Skip to main content
A new proof written, covering the general case
Source Link
Mikhail Borovoi
  • 14.1k
  • 2
  • 31
  • 71

WriteEDIT: I treat the general case. Write ${\Bbb F}={\Bbb F}_q$ where $q=p^l$ for some natural $l$. I can treat only the case $l=1$, that is,Then $q=p$${{\Bbb F}}_q\supseteq {{\Bbb F}}_p={\Bbb Z}/p{\Bbb Z}$.

It follows from the definitionsUsing a comment of @DavidLoeffler, we obtain that for an ${\Bbb F}$-vector space $W$, we can identify \begin{multline*} W^*={\rm Hom}_{{\Bbb F}}(W, {\Bbb F})\cong {\rm Hom}_{{{\Bbb F}_p}}(W, {\Bbb F}_p) ={\rm Hom}(W, {\Bbb Z}/p{\Bbb Z})\\ \cong{\rm Hom}(W, \tfrac1p{\Bbb Z}/{\Bbb Z}) ={\rm Hom}(W, {\Bbb Q}/{\Bbb Z})=W^\vee. \end{multline*}

By definition, our $V$ is a finite $G_K$-module and that $$ V^*(1)={\rm Hom}(V,\mu_p)={\rm Hom}(V, \overline K^\times)=V'$$$$ V^*(1)\cong{\rm Hom}_{{\Bbb F}}(V,{{\Bbb F}})\otimes_{{{{\Bbb F}_p}}}\mu_p$$ where $\mu_p$ denotes the group of roots of unity of degree dividing $p$ in    $\overline K^\times$.

Moreover, for an ${\Bbb F_p}$-vector space $W$, we We can identify $$ W^*={\rm Hom}(W, {\Bbb F})={\rm Hom}(W, {\Bbb Z}/p{\Bbb Z}) ={\rm Hom}(W, \tfrac1p{\Bbb Z}/{\Bbb Z}) ={\rm Hom}(W, {\Bbb Q}/{\Bbb Z})=W^\vee. $$\begin{multline*} V^*(1)\cong{\rm Hom}_{{\Bbb F}}(V,{{\Bbb F}})\otimes_{{{{\Bbb F}_p}}}\mu_p\cong {\rm Hom}(V,{\Bbb Z}/p{\Bbb Z})\otimes_{{{\Bbb F}_p}} \mu_p\\ \cong {\rm Hom}(V,\mu_p)={\rm Hom}(V, \overline K^\times)=V'. \end{multline*}

We conclude that $$H^{2-i}(G_K,V^*(1))^*=H^{2-i}(G_K, V')^\vee. $$$$H^{2-i}(G_K,V^*(1))^*\cong H^{2-i}(G_K, V')^\vee. $$ Now we see that when $q=p$, the the second assertion of the question is a special case of the first one.

Write ${\Bbb F}={\Bbb F}_q$ where $q=p^l$ for some natural $l$. I can treat only the case $l=1$, that is, $q=p$.

It follows from the definitions that $V$ is a finite $G_K$-module and that $$ V^*(1)={\rm Hom}(V,\mu_p)={\rm Hom}(V, \overline K^\times)=V'$$ where $\mu_p$ denotes the group of roots of unity of degree dividing $p$ in  $\overline K^\times$.

Moreover, for an ${\Bbb F_p}$-vector space $W$, we can identify $$ W^*={\rm Hom}(W, {\Bbb F})={\rm Hom}(W, {\Bbb Z}/p{\Bbb Z}) ={\rm Hom}(W, \tfrac1p{\Bbb Z}/{\Bbb Z}) ={\rm Hom}(W, {\Bbb Q}/{\Bbb Z})=W^\vee. $$

We conclude that $$H^{2-i}(G_K,V^*(1))^*=H^{2-i}(G_K, V')^\vee. $$ Now we see that when $q=p$, the second assertion is a special case of the first one.

EDIT: I treat the general case. Write ${\Bbb F}={\Bbb F}_q$ where $q=p^l$ for some natural $l$. Then ${{\Bbb F}}_q\supseteq {{\Bbb F}}_p={\Bbb Z}/p{\Bbb Z}$.

