# Splitting field of $\sqrt{\vphantom{\sum}1+{\sqrt2}}$ and Galois group

Let $$\alpha= \sqrt{\vphantom{\sum}1+{\sqrt2}}$$.

(a) Let $$p(x)$$ be the minimal polynomial of $$\alpha$$. Find $$p(x)$$. Let K be the splitting field of $$p(x)$$

(b)Let $$E= \mathbb{Q}(i, \sqrt2)$$ Show that $$E \subseteq K$$

(c) Show that $$K=E\sqrt{\vphantom{\sum}1+{\sqrt2}}$$

(d) Show that there exists $$j:E \longrightarrow E$$ such that $$j(i)=-i$$ and $$j(\sqrt2)=-(\sqrt2)$$

(e) Show that there exists $$f:K \longrightarrow K$$ which extends $$j$$ such that $$f(\sqrt{\vphantom{\sum}1+{\sqrt2}})=\sqrt{\vphantom{\sum}1-{\sqrt2}}$$

(f)Show that $$f$$ has order 4

(g) Let $$g \in$$ Gal$$(K/ \mathbb{Q})$$ be a complex conjugation. Then show that $$fg=gf^3$$

(h) Conclude that Gal $$(K/ \mathbb{Q})$$ is isomorphic to $$D_8$$

Attempt

Finding the minimal polynomial is the only easy part here $$(x^2 -1)^2 -2= x^4 -2x^2 -1$$

Also I calculated the splitting field which I got to be the Galois closure as well which is done by computing $$x=-i\sqrt{\vphantom{\sum}-1+{\sqrt2}}$$

$$x=i\sqrt{\vphantom{\sum}-1+{\sqrt2}}$$

$$x=-\sqrt{\vphantom{\sum}1+{\sqrt2}}$$

$$x=\sqrt{\vphantom{\sum}1+{\sqrt2}}$$

So we just get $$\mathbb{Q}(i,\sqrt{\vphantom{\sum}1+{\sqrt2}})$$

(h) I have no idea how to proceed but I have found the following facts in the dummit and foote

1) The Galois group of the splitting field of an irreducible polynomial f acts transitively on the roots of f

2)The order of the Galois group of the splitting field of an irreducible polynomial f is divisible by degree of f

Not sure how I'm supposed to use these hints

(b) Look at this it's obvious to see that the 'elements' that K and E have in common is $$i$$ but what is the mechanism for showing such? I'm not sure how to proceed here.

(c) For this part is it okay to just take $$\mathbb{Q}(i,\sqrt{\vphantom{\sum}1+{\sqrt2}})$$ = $$\mathbb{Q}(i,\sqrt2)$$ x $$\sqrt{\vphantom{\sum}1+{\sqrt2}})$$ and proceed by normal multiplication?

(f) It's not hard to see from the splitting field $$x^4 -2x^2 -1$$ that it is irreducible over $$\mathbb{Q}$$ and the degree of the extension is 4.

for (d) and (e) i'm drawing a blank but i'm thinking they are related and somehow the minimal polynomial of $$\sqrt{\vphantom{\sum}1+{\sqrt2}})$$ and $$\sqrt{\vphantom{\sum}1-{\sqrt2}})$$ are some how related by the function $$j$$ but I just can't quite see it.

Please note I don't have much experience in this field and I'm currently using the dummit and foote.

Can anyone assist me.

• If the hints don't help you, you can look at the tower of quadratic extensions $\Bbb{Q}(ai,b\sqrt{2},c\sqrt{1+b\sqrt{2}})/\Bbb{Q}(ai,b\sqrt{2})/\Bbb{Q}(ai)/\Bbb{Q}$ where $a = \pm 1,b = \pm 1,c=\pm 1$ and see how the order $2$ Galois groups affect $a,b,c$. – reuns Apr 15 at 3:31

(b) take products and sums of the generators in $$K$$ to get elements in $$E$$. E.g $$\sqrt{1 + \sqrt{2}}$$ is contained in $$K$$, so $$1 + \sqrt{2}$$ is. Similarly $$1 - \sqrt{2}$$ is. Subtracting these two elements $$1$$ and $$\sqrt{2}$$ are both contained in $$K$$.

