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In an exercise in the textbook you are asked to describe the Galois Group and the intermediate fields of the extension $$ L=\newcommand{\Q}{\mathbb Q}\Q(\sqrt 2,\sqrt 3)\supset\Q $$ I have noted that $L$ is Galois:

  • It is normal as it is a splitting field for both of the defining polynomials $$ \begin{align} X^2-2&=(X+\sqrt 2)(X-\sqrt 2)\\ X^2-3&=(X+\sqrt 3)(X-\sqrt 3) \end{align} $$
  • It is separable since we are in characteristic zero
  • It is clearly algebraic and finite having degree 4

Any automorphism $\varphi$ of $\mbox{Gal}(L/\Q)$ will be determined by the pair $\left(\varphi(\sqrt 2),\varphi(\sqrt 3)\right)$ so by permuting roots in the defining polynomials we obtain the following table $$ \begin{array}{c|c:c:c:c} \varphi&\ 1\ &\ \varphi_1\ &\ \varphi_2\ &\ \varphi_3\ \\ \hline \varphi(\sqrt 2)&\sqrt 2&-\sqrt 2&-\sqrt 2&\sqrt 2\\ \varphi(\sqrt 3)&\sqrt 3&-\sqrt 3&\sqrt 3&-\sqrt 3 \end{array} $$ so the Galos Group is $\{1,\varphi_1,\varphi_2,\varphi_3\}$. The table also shows that $\sqrt 2+\sqrt 3$ maps to four distinct elements so that this must be a primitive element of the extension $L\supset\Q$. It is immediate from the the table that $\varphi_i^2=1$ for $i=1,2,3$ so we have $$ \mbox{Gal}(L/\Q)\simeq\newcommand{\Z}{\mathbb Z}\Z_2\times\Z_2 $$ The subgroups of $\mbox{Gal}(L/\Q)$ appears to be like in this (primitive) diagram $$ \begin{matrix} &&\{1\}&&\\ &\swarrow&\downarrow&\searrow\\ \langle \varphi_1\rangle&&\langle\varphi_2\rangle&&\langle\varphi_3\rangle\\ &\searrow&\downarrow&\swarrow\\ &&\mbox{Gal}(L/\Q)&& \end{matrix} $$ and based on that I have found that the intermediate fields are as follows $$ \begin{matrix} &&L&&\\ &\swarrow&\downarrow&\searrow\\ \Q(\sqrt 6)&&\Q(\sqrt 3)&&\langle\Q(\sqrt 2)\\ &\searrow&\downarrow&\swarrow\\ &&\Q&& \end{matrix} $$

Question: Is there a system for the Galois Group and the Intermediat Fields?

Whereas all of this makes sense to me, I still feel that I have no method for determining the structure of the subgroups of the Galois Group, and also that I have no particular system for finding the generators $\sqrt 6,\sqrt 3$ and $\sqrt 2$ for the intermediate fields. It took me some minutes to figure out $\sqrt 6$ in particular.

Is there any nice and systematic way to deal with this. In particular for slightly more complex examples such as $\Q(\sqrt[4] 3,i)\supset\Q$ for instance.

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    $\begingroup$ Mostly, there's a bag of increasingly clever tricks to deal with increasingly complicated situations. It's probably best just to look at lots of worked examples, and pick up the tricks on the way. Some will be done in any textbook that does Galois Theory, and some have been worked out on this very website --- have a look through the questions with the galois-theory tag. $\endgroup$ – Gerry Myerson Jan 19 '14 at 20:07
  • $\begingroup$ @GerryMyerson: Ok, thank you! $\endgroup$ – String Jan 19 '14 at 20:08
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    $\begingroup$ Determining the subgroups themselves is a matter of group theory (and not trivial in general). To determine the fixed fields, it is often helpful to condider all products of powers of generators - which in this example included $\sqrt 6$ and in your other example would include $\sqrt 3$ and $i\sqrt[4]{27}$ and ... $\endgroup$ – Hagen von Eitzen Jan 19 '14 at 20:12
  • $\begingroup$ @HagenvonEitzen: Ok. Can you also say that by multiplying degree $2$ elements like here the product will itself be of degree $2$? In this case it does, but should we expect that in general... $\endgroup$ – String Jan 19 '14 at 21:38
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    $\begingroup$ Represent elements of $L$ as vectors with rational coordinates with respect to basis elements of the form proposed by Hagen. Then, given some generators $\varphi_j$ of a subgroup $H$ of the Galois group $G$, consider how $\varphi_j$ changes the coordinates of their argument. Then demand that all those changes shall have no effect. This results in linear relations among the coordinates. Those relations describe the intermediate field $K$ fixed by $H$, and the degree of $K$ over $\mathbb{Q}$ is determined by the index of $H$ in $G$. $\endgroup$ – ccorn Jan 19 '14 at 23:59

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