Showing $\mathbb{R} \otimes_{\mathbb{Q}} K \cong \mathbb{R} \times \mathbb{R}$

Question

Let $K = \mathbb{Q}[x]/(x^2 - 2)$. I want to show that $\mathbb{R} \otimes_{\mathbb{Q}} K \cong \mathbb{R}^2$. Then $\mathbb{R} \otimes_{\mathbb{Q}} K \cong \mathbb{R} \otimes_{\mathbb{Q}} \mathbb{Q}[x]/(x^2 - 2) \cong \mathbb{R}[x]/(x-\sqrt2)(x+ \sqrt2) \cong \mathbb{R}[x]/(x - \sqrt2) \times \mathbb{R}[x]/(x + \sqrt2)$ by the Chinese remainder theorem of modules.

Now this is where I am confused. Is this then isomorphic to $\mathbb{R} \times \mathbb{R}$? I know they are isomorphic as rings, but I'm not sure about as $\mathbb{R}$-modules since I know two simple modules are only isomorphic if they have the same annihilator. Thanks!

Answer 1

Since $\mathbb R$ is a field, an $\mathbb R$-module is exactly the same thing as a vector space over $\mathbb R$. The only isomorphism invariant of a vector space over a field is its dimension over that field. So all you have to prove is that $\mathbb R \otimes_{\mathbb Q} K$ has dimension $2$ over $\mathbb R$. And this should be a simple consequence of the fact that $K$ has dimension $2$ over $\mathbb Q$.

Having said that, there's a question of what "kind" of isomorphism the problem asks for. But $\mathbb R \times \mathbb R$ does not ordinarily come equipped with any multiplication, it's only structure is really just its bare vector space structure over $\mathbb R$. So I think that the "ring" structure is not relevant to this question, although since I cannot see the original problem this is not entirely clear to me.

Answer 2

The isomorphism $\Bbb{R}^2 \to \Bbb{R}[x]/(x^2-2)$ is $(a,b) \to a \frac{x-\sqrt{2}}{-2 \sqrt{2}} + b\frac{x+\sqrt{2}}{2 \sqrt{2}}$

and the isomorphism $ \Bbb{R}[x]/(x^2-2)\to \Bbb{R} \otimes_{\Bbb{Q}}\Bbb{Q}[x]/(x^2-2)$ is $c+dx \to c \otimes_{\Bbb{Q}} 1 + d \otimes_{\Bbb{Q}} x$