Determining isomorphism between field extensions

Question

Let $f \in \mathbb{Q}[x]$ be irreducible. Assume $\mathbb{Q} \subset E$ is a finite galois extension and $a, b \in \mathbb{C}$ are roots of $f$.

1. Prove that $\mathbb{Q}(a) \cap E \cong \mathbb{Q}(b) \cap E$
2. If $E = \mathbb{Q}(\zeta_n)$, where $\zeta_n$ is the primitive $n$-th root of unity, then the isomorphism is equality.

I started by assuming $f$ has a root in $E$. Since $E$ is a finite galois extension, $f$ reduces over $E$. This gives us an isomorphism pretty quickly. I am stuck on the case where $f$ does not have a root over $E$. At first glance I thought that we could just let the isomorphism be the isomorphism between two extensions of the roots but not sure if that is allowed. Similarly, I thought of treating each field extension as a $\mathbb{Q}$-vector space and since $dim(\mathbb{Q}(a)) = dim(\mathbb{Q}(b))$ we get an isomorphism.

Did not know how to start 2 if I assumed 1.

Answer 1

• Find $\alpha_1,\ldots,\alpha_k$ such that $E \cap \mathbb{Q}(a) = \mathbb{Q}(\alpha_1,\ldots,\alpha_k)$. Since $\varphi$ is an isomorphism, $\alpha_m$ and $\varphi(\alpha_m)$ have the same minimal polynomial ($\sum_{l=0}^d c_l \alpha_m^l = 0$ iff $\varphi(\sum_{l=0}^d c_l \alpha_m^l)=\sum_{l=0}^d c_l \varphi(\alpha_m)^l = 0$)

  Since $E/\mathbb{Q}$ is a Galois extension, if $u,v$ have the same minimal polynomial then $u \in E$ iff $v \in E$. Thus $\varphi(\alpha_1),\ldots,\varphi(\alpha_k) \in E$ and $$\varphi(E \cap \mathbb{Q}(a))=\mathbb{Q}(\varphi(\alpha_1),\ldots,\varphi(\alpha_k)) \subseteq E$$ and since $\varphi(E \cap \mathbb{Q}(a)) \subseteq \varphi(\mathbb{Q}(a))=\mathbb{Q}(b)$ we obtain $$\varphi(E \cap \mathbb{Q}(a)) \subseteq E \cap \mathbb{Q}(b)$$

  Repeat exactly the same argument with $\varphi^{-1}$ and $E \cap \mathbb{Q}(b) = \mathbb{Q}(\beta_1,\ldots,\beta_K)$ to obtain $$\varphi^{-1}(E \cap \mathbb{Q}(b)) \subseteq E \cap \mathbb{Q}(a) \quad \implies \quad E \cap \mathbb{Q}(b)= \varphi(\varphi^{-1}(E \cap \mathbb{Q}(b))) \subseteq \varphi(E \cap \mathbb{Q}(a))$$ ie $$E \cap \mathbb{Q}(b) = \varphi(E \cap \mathbb{Q}(a)) \quad \implies \quad E \cap \mathbb{Q}(b) \cong E \cap \mathbb{Q}(a)$$

• If $E = \mathbb{Q}(\zeta_n)$ then $E/\mathbb{Q}$ is an abelian extension so that all its subfields are abelian extensions, thus $E \cap \mathbb{Q}(a) / \mathbb{Q}$ is a Galois extension and $\varphi(\alpha_m) \in E \cap \mathbb{Q}(a)$ which means $\varphi(E \cap \mathbb{Q}(a))\subseteq E \cap \mathbb{Q}(a)$ and $$E \cap \mathbb{Q}(a) = E \cap \mathbb{Q}(b)$$

Answer 2

Here is a simple argument. There is an isomorphism $$\varphi:\mathbb{Q}(a)\rightarrow \mathbb{Q}(b)$$

we show that the image under $\varphi$ of $E\cap \mathbb{Q}(a)$ is $E\cap \mathbb{Q}(b)$.

Extend $\varphi$ to an automorphism of some Galois $K$ containing all these fields.

Let $c\in E\cap \mathbb{Q}(a)$ then $\varphi (c) \in E$ since $E\subseteq K$ is Galois. Thus $\varphi (c) \in E\cap \mathbb{Q}(b)$.