Computing the minimal polynomial (field extension)

Question

How to compute the minimal polynomial of $\sqrt{3}+\sqrt{7}$ over $\mathbb{Q}$?

My way was:

Since $(\sqrt{3}+\sqrt{7})^2=10+2\sqrt{21} \notin \mathbb{Q}$ and $\sqrt{21} \neq a(\sqrt{3}+\sqrt{7})+b$ for $a,b \in \mathbb{Q}$, the polynomial cannot have degree $2$.

$(\sqrt{3}+\sqrt{7})^2-10=2\sqrt{21}$

$\Rightarrow ((\sqrt{3}+\sqrt{7})^2-10)^2=84$

$ \Rightarrow (\sqrt{3}+\sqrt{7})^4-20(\sqrt{3}+\sqrt{7})^2+16=0$

So $\sqrt{3}+\sqrt{7}$ is a root of the polynomial $x^4-20x^2+16=:f(x) \in \mathbb{Q}[x]$.

Since $\mathbb{Q}(\sqrt{3}+\sqrt{7})=\mathbb{Q}(\sqrt{3},\sqrt{7})$, there does not exist a polynomial of degree $3$ with $\sqrt{3}+\sqrt{7}$ as a root.

$[\mathbb{Q}(\sqrt{3}+\sqrt{7}):\mathbb{Q}]\mid [\mathbb{Q}(\sqrt{3},\sqrt{7}):\mathbb{Q}]=4$

So $[\mathbb{Q}(\sqrt{3}+\sqrt{7}):\mathbb{Q}]=3$ is not possible and $x^4-20x^2+16=:f(x) \in \mathbb{Q}[x]$ is the minimal polynomial of $a=\sqrt{3}+\sqrt{7}$ over $\mathbb{Q}$.

Is this proof right? Or is something wrong or missing?

Also, I'm not sure how to show that $\mathbb{Q}(\sqrt{3}+\sqrt{7})=\mathbb{Q}(\sqrt{3},\sqrt{7})$.

Answer 1

Yes, your proof is correct.

The way to show that $\mathbb{Q}(\sqrt{3}+\sqrt{7})=\mathbb{Q}(\sqrt{3},\sqrt{7})$ is by saying that the former clearly is contained in the latter (since $\sqrt{3}+\sqrt{7}\in\mathbb{Q}(\sqrt{3},\sqrt{7})$).

On the other hand, since the extensions $\Bbb Q(\sqrt 3)$ and $\Bbb Q(\sqrt 7)$ are quadratic extensions of $\Bbb Q$, then $\Bbb Q(\sqrt 3,\sqrt 7)$ is at most a quadratic extension of $\Bbb Q(\sqrt 3)$, therefore at most a degree-4 extension of $\Bbb Q$. Since it contains the degree-4 extension $\mathbb{Q}(\sqrt{3}+\sqrt{7})$, then it must coincide with that extension.

Answer 2

Essentially correct.

A simpler argument for the end: if the minimal polynomial of $\sqrt 3+\sqrt 7$ were a cubic polynomial, it would divide the quartic polynomial $\;x^4-20x^2+16$, which would therefore have a rational root – necessarily a power of $2 \le 2^4$, which is impossible as it would imply $2\mid 5$.