Degree of splitting field of $x^5-1$ over $\mathbb{Q}$ and the order of its Galois group?

Question

If I understand the result here correctly, the order of the Galois group is equal to the degree of the extension. That said, I am fairly certain that the splitting field $E$ of $x^5-1$ over $\mathbb{Q}$ is $$E=\{a_0+a_1\omega+a_2\omega^2+a_3\omega^3+a_4\omega^4\}$$ where $\omega$ is a complex root of unity. $E$, as a $\mathbb{Q}$-vector space, has dimension 5.

However, if I understand this correctly, the Galois group for this polynomial is $\mathbb{Z}/5\mathbb{Z}^*$, which has order 4.

What is it that I'm missing? Am I misinterpreting or miscomputing the dimension of $E$ as a $\mathbb{Q}$-vector space, or am I misinterpreting the post and the Galois group is in fact not the aforementioned group of units?

Answer 1

The polynomial is reducible: $x^5-1=(x-1)(x^4+x^3+x^2+x+1)$. You do not need to add the element $1$ to the field (it is already there), only a root of the quaternary polynomial (which generates the rest of the roots). That is why the degree is $4$.

Answer 2

While your $E$ is the right space, the set $\{1,\omega,\omega^2,\omega^3,\omega^4\}$ is not a basis for it, because that set isn't linearly independent over $\mathbb{Q}$. We have the dependence relation $1+\omega+\omega^2+\omega^3+\omega^4$ (and nothing else), so any four of those suffice as a basis. As a vector space over $\mathbb{Q}$, $E$ has dimension $4$ - so it's not a surprise when the Galois group has order $4$.

Answer 3

Order of Galois extension need not equal to degree of extension it is less than equal to degree of extension example is |Aut (Q(2^{1/3})/Q)| =1 but [ Q(2^{1/3}) : Q ] is 3 and equality hold when extension is splitting field of separable polynomial.