cohomology of number fields, printed version, Theorem 2.1.11) involves the corestriction map (loc cit Theorem 2.1.12). The resulting five term exact sequence answers the question. Let $K/k$ be a finite extension with Galois group $G$ and let $M$ be a Galois module for $\operatorname{Gal}(\bar k/k)$ whose order is a unit in $k$. Then this is the dual of the said 5-term exact sequence: $$ 0 \leftarrow H_1\bigl(G, H^2(K,M)\bigr) \leftarrow H^1(k,M)\overset{\operatorname{cor}}{\leftarrow} H_0\bigl(G,H^1(K,M)\bigr) \leftarrow H_2\bigl(G,H^2(K,M)\bigr) \leftarrow H^0(k,M) $$ This holds if $H^i(k,M)=0$ for all $i>2$. If $k$ is local, then this condition holds for all $M$ as $\operatorname{scd}(k)=2$ by Corollary 7.2.5.

In the case of a global field, I would switch to the restricted cohomology with respect to a finite set $S$ of places such that $K/k$ and $M$ are unramified outside $S$ and $S$ contains all places above primes dividing the order of $M$ and all infinite places. Write $G_S(k)$ for the Galois group of the maximal extension of $k$ which is unramified outside $k$. One obtains the above 5-terms sequence with cohomology with respect for $G_S(K)$ and $G_S(k)$ as long as $H^i\bigl(G_S(k), M)=0$ for all $i>2$. This condition holds for all finite $M$ as $\operatorname{cd}(G_S(k))=2$, neglecting issues for $p=2$ by adding the assumption that $k$ has no real places. One expects that $\operatorname{scd}(G_S(k))=2$ and this is known if the Leopoldt conjecture holds for $k$ as explained in Corollary 10.3.8. For an elliptic curve $E$ defined over $k$, there is a surjective map $H^3\bigl( G_S(k), E[m]\bigr) \to H^3\bigl( G_S(k), E\bigr)[m]$ for all $m$. Hence for odd $m$ we also have the required vanishing of $M=E(\bar k)$ and for even $m$ one needs to add the hypothesis that $k$ has no real place, to impose local conditions at these infinite places.

For infinite $G$, one has to replace the inner cohomology groups by limits taken over corestriction maps for finite subextensions as in loc.cit.

(All references in Cohomology of number fields are to the printed version on my desk, the numbering of theorems in the online version may be different.)

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    $\begingroup$ Thank you very much. But is your $M$ finite? In the case, like $M=E(\overline{\Bbb{Q}})$, does the same kind of exact sequence exist? $\endgroup$
    – Duality
    Commented Jun 17, 2023 at 17:12
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    $\begingroup$ I have added to be more precise in the case $M$ is not finite. $\endgroup$ Commented Jun 17, 2023 at 18:18
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    $\begingroup$ Maybe I should also comment on the less interesting particular case in the question. Let $W'$ be the set of elements of odd order in $H^1(K,E)$ where $K/\mathbb{Q}$ is quadratic. Then $W' = W^{+} \oplus W^{-}$ where the Galois group acts as $+1$ or $-1$. Of course $W^{+}$ is the odd part of $H^1(\mathbb{Q}, E)$. Then the kernel of corestriction on $W'$ is $W^{-}$ and that is huge as it is the odd part of $H^1(\mathbb{Q}, E^D)$ where $E^D$ is the quadratic twist. Similarly all elements of order $2$ in the image of restriction from the $2$-primary part of $H^1(\mathbb{Q}, E)$ are in this kernel. $\endgroup$ Commented Jun 18, 2023 at 10:55

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