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In order to construct an example of Herbrand quotient, I want to know the cohomology group of $H^i(\mathbb{Z}/p\mathbb{Z}, \mathbb{Z})$ and $H^i(\mathbb{Z}/p\mathbb{Z}, \mathbb{Z}/p\mathbb{Z})$ for $i = 0, 1$.

When $i = 0$, I know $H^0(G, M) = M^G$.

Therefore, I guess $H^0(\mathbb{Z}/p\mathbb{Z}, \mathbb{Z}) = \{x \in \mathbb{Z} \mid \sigma + x = x, x \in \mathbb{Z}/p\mathbb{Z}\} = p\mathbb{Z}$, and $H^0(\mathbb{Z}/p\mathbb{Z}, \mathbb{Z}/p\mathbb{Z}) = \{x \in \mathbb{Z}/p\mathbb{Z} \mid \sigma + x = x, x \in \mathbb{Z}/p\mathbb{Z}\} = p\mathbb{Z}$. Is this right?

But I cannot calculate the case of $i = 1$.

I would appreciate if you could help me calculating $H^1(\mathbb{Z}/p\mathbb{Z}, \mathbb{Z}/p\mathbb{Z})$ and $H^1(\mathbb{Z}/p\mathbb{Z}, \mathbb{Z})$. Thank you.

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    In future, you should use MathJax to format your questions. – Michael Albanese Dec 29 '19 at 04:19
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    I am not used to working with Mathjax, but I will do my best to gradually write in a good format. Thank you very much for your help. –  Dec 29 '19 at 04:21
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    $\mathbb{Z}/p\mathbb{Z}$ should act trivially on $\mathbb{Z}$. First of all "${x \in \mathbb{Z} \mid \ldots, x \in \mathbb{Z}/p\mathbb{Z}}$" doesn't make sense; you meant $\sigma \in \mathbb{Z}/p\mathbb{Z}$—but then $\sigma + x$ doesn't make sense. – Trevor Gunn Dec 29 '19 at 04:46
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    Also I don't think $\mathbb{Z}/p\mathbb{Z}$ acts non-trivially on $\mathbb{Z}/p\mathbb{Z}$. If the "action" is by multiplication that doesn't work because multiplication by $0$ is not an automorphism. If the "action" is by addition, that doesn't work because the action needs to fix the identity element. – Trevor Gunn Dec 29 '19 at 05:00

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You need to specify the action of $C_p=\Bbb Z/p\Bbb Z$ on $M$ in order to make $H^i(C_p,M)$ meaningful. That is you need an automorphism $\alpha$ of $M$ with $\alpha^p$ equalling the identity. Then, for a generator $g$ of $C_p$ you define $m\cdot g=\alpha(m)$.

In any case, one can take any Abelian group $M$ and the trivial action of $C_p$, that is $m\cdot g=m$. Even in this case, the cohomology is interesting. In this case $H^0(C_p,M)=M^G=M$. The Tate cohomology (with occurs in the Herbrand quotient) is slightly more interesting: $\hat H^0(C_p,M)=M^G/T(M)$ where $T$ is the trace map: $$T(m)=\sum_{k=0}^{p-1}m\cdot g^k.$$ When the action is trivial, then $T(m)=pm$, and so $\hat H^0(C_p,M)=M/pM$. In particular, $\hat H^0(C_p,\Bbb Z)\cong \Bbb Z/p\Bbb Z$ and $\hat H^0(C_p,\Bbb Z/p\Bbb Z)\cong \Bbb Z/p\Bbb Z$.

What about $H^1$. For the cyclic group $C_p$, $$H_1(C_p,M)\cong\frac{\ker T}{\{m-m\cdot g:m\in M\}}.$$ When the action on $M$ is trivial, the denominator vanishes and $T$ is multiplication by $p$ so that $$H_1(C_p,M)\cong\{m\in M:pm=0\}.$$ In particular, $H^1(C_p,\Bbb Z)=\{0\}$ and $H^1(C_p,\Bbb Z/p\Bbb Z)\cong \Bbb Z/p\Bbb Z$.

Angina Seng
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  • Thank you very much! I'll try to understand your answer. Maybe I ask some question about this answer in this comment space, thank you! –  Dec 29 '19 at 05:27
  • Sorry,https://math.stackexchange.com/questions/1217674/values-of-the-herbrand-quotient This answer's 2 looks like contradicting your answer（p＝２）? Why? –  Dec 30 '19 at 15:06
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    @buoyant That's because in that answer the respondent is considering a nontrivial action of $C_2$. – Angina Seng Dec 30 '19 at 19:25
  • Then, the notational system H^i(G,N) is not well defind unless we specify the action of G ？ –  Dec 31 '19 at 04:14
  • Not welldefind means ambiguous here.Ｉ am confused about the different value of the same appearance H^i(G,N) –  Dec 31 '19 at 04:22