Understanding cyclotomic extensions: $\mathbb{Q}\subseteq \mathbb{Q}(\xi)$, where $\xi = e^{\frac{2\pi i}{3}}$.

Question

In class, we have seen the following result:

Let $K\subseteq K(\xi)$ a cyclotomic extension (i.e. $\xi$ is a primitive root of unit). Then, $K\subseteq K(\xi)$ is a Galois extension and $\text{Gal}(K(\xi)/K)\cong H \leq \mathcal{U}(\mathbb{Z}/(n))$ where $n$ is the order of $\xi$.

I am trying to use this result to prove that there exists an extension $\mathbb{Q}\subseteq L$ of order 3 such that $\text{Gal}(L/\mathbb{Q})$ is a cubic cyclic group. Then, I have write an example that looks like a contradiction to me (I guess I am considering something wrong).

If I consider $\xi = e^{\frac{2\pi i}{3}}$, then I know its order is $3$. Thus, if I consider $\mathbb{Q}\subseteq \mathbb{Q}(\xi)$, I would have $$\text{Gal}(\mathbb{Q}(\xi)/\mathbb{Q}) \cong H \leq \mathcal{U}(\mathbb{Z}/(3)).$$

The result is telling me that my extension is a Galois extension. Thus, I can conclude that $$|\text{Gal}(L/\mathbb{Q})| = [\mathbb{Q}(\xi) : \mathbb{Q}]= 3$$ since $p(x)=x^3-1$ is the irreducible polynomial of $\xi$ over $\mathbb{Q}$.

Now, if I am not wrong, $\mathcal{U}(\mathbb{Z}/(3)) \cong \mathbb{Z}/(2)$. From there, I would get a contradiction since an order 3 group can not be contained in an order 2 group.

QUESTION

1. What am I supposing wrong?

Answer 1

Your polynomial $X^3-1$ is NOT irreducible since $X^3-1=(X-1)(X^2+X+1)$.

In fact, the irreducible polynomial of $\xi$ over $\mathbb{Q}$ is $X^2+X+1$, which has degree $2$, and you get no contradiction.