(c) When you say $$E \sqrt{1 + \sqrt{2}}$$ you must mean $$L = E ( \sqrt{ 1 + \sqrt{2}})$$. Since $$E \subset K$$, we can use degrees here to show they are the same (if a finite field extension is contained in another finite field extension of the same degree then they are the same. $$L/E$$ has degree $$2$$ and $$E/\mathbb{Q}$$ has degree $$4$$, as $$E / \mathbb{Q}(\sqrt{2})$$ has degree $$2$$ and $$\mathbb{Q}(\sqrt{2}) / \mathbb{Q}$$ has degree $$2$$. Now show that $$K$$ has degree $$8$$ using the tower $$K / \mathbb{Q}(\sqrt{1 + \sqrt{2}})/\mathbb{Q}(\sqrt{2}) / \mathbb{Q}$$.

(d) For a simple extension $$L(a)/L$$ of fields, with $$a$$ having minimal polynomial $$f(x) \in L[x]$$, there is always an automorphism permuting any two roots of $$f$$ contained in $$L[x]$$. To get one of the desired maps in the problem, take the map $$\mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q}(\sqrt{2})$$ sending the root $$\sqrt{2}$$ of the minimal polynomial $$x^2 - 2$$ to the other root $$-\sqrt{2}$$ of the minimal polynoimal. This gives an automorphism of $$\mathbb{Q}(\sqrt{2})$$. This gives a map $$\mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q}(\sqrt{2}, i)$$, which extends to a map $$\mathbb{Q}(\sqrt{2}, i)$$ sending $$i$$ to $$i$$. The reason this extension exists is that $$\mathbb{Q}(\sqrt{2}, i) / \mathbb{Q}(\sqrt{2})$$ is a simple extension generated by $$i$$, with minimal polynomial $$x^2 + 1$$, and $$x^2 + 1$$ has a root in the target of the embedding $$\mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q}(\sqrt{2}, i)$$.

So the two general facts we used were

1. Let $$L(a)/L$$ be a finite simple extension of fields. For any root $$b$$ of the minimal polynomial $$f$$ in $$L(a) / L$$, there is an $$L$$-automorphism of $$L(a)$$ sending $$a$$ to $$b$$.

2. Let $$L(a)$$ be a finite simple extension of fields. If $$M/L$$ contains a root $$b$$ of the minimal polynoimal $$f(x)$$ of $$a$$ in $$L[x]$$, then there is an embedding $$L(a) \rightarrow M$$ sending $$a$$ to $$b$$.

(e) this uses the same methods discussed in the last one. Here $$K/E$$ is a simple extension with $$a = \sqrt{1 + \sqrt{2}}$$ its generator. To proceed, calculate the minimal polynomial $$f(x)$$ of $$a$$ over $$E$$. There is an $$E$$-automorphism of $$K$$ sending $$a$$ to the other root of $$f(x)$$ in $$K$$.

(f) you got it!

(g) $$fg$$ and $$gf^3$$ are determined by what they do to the roots of certain polynomials. This is nice for us, since we have identified finitely many of these, and automorphisms send roots to other roots. So it is simply a matter of evaluating what they do to the roots. E.g. $$f g ( \sqrt{1 + \sqrt{2} } ) = f ( \sqrt{1 + \sqrt{2}} ) = \sqrt{1 - \sqrt{2}}$$. Do all the calculations of this kind, observe that they are the same for $$fg$$ and $$gf^3$$, and you'll be done.

When we found roots of the minimal polynomials and observed how each automorphism permutes them, we have really reduced an "intractible" question of automorphisms of fields to a tractible question of a finite group acting on the roots; automorphisms are determined by how they act on the roots of certain polynomials, and they permute those roots. In this way, instead of automorphisms acting on a field we have automorphisms acting on a finite set. The second should seem a lot easier to answer, since it is about something finite- in the worst case, we have a brute force approach.

(h) This is potentially the most tricky part, but it's also where all this computation has been leading us. $$D_8$$ is given by generators and relations as $$\langle x, y | x^4, y^2, xy = yx^3 \rangle$$. To give an isomorphism, let's start with a map from the free group $$\langle x, y \rangle$$ to the galois group in question sending $$x$$ to $$f$$ and $$y$$ to $$g$$. We must show a few things to establish an isomorphism:

1. $$x^4$$ is sent to $$\text{Id}$$, and so are $$y^2$$ and $$x^{-3}y^{-1}xy$$. You showed the last one already, but the first two are not so different!

2. All automorphisms in the galois group are some combination of $$f$$ and $$g$$.

3. The order of the galois group is $$8$$.

I have to go for now, but I hope this is enough.