Prove some number is algebraic over a field

Question

How do you prove (without calculating the minimum polynomial) that $\sqrt{3}$ + $\sqrt[]{5}$ is algebraic over $\mathbb{Q}$. Also prove that $\left(\mathbb{Q}(\sqrt{3} + \sqrt[]{5}\right):\mathbb{Q}) \neq 2$ and that $\mathbb{Q}(\sqrt{3},\sqrt[]{5})=\mathbb{Q}(\sqrt{3} + \sqrt[]{5})$. I don't really have an idea on how to solve this.

Answer 1

Your questions are related to each oher.

Since, $\sqrt{3}+\sqrt[]{5}\in \mathbb{Q}(\sqrt{3},\sqrt[]{5})$ then it must be algebraic.

Assume, $|\mathbb{Q}(\sqrt{3}+\sqrt[]{5}):\mathbb Q|=2$ which means that $\sqrt{3}+\sqrt[]{5}$ is a root of second degree polynomial $x^2+bx+c$, by this you can easily reach contradiction.

Now, you know that $\mathbb{Q}(\sqrt{3}+\sqrt[]{5})\subseteq \mathbb{Q}(\sqrt{3},\sqrt[]{5})$ and $|\mathbb{Q}(\sqrt{3},\sqrt[]{5}):\mathbb Q|=4$ then we must have $|\mathbb{Q}(\sqrt{3}+\sqrt[]{5}):\mathbb Q|=4$ as it cannot be two, since their dimensions are equal, we must have equality, $$\mathbb{Q}(\sqrt{3},\sqrt[]{5})=\mathbb{Q}(\sqrt{3} + \sqrt[]{5})$$

Answer 2

Hint: Consider $f:\Bbb Q[\sqrt{2},\sqrt{3}] \to \Bbb Q[\sqrt{2},\sqrt{3}] $, $f(x) = (\sqrt{2}+\sqrt{3} )x$.

$f$ is linear. The the characteristic polynomial $P$ of $f$ is such as $P(\sqrt{2}+\sqrt{3})=0$.

Answer 3

Let me give an answer to the second half, that $(\mathbb Q(\sqrt{3}+\sqrt{5}):\mathbb Q)\ne 2$.

If $(\mathbb Q(\alpha):\mathbb Q)= m$, then $1,\alpha,\ldots,\alpha^{m}$ are linearly dependent over $\mathbb Q$.

You need to show that, $1, \sqrt{3}+\sqrt{5}$ and $(\sqrt{3}+\sqrt{5})^2$ are NOT linearly depenendent over $\mathbb Q$, or equivalent, that $1,\sqrt{3},\sqrt{5}$ and $\sqrt{15}$ are NOT linearly depenendent over $\mathbb Q$.