3

The $4$th edition of Ian Stewart's Galois Theory is what I am looking at where there is an example stated in the following manner

Let the field extension be $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}$. We have previously seen that $t^2-5$ is irreducible over $\mathbb{Q}(\sqrt{2},\sqrt{3})$. Similarly, $t^2-2$ i irreducible over $\mathbb{Q}(\sqrt{3},\sqrt{5})$ and $t^2-3$ is irreducible over $\mathbb{Q}(\sqrt{2},\sqrt{5})$. Thus there are three automorphisms of $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ which are,

\begin{equation} \begin{split} \rho_1:& \sqrt{2} \rightarrow -\sqrt{2} \text{ , }\sqrt{3} \rightarrow \sqrt{3} \text{ , } \sqrt{5} \rightarrow \sqrt{5}\\ \rho_2:& \sqrt{2} \rightarrow \sqrt{2} \text{ , }\sqrt{3} \rightarrow -\sqrt{3} \text{ , } \sqrt{5} \rightarrow \sqrt{5}\\ \rho_3:& \sqrt{2} \rightarrow \sqrt{2} \text{ , }\sqrt{3} \rightarrow \sqrt{3} \text{ , } \sqrt{5} \rightarrow -\sqrt{5} \end{split} \end{equation}

It is easy to see that the maps commute so it generates the group $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 $.

Okay, my question.

$1.$ First, what are the elements of $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$? Tower law leads me to say the degree of extension $[\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}]=8$. So, there are $8$ elements as bases, yes? I have found these to be $\beta=\{1,\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{6},\sqrt{10},\sqrt{15},\sqrt{30}\}$. Now, all $\rho$ up there, seem only to care about $\sqrt{2},\sqrt{3},\sqrt{5}$.Since we are looking for $\mathbb{Q}$-automorphisms over $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$, sure $\rho$ will not touch on any rationals. But why do they ignore, say a map like $\sqrt{30} \rightarrow -\sqrt{30}$(while all other radicals are mapped to itself for instance). SHouldn't we consider "all" automorphisms like $\rho$? Then surely there are more than just $3$?

$2.$ And okay, say you managed to convince me that there are only those $3$ $\rho$s up there. But the next confusion is, how do these generate $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 $? If I understand the notation correctly, $\mathbb{Z}_2 =\{e,x\}$ where $e$ is the identity and $x$ is some non-identity. Commonly, I see it as $\mathbb{Z}_2 = \{\bar{0},\bar{1}\}$. Then, taking direct products should give me a set of $3$-tuples $\{(a,b,c): a,b,c \in \mathbb{Z}_2=\{\bar{0},\bar{1}\}\}$ so something like $(\bar{0},\bar{1},\bar{1})$ etc. I believe there are $8$ distinct such t$3$-tuples. My understanding from the example is that the Galois Group of $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}$ is $\{id, \rho_1,\rho_2,\rho_3\}$ so $4$ elements. How is this going to generate $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$? I am very well confused.

Overall, I am only starting to learn all of this so my experience is very limited. I may be confusing some mathematical concept with another(afterall, math does jump from one and another definition "depending on the context") or simply misunderstanding an idea, notion, theorem whatever. Tricky wordings like $K$-autmorphisms "over" $L$ gets me to stop for a second and check in my head what exactly the properties are and what it's meant. Perhaps I could have some error there.

John Trail
  • 3,279
  • For (1), once it is specified that roots of 2,3,5 go to themselves, that implies the root of 30 also goes to itself, not its negative. – coffeemath Mar 06 '16 at 16:21

2 Answers2

5

(1) The elements of $K := \Bbb Q[\sqrt{2}, \sqrt{3}, \sqrt{5}]$ are those of the form $$a + b \sqrt{2} + c \sqrt{3} + d \sqrt{5} + e \sqrt{6} + f \sqrt{10} + g \sqrt{15} + h \sqrt{30},$$ corresponding to the fact that $[K : \Bbb Q] = 8$, that is, that $K$ is an $8$-dimensional vector space over $\Bbb Q$. But as a field $K$ is generated over $\Bbb Q$ by just $3$ elements, namely (for example) $\sqrt{2}, \sqrt{3}, \sqrt{5}$, and the action of any automorphism $\sigma \in \operatorname{Gal}(K / \Bbb Q)$ is determined by its action on these. For example, $\sigma(\sqrt{30}) = \sigma(\sqrt{2} \sqrt{3} \sqrt{5}) = \sigma(\sqrt{2}) \sigma(\sqrt{3}) \sigma(\sqrt{5}).$

(2) The additional automorphisms come from compositions of $\rho_1, \rho_2, \rho_3$. For example, $\tau := \rho_1 \circ \rho_2$ is characterized by $\tau(\sqrt{2}) = - \sqrt{2}$, $\tau(\sqrt{3}) = - \sqrt{3}$, $\tau(\sqrt{5}) = \sqrt{5})$. Since all three generators $\rho_i$ have order $2$ and commute with each other, we can write any automorphism as $\rho_1^{a_1} \rho_2^{a_2} \rho_3^{a_3}$, $a_1, a_2, a_3 \in \{0, 1\}$, yielding eight automorphisms in total as claimed.

Travis Willse
  • 99,363
  • Hi, I came back to this since I wanted to double check I understood it, but one more question; How does the "irreducibility" of certain polynomials as given in the question box relate to the number of automorphisms available? It discusses the irreducible polynomial in each extension, and then straight goes to say "thus there are ..." – John Trail May 13 '16 at 11:40
  • 1
    If $p \in \Bbb F[x]$ is irreducible with root $\alpha$, then the Galois group $\operatorname{Gal}(\Bbb F(\alpha) / \Bbb F)$ acts transitively on the roots of $\Bbb F[x]$. In particular, if $p$ is quadratic, as in our case, then the Galois group has order $2$, and its nontrivial element is the automorphism characterized by interchanging the two roots. For a general polynomial of degree $n$, since automorphisms of the splitting field are characterized by their action on the roots, the Galois group some subgroup of $S_n$, and hence there are at most $|S_n| = n!$ automorphisms. – Travis Willse May 13 '16 at 11:55
3

Although it is true that $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ has those 8 elements as a base as you have listed them, that only tells you about the vector space structure.

The ring structure is also important. As a ring the three elements $\sqrt{2},\sqrt{3},\sqrt{5}$ are generators of $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$. An automorphism is therefore uniquely determined by the images of those three generators. You can compute the image of $\sqrt{30}$ under each of $\rho_1,\rho_2$ and $\rho_3$ just using the given information that tells where each of these automorphisms takes each of the generators. For example, $$\rho_1(\sqrt{30}) = \rho_1(\sqrt{2}\sqrt{3}\sqrt{5}) = \rho_1(\sqrt{2}) \rho_1(\sqrt{3}) \rho_1(\sqrt{5}) = (-\sqrt{2})(\sqrt{3})(\sqrt{5}) = - \sqrt{30} $$

For your second question, the statement is that $\rho_1,\rho_2,\rho_3$ are automorphisms, not that they are the only automorphisms. In fact each $\rho_i$ is the generator of an order 2 cyclic subgroup of the Galois group, namely $\{id,\rho_i\}$ (you should definitely not saddle yourself with elementary notation about order 2 cyclic groups). Furthermore the entire Galois group is the direct product of those three order 2 cyclic subgroups (which still has to be proved), and hence the entire Galois group consists of the elements $$\{id, \rho_1, \rho_2, \rho_3, \rho_1 \rho_2, \rho_1 \rho_3, \rho_2 \rho_3, \rho_1 \rho_2 \rho_3\} $$

Lee Mosher
  • 120,280