For $H$ = {$a_1$, $a_2$, $a_3$}, let $S_H$ be the symmetric group on $H$.

Identifying $Gal(L/\mathbb{Q})$ with a subgroup of $S_H$, how do we know if the transposition ($a_2$ $a_3$) the same as complex conjugation?

  • $\begingroup$ Just because knowing $a_1$ is real then the two complex roots are complex conjugates $\endgroup$ – marwalix Jul 31 '16 at 17:03
  • $\begingroup$ For example, if $a_2 = 1+i$ and $a_3 = a-i$, then the transposition $(a_2 a_2)$ will map $1+i$ to $1-i$ and map $1-i$ to $1+i$. $\endgroup$ – 3x89g2 Jul 31 '16 at 17:18

First you should think about what you mean by 'it is the same as complex conjugation', so that the proof becomes evident. If $\sigma \in H$ is complex conjugation this means that $\sigma=\tau|_{\Omega}$, where $\tau: \mathbb{C} \rightarrow \mathbb{C}$ is complex conjugation, and $\Omega \subset \mathbb{C}$ is the splitting field of $f$. Now, a field homomorphism $\varphi: \Omega \rightarrow \mathbb{C}$ is completely determined by the images of $a_1$, $a_2$ and $a_3$. $\tau|_{\Omega}$ clearly is such a homomorphism, and even more clearly $(a_2 a_3)$ is such a homomorphism (by assumption). $\tau(a_1)=a_1=(a_2 a_3) (a_1)$, $\tau(a_2)=\overline{a_2}=a_3=(a_2 a_3) (a_2)$ and $\tau(a_3)=a_2=(a_2 a_3) (a_ 3)$, so both homomorphisms coincide on $\{a_1, a_2, a_3 \}$, so they must be the same.


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