Decomposition of $\mathbb{Q}(\sqrt{2})\otimes_{\mathbb{Q}}\mathbb{Q}(\sqrt{3})$ and $\textbf{F}_q(t)\otimes_{\textbf{F}_q(t^q)}\textbf{F}_q(t)$

Question

I have two questions about splitting of the tensor product into the product of fields

1. How can one find a decomposition of

$$\mathbb{Q}(\sqrt{2})\otimes_{\mathbb{Q}}\mathbb{Q}(\sqrt{2})$$

and

$$\mathbb{Q}(\sqrt{2})\otimes_{\mathbb{Q}}\mathbb{Q}(\sqrt{3})$$

into the direct product of fields?

2. Let $\textbf{F}_q$ be a field with $q$ elements. Is it possible to decompose $$\textbf{F}_q(t)\otimes_{\textbf{F}_q(t^q)}\textbf{F}_q(t)$$

into the direct product of fields?

Thank you.

$\textbf{Update}.$

To answer the first question I have the following argument. Since $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ are the field extensions of $\mathbb{Q}$ I have

$$\mathbb{Q}(\sqrt{2})\otimes_{\mathbb{Q}}\mathbb{Q}(\sqrt{2})\cong\mathbb{Q}[x]/(x^2-2)\otimes_\mathbb{Q}\mathbb{Q}(\sqrt{2})\cong\mathbb{Q}(\sqrt{2})[x]/(x^2-2)\cong\mathbb{Q}(\sqrt{2}),$$

and

$$\mathbb{Q}(\sqrt{2})\otimes_{\mathbb{Q}}\mathbb{Q}(\sqrt{3})\cong\mathbb{Q}[x]/(x^2-2)\otimes_\mathbb{Q}(\sqrt{3})\cong\mathbb{Q}(\sqrt{3})[x]/(x^2-2)\cong\mathbb{Q}(\sqrt{2}, \sqrt{3}).$$

Is this correct?

Answer 1

Note that $[F_q(t) : F_q(t^q)] = q$. So, using a dimension argument over $F_q(t^q)$ you can see that $$\dim F_q(t) \otimes F_q(t) = q^2$$ while $$\dim F_q(t) \oplus F_q(t) = 2q $$ so they are not isomorphic for $q \neq 2$.