What is the easiest description, if any, of the tensor product $K \otimes_\mathbb{Q} \mathbb{R}$? The question is on the title, with some clarification as followed.
Firstly, $K$ is a simple and algebraic field extension of $\mathbb{Q}$, say $\mathbb{Q}(\alpha)$ with some algebraic $\alpha$. So, $K$ and $\mathbb{R}$ here are considered as vector spaces of $\mathbb{Q}$.
Secondly, I think I have known quite a bit about tensor product. It's not deep, but at least I've understood definitions and somewhat of its universal property. However, I still don't know how to visualize some particular cases in my head. For now, $K \otimes_\mathbb{Q} \mathbb{R}$ is my only concern, and I'm trying to find a way to beautifully describe it.
Thanks in advance.
P.S: I know this is not a rigorous question, but I'm in a dead end, so it's worth a try.
 A: There is a very simple description.  Let $f(x)$ be the minimal polynomial of $\alpha$, so $K\cong\mathbb{Q}[x]/(f(x))$.  There is then an isomorphism $$K\otimes\mathbb{R}\cong \mathbb{Q}[x]/(f(x))\otimes\mathbb{R}\cong\mathbb{R}[x]/(f(x)).$$  To identify the ring $\mathbb{R}[x]/(f(x))$, we just have to factor $f(x)$ over $\mathbb{R}$.  Since $f(x)$ is irreducible over $\mathbb{Q}$, it is squarefree over $\mathbb{R}$, so $f(x)$ factors over $\mathbb{R}$ as a product of distinct irreducible polynomials $f_1(x)f_2(x)\dots f_n(x)$.  By the Chinese remainder theorem, $$\mathbb{R}[x]/(f(x))\cong\prod_i\mathbb{R}[x]/(f_i(x)).$$  To identify $\mathbb{R}[x]/(f_i(x))$, note that any irreducible polynomial over $\mathbb{R}$ is either linear (in which case the quotient is $\mathbb{R}$) or quadratic (in which case the quotient is $\mathbb{C}$).  So $K\otimes\mathbb{R}$ is just a product of copies of $\mathbb{R}$ and $\mathbb{C}$, with a copy of $\mathbb{R}$ for each linear factor of $f(x)$ over $\mathbb{R}$ and a copