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Let's say we have field extensions $F(\alpha)$, $F(\beta)$ and $F(\alpha, \beta)$of a field $F$. My question is this: if $\sigma \in {\rm Gal}(F(\alpha)/F)$, is it correct to say that $\sigma \in {\rm Gal}(F(\alpha, \beta)/F)$ as well? And vice versa.

As an example I had a look at the extensions $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt{3})$ and $\mathbb{Q}(\sqrt{2},\sqrt{3})$ of $\mathbb{Q}$. The elements of ${\rm Gal}(\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q})$ are: \begin{align*} \sigma_1 &: \sqrt{2}\rightarrow\sqrt{2}, \: \sqrt{3}\rightarrow\sqrt{3} \\ \sigma_2 &: \sqrt{2}\rightarrow\sqrt{2}, \: \sqrt{3}\rightarrow -\sqrt{3}\\ \sigma_3 &: \sqrt{2}\rightarrow -\sqrt{2}, \: \sqrt{3}\rightarrow\sqrt{3}\\ \sigma_4 &: \sqrt{2}\rightarrow -\sqrt{2}, \: \sqrt{3}\rightarrow -\sqrt{3}\\ \end{align*} Would it be correct here to say that for example $\sigma_2 \in {\rm Gal}(\mathbb{Q}(\sqrt{3})/\mathbb{Q})$, since it leaves $\sqrt{2}$ fixed? Can I also say that any $\sigma \in {\rm Gal}(\mathbb{Q}(\sqrt{2})/\mathbb{Q})$ and any $\tau \in {\rm Gal}(\mathbb{Q}(\sqrt{3})/\mathbb{Q})$ will also be in ${\rm Gal}(\mathbb{Q}(\sqrt{2},\sqrt{3})/\mathbb{Q})$, as well as their products? If I were to "move up" such a $\sigma$, I would then just define it to leave $\sqrt{3}$ fixed (similarly for a $\tau$), and their product would just be combining their actions on $\sqrt{2}$ and $\sqrt{3}$. Is this valid, or will it fail in some cases? In the more general case in my first paragraph, will it hold there as well?

The reason I'm wondering is because I'm doing an exercise with field extensions as in my first paragraph, and ${\rm Gal}(F(\alpha)/F)$ and ${\rm Gal}(F(\beta)/F)$ are given to be abelian. I'm to show that ${\rm Gal}(F(\alpha, \beta)/F)$ is abelian, and I thought if the reasoning I did here is valid, I could more easily study the automorphisms in my problem.

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  • $\begingroup$ Why do you think that an element of the galois group of Q(sqrt3)/Q would fix the square root of 2?You need to extend the automorphism to the bigger field first, since that element is reserved only to the smaller field. $\endgroup$ – Bogdan Simeonov May 26 '16 at 10:22
  • $\begingroup$ @BogdanSimeonov That's what I did. Take a $\sigma \in {\rm Gal}(\mathbb{Q}(\sqrt{3})/\mathbb{Q})$, and extend it to $\mathbb{Q}(\sqrt{2},\sqrt{3})$ by letting it leave $\sqrt{2}$ alone. $\endgroup$ – Auclair May 26 '16 at 10:26
  • $\begingroup$ I suggest you take a look at Daniel Marcus's number fields book ,mainly the parts about embeddings, automorphisms (also look at appendix 2)wstein.org/edu/2010/581b/books/marcus-number_fields.pdf $\endgroup$ – Bogdan Simeonov May 26 '16 at 10:59
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It is true that any automorphism $\sigma: F(\alpha) \rightarrow F(\alpha)$ can be extended to an automorphism $\tilde{\sigma} :F(\alpha, \beta) \rightarrow F(\alpha, \beta)$. Briefly this is because $\sigma$ induces an automorphism from $F(\alpha)[X]/(f)$ to itself, where $f$ is the minimal polynomial of $\beta$ over $F(\alpha)$. As for your exercise, I do not think that going from automorphisms of $F(\alpha)$ to automorphisms of $F(\alpha, \beta)$ will help you, because you begin with $\psi, \varphi \in \text{Gal}(F(\alpha, \beta)/F)$, and you want to show they commute. So in some sense it's the other way around. If you show the restriction of $\psi$ and $\varphi$ on $F(\alpha)$ and on $F(\beta)$ are in $\text{Gal}(F(\alpha)/F)$ and $\text{Gal}(F(\beta)/F)$ respectively, then you are almost there. I know this is more of a hint than an answer, but I do not know whether you want a full answer to your exercise because your question is more general.

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  • $\begingroup$ Thanks! What I did was form the "product" of two automorphisms of $F(\alpha)$ and $F(\beta)$ by just combining their actions. For example, combining $\sigma: \sqrt{2}\rightarrow -\sqrt{2}$ and $\tau: \sqrt{3}\rightarrow -\sqrt{3}$ from $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ would give me $\sigma_4$ from my post. Isn't this almost equivalent to restricting automorphisms $\psi$ and $\phi$ down to $F(\alpha)$ and $F(\beta)$ in your comment? $\endgroup$ – Auclair May 26 '16 at 10:37
  • $\begingroup$ You first extend $\sigma$ and $\tau$ to the whole field and then you compose them. That gives you the same result I think. $\endgroup$ – M. Van May 26 '16 at 16:33

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