Using a comment of @DavidLoeffler, we obtain that for an ${\Bbb F}$-vector space $W$, we can identify \begin{multline*} W^*={\rm Hom}_{{\Bbb F}}(W, {\Bbb F})\cong {\rm Hom}_{{{\Bbb F}_p}}(W, {\Bbb F}_p) ={\rm Hom}(W, {\Bbb Z}/p{\Bbb Z})\\ \cong{\rm Hom}(W, \tfrac1p{\Bbb Z}/{\Bbb Z}) ={\rm Hom}(W, {\Bbb Q}/{\Bbb Z})=W^\vee. \end{multline*}

By definition, our $V$ is a finite $G_K$-module and $$ V^*(1)\cong{\rm Hom}_{{\Bbb F}}(V,{{\Bbb F}})\otimes_{{{{\Bbb F}_p}}}\mu_p$$ where $\mu_p$ denotes the group of roots of unity of degree dividing $p$ in  $\overline K^\times$. We can identify \begin{multline*} V^*(1)\cong{\rm Hom}_{{\Bbb F}}(V,{{\Bbb F}})\otimes_{{{{\Bbb F}_p}}}\mu_p\cong {\rm Hom}(V,{\Bbb Z}/p{\Bbb Z})\otimes_{{{\Bbb F}_p}} \mu_p\\ \cong {\rm Hom}(V,\mu_p)={\rm Hom}(V, \overline K^\times)=V'. \end{multline*}

We conclude that $$H^{2-i}(G_K,V^*(1))^*\cong H^{2-i}(G_K, V')^\vee. $$ Now we see that the second assertion of the question is a special case of the first one.

added 11 characters in body
Source Link
Mikhail Borovoi
  • 14.1k
  • 2
  • 31
  • 71

Write ${\Bbb F}={\Bbb F}_q$ where $q=p^l$ for some natural $l$. First I assume that I can treat only the case $l=1$, that is, $q=p$. It

It follows from the definitions that $V$ is a finite $G_K$-module and that $$ V^*(1)={\rm Hom}(V,\mu_p)={\rm Hom}(V, \overline K^\times)=V'$$ where $\mu_p$ denotes the group of roots of unity of degree dividing $p$ in $\overline K^\times$.

Moreover, for an ${\Bbb F_p}$-vector space $W$, we can identify $$ W^*={\rm Hom}(W, {\Bbb F})={\rm Hom}(W, {\Bbb Z}/p{\Bbb Z}) ={\rm Hom}(W, \tfrac1p{\Bbb Z}/{\Bbb Z}) ={\rm Hom}(W, {\Bbb Q}/{\Bbb Z})=W^\vee. $$ We

We conclude that $$H^{2-i}(G_K,V^*(1))^*=H^{2-i}(G_K, V')^\vee. $$ Now we see that when $q=p$, the second assertion is a special case of the first one.

Write ${\Bbb F}={\Bbb F}_q$ where $q=p^l$ for some natural $l$. First I assume that $l=1$, that is, $q=p$. It follows from the definitions that $V$ is a finite $G_K$-module and that $$ V^*(1)={\rm Hom}(V,\mu_p)={\rm Hom}(V, \overline K^\times)=V'$$ where $\mu_p$ denotes the group of roots of unity of degree dividing $p$ in $\overline K^\times$.

Moreover, for an ${\Bbb F_p}$-vector space $W$, we can identify $$ W^*={\rm Hom}(W, {\Bbb F})={\rm Hom}(W, {\Bbb Z}/p{\Bbb Z}) ={\rm Hom}(W, \tfrac1p{\Bbb Z}/{\Bbb Z}) ={\rm Hom}(W, {\Bbb Q}/{\Bbb Z})=W^\vee. $$ We conclude that $$H^{2-i}(G_K,V^*(1))^*=H^{2-i}(G_K, V')^\vee. $$ Now we see that when $q=p$, the second assertion is a special case of the first one.

Write ${\Bbb F}={\Bbb F}_q$ where $q=p^l$ for some natural $l$. I can treat only the case $l=1$, that is, $q=p$.

It follows from the definitions that $V$ is a finite $G_K$-module and that $$ V^*(1)={\rm Hom}(V,\mu_p)={\rm Hom}(V, \overline K^\times)=V'$$ where $\mu_p$ denotes the group of roots of unity of degree dividing $p$ in $\overline K^\times$.

