A question about the embedding from $\mathbb{Q}(\sqrt{2},\sqrt{3})$ to an algebraic closure of $\mathbb{Q}$

Question

I am now just beginning my study in field theory and I am trying to find all embeddings from $\mathbb{Q}(\sqrt{2},\sqrt{3})$ to $\bar{\mathbb{Q}}$ (an algebraic closure of $\mathbb{Q}$).

Here, an embedding from a field $K$ to another field $F$ refers to an injective homomorphism from $K$ to $F$.

I asked my instructor for hint, and he told me that, first there are $4$ embeddings from $\mathbb{Q}(\sqrt{2},\sqrt{3})$ to $\bar{\mathbb{Q}}$, because $[\mathbb{Q}(\sqrt{2},\sqrt{3}) : \mathbb{Q}] = 4$, and $\sqrt{2}$ has to be mapped to $\sqrt{2}$ or $-\sqrt{2}$, and the case for $\sqrt{3}$ is similar.

$\textbf{Question:}$

$1.$ I am wondering why $[\mathbb{Q}(\sqrt{2},\sqrt{3}) : \mathbb{Q}] = 4$ implies that there are 4 embeddings.

$2.$ I am wondering why $\sqrt{2}$ has to be mapped to $\sqrt{2}$ or $-\sqrt{2}$.

$3.$ If $K = \mathbb{Q}(\sqrt[3]{2})$, then for the embeddings from $K$ to $\bar{\mathbb{Q}}$, what should $\sqrt[3]{2}$ be mapped to?

Thanks!

Answer 1

1. It follows from a more general result: If $[K:F]<\infty$ and $K/F$ is a separable extension, then there are exactly $[K:F]$ distinct embeddings of $K$ to $\overline{F}$. To show this, first establish the primitive element theorem, i.e. $K=F[\alpha]\simeq F[x]/(p(x))$ where $p(x)$ is the minimal polynomial of $\alpha$. And a homomorphism $\rho$ from $F[x]/(p(x))$ to $\overline{F}$ is completely characterized by the image of $x$ with the only condition that $\rho(x)$ is a root of $p(x)$. Since $p(x)$ has exactly $\deg p(x)=[K:F]$ distinct roots in $\overline{F}$, there are exactly $[K:F]$ distinct embeddings.

2. Because any embedding $\rho: \mathbb Q(\sqrt 2, \sqrt 3)\rightarrow\overline{\mathbb Q}$ must keep $\mathbb Q$ preserved, so $\rho(2)=2$, and then $\rho(\sqrt 2)^2=\rho((\sqrt 2)^2)=\rho(2)=2$, so $\rho(\sqrt 2)$ is a square-root of $2$ in $\overline{\mathbb Q}$, which can only be $\pm \sqrt{2}$. (And the irreducibility of $x^2-2$ implies both cases can be achieved). In another way, $\sqrt 2$ satisfies $x^2-2=0$, so $\rho(x^2-2)=\rho(x)^2-\rho(2)=\rho(x)^2-2=0$, that is $\rho(x)$ is also a root of $x^2-2$.

3. The minimal polynomial of $\sqrt[3]{2}$ over $\mathbb Q$ is just $x^3-2$, so $\sqrt[3]{2}$ can be mapped to any other root of the polynomial, and they are $\sqrt[3]{2}, \omega \sqrt[3]{2}, \overline{\omega}\sqrt[3]{2}$, where $\omega=\frac{-1+\sqrt{3} i}{2}$ is a primitive 3rd root of unity.

Note that this is a standard technique: If $\alpha$ satisfies $p(x)$, then by $\rho$ being a homomorphism that preserves the coefficients of $p$, we have $\rho(\alpha)$ must also be a root of $p(x)$. On the other hand if further $p(x)$ is irreducible, then $\alpha$ can be mapped into any root of $p(x)$, because $F[\alpha]\simeq F[x]/(p(x))$.

Intuitively, it's easy to see that $\sqrt{2}$ (resp. $\sqrt{3}$) can only be mapped to $\pm \sqrt{2}$ (resp. $\pm \sqrt{3}$). But we cannot conclude there are four embeddings yet, because we don't know if fixing the image of $\sqrt 2$ may limit our choices for the image of $\sqrt 3$. That is, $\sqrt 2$ and $\sqrt 3$ might be "related" in a certain way, and the fact $[\mathbb Q(\sqrt 2):\mathbb Q][\mathbb Q(\sqrt 3):\mathbb Q]=[\mathbb Q(\sqrt 2, \sqrt 3):\mathbb Q]$ tells us there is no such hidden correlation.