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I want to understand one of the properties of the action of a group ring on cyclotomic Galois extension.

If $\zeta_n$ is primitive root of unity, $G = Gal(\mathbb{Q}(\zeta_n) / \mathbb{Q})$ and $f : G \rightarrow \mathbb{Z}$ is an element of the group ring $\mathbb{Z}[G]$, the action of $f$ on an element $x \in \mathbb{Q}(\zeta_n)$ is defined as: $$x^f = \prod_{\sigma \in G} \sigma(x)^{f(\sigma)}$$

The property I'm trying to prove is:

$$(x_1 + x_2)^f = x_1^f + x_2^f$$

My calculations go like this:

$$(x_1 + x_2)^f = \prod_{\sigma \in G}\sigma(x_1 + x_2)^{f(\sigma)} = \prod_{\sigma \in G}(\sigma(x_1) + \sigma(x_2))^{f(\sigma)}$$

$$x_1^f + x_2^f = \prod_{\sigma \in G} \sigma(x_1)^{f(\sigma)} + \prod_{\sigma \in G} \sigma(x_2)^{f(\sigma)}$$

Don't know how to finish the proof. Any help with this? Thanks in advance!

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    $\begingroup$ As you see it is not true. $x^f = \prod_{\sigma \in G} \sigma(x)^{f(\sigma)}$ is an action of $\mathbb{Z}[G]$ on the multiplicative group $\mathbb{Q}(\zeta_n)^*$ $\endgroup$ – reuns Jun 8 '17 at 2:24
  • $\begingroup$ What user1952009 says! $\Bbb{Q}(\zeta_n)^*$ becomes a $G$-module, but the "addition" operation of this module is the multiplication of $\Bbb{Q}(\zeta_n)^*$. Your identity cannot possible hold. For an obvious example consider the case of $f=2*1_G$, when $x^f=x^2$. $\endgroup$ – Jyrki Lahtonen Jun 8 '17 at 6:03
  • $\begingroup$ @user1952009 thank you for the answer. I saw the statement in this article (on the 3rd page) and in the Henri Cohen's book A course in computational Algebraic number theory (447 page). Is the statement really completely wrong? $\endgroup$ – brick Jun 8 '17 at 12:53
  • $\begingroup$ Did you read what Jyrki wrote ? It is true if you replace $+$ by $\times$ the multiplication of $\mathbb{Q}(\zeta_n)^*$. $\endgroup$ – reuns Jun 8 '17 at 13:17
  • $\begingroup$ Yeah, @user1952009, I read it. But did you saw the article I linked in my previous comment? There are given both of the properties $(x_1 + x_2)^f = x_1^f + x_2^f$ and $(x_1x_2)^f = x_1^fx_2^f$. I agree that the second one is OK, but the first one... $\endgroup$ – brick Jun 8 '17 at 13:27

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