Moreover, for an ${\Bbb F_p}$-vector space $W$, we can identify $$ W^*={\rm Hom}(W, {\Bbb F})={\rm Hom}(W, {\Bbb Z}/p{\Bbb Z}) ={\rm Hom}(W, \tfrac1p{\Bbb Z}/{\Bbb Z}) ={\rm Hom}(W, {\Bbb Q}/{\Bbb Z})=W^\vee. $$

We conclude that $$H^{2-i}(G_K,V^*(1))^*=H^{2-i}(G_K, V')^\vee. $$ Now we see that when $q=p$, the second assertion is a special case of the first one.

deleted 54 characters in body
Source Link
Mikhail Borovoi
  • 14.1k
  • 2
  • 31
  • 71

Write ${\Bbb F}={\Bbb F}_q$ where $q=p^l$ for some natural $l$. First I assume that $l=1$, that is, $q=p$. It follows from the definitions that $V$ is a finite $G_K$-module and that $$ V^*(1)={\rm Hom}(V,\mu_p)={\rm Hom}(V, \overline K^\times)=V'$$ where $\mu_p$ denotes the group of roots of unity of degree dividing $p$ in $\overline K^\times$.

Moreover, for an ${\Bbb F_p}$-vector space $W$, we can identify $$ W^*={\rm Hom}(W, {\Bbb F})={\rm Hom}(W, {\Bbb Z}/p{\Bbb Z}) ={\rm Hom}(W, \tfrac1p{\Bbb Z}/{\Bbb Z}) ={\rm Hom}(W, {\Bbb Q}/{\Bbb Z})=W^\vee. $$ We conclude that $$H^{2-i}(G_K,V^*(1))^*=H^{2-i}(G_K, V')^\vee. $$ Now we see that when $q=p$, the second assertion is a special case of the first one. I will try to type the case of arbitrary $l$ tomorrow,

Write ${\Bbb F}={\Bbb F}_q$ where $q=p^l$ for some natural $l$. First I assume that $l=1$, that is, $q=p$. It follows from the definitions that $V$ is a finite $G_K$-module and that $$ V^*(1)={\rm Hom}(V,\mu_p)={\rm Hom}(V, \overline K^\times)=V'$$ where $\mu_p$ denotes the group of roots of unity of degree dividing $p$ in $\overline K^\times$.

Moreover, for an ${\Bbb F_p}$-vector space $W$, we can identify $$ W^*={\rm Hom}(W, {\Bbb F})={\rm Hom}(W, {\Bbb Z}/p{\Bbb Z}) ={\rm Hom}(W, \tfrac1p{\Bbb Z}/{\Bbb Z}) ={\rm Hom}(W, {\Bbb Q}/{\Bbb Z})=W^\vee. $$ We conclude that $$H^{2-i}(G_K,V^*(1))^*=H^{2-i}(G_K, V')^\vee. $$ Now we see that when $q=p$, the second assertion is a special case of the first one. I will try to type the case of arbitrary $l$ tomorrow,

Write ${\Bbb F}={\Bbb F}_q$ where $q=p^l$ for some natural $l$. First I assume that $l=1$, that is, $q=p$. It follows from the definitions that $V$ is a finite $G_K$-module and that $$ V^*(1)={\rm Hom}(V,\mu_p)={\rm Hom}(V, \overline K^\times)=V'$$ where $\mu_p$ denotes the group of roots of unity of degree dividing $p$ in $\overline K^\times$.

Moreover, for an ${\Bbb F_p}$-vector space $W$, we can identify $$ W^*={\rm Hom}(W, {\Bbb F})={\rm Hom}(W, {\Bbb Z}/p{\Bbb Z}) ={\rm Hom}(W, \tfrac1p{\Bbb Z}/{\Bbb Z}) ={\rm Hom}(W, {\Bbb Q}/{\Bbb Z})=W^\vee. $$ We conclude that $$H^{2-i}(G_K,V^*(1))^*=H^{2-i}(G_K, V')^\vee. $$ Now we see that when $q=p$, the second assertion is a special case of the first one.

Post Undeleted by Mikhail Borovoi
added 350 characters in body
Source Link
Mikhail Borovoi
  • 14.1k
  • 2
  • 31
  • 71
Loading
Post Deleted by Mikhail Borovoi
Source Link
Mikhail Borovoi
  • 14.1k
  • 2
  • 31
  • 71
Loading