An Example of Galois Group

Question

I wish to classify the Galois group of $\mathbb{Q}(e^{i\pi/4})/\mathbb{Q}$. Let me denote the eighth root of unity as $\epsilon$. I see that $1, \epsilon, \epsilon^2, \epsilon^3$ are linearly independent over $\mathbb{Q}$, although I do not know how to rigorously prove this (can anyone show me?). Since $\{\epsilon, \epsilon^2, \epsilon^3\}$ are linearly independent does that mean $[\mathbb{Q}(e^{i\pi/4}):\mathbb{Q}] = 8$, or is it some other value (if it is please explain why)? Are my observations correct so far?

Answer 1

The minimal polynomial of $\epsilon = e^{\pi i\over 4}$ is $x^4 + 1 = 0$. To see this, note that it is easy to see from Euler's formula that $x^4 = e^{i \pi} = -1$. On the other hand, $x^4+1$ is irreducible over $\mathbb Q$ (Apply Eisenstein to $(x+1)^4+1$).

Now, $1,\epsilon, \epsilon^2,\epsilon^3$ are linearly independent over $\mathbb Q$ , for if not then any linear combination of theirs is a cubic polynomial with rational coefficients in $\epsilon$. If it equaled zero, then there is a polynomial of degree $4$ and degree less than $3$ which $\epsilon$ satisfies. So $\epsilon$ must satisfy the $\gcd$ of these polynomials. What can that $\gcd$ be? Get a contradiction.

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From the fact that the degree of the minimal polynomial equals the degree of the extension, we get that $[\mathbb Q(\epsilon) : \mathbb Q] = 4$. (Hint : Suppose you did not know this fact : we know that $1,\epsilon,\epsilon^2,\epsilon^3$ are l.i. : use $\epsilon^4 = -1$ to reduce any polynomial in $\epsilon$ to one involving only these four powers. That will show the spanning nature of these numbers and therefore the result as well.)

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The Galois group of this extension is $\mathbb Z_2 \times \mathbb Z_2$. To see this, we look at a specific member of the automorphism , the one sending $\epsilon$ to $\epsilon^3$.

What does that mean? Well, the automorphism is multiplicative and additive. So, it sends $$\epsilon^2 = \epsilon \times \epsilon \to \epsilon^3 \times \epsilon^3 = \epsilon^6 = -\epsilon^2 \\ \epsilon^3 \to \epsilon^9 = \epsilon \\ \epsilon^{-1} \to (\epsilon^3)^{-1} = \epsilon^{-3} \\ \epsilon^{-3} \to \epsilon^{-9} = -\epsilon$$

So basically, any element of the form $a+b\epsilon + c \epsilon^2 + d \epsilon^3$ (which is the general form of an element of this extension) goes to $a + b\epsilon^3 - c \epsilon^2 + d \epsilon$.

This has order $2$, as can be seen easily.

See, the roots of $x^4+1 = 0$ are $\epsilon, - \epsilon , i \epsilon , -i\epsilon$. So the four automorphisms are taking $\epsilon$ to each of those above (one of them is secretly $\epsilon^3$, so we have already seen that above. Which one?). One can check this.

Answer 2

Let $\epsilon:=e^{i\pi/4}$ be the primitive 8th root of unity.

Then $\epsilon$ is an $8$-th root but not a $4$-th root, so is a root of $(X^8 -1)/(X^4-1)$.

That is $\epsilon$ is a root of $X^4 +1$.

Now for $X$ real $X^4+1>1$; so $X^4+1$ has no real roots, and so no rational roots.

Hence if it factorises it factorises in $\mathbb{Q}[X]$ then it factorises as the product of two rational quadratics. However in $\mathbb{R}[X]$ we have $X^4+1=(X^2 +\sqrt{2}X+1)(X^2-\sqrt{2}X+1)$, and as these factors are not rational we see that $X^4+1$ is irreducible in $\mathbb{Q}[X]$.

We therefore have that $1, \epsilon,\epsilon^2,\epsilon^3$ are linearly independent over $\mathbb{Q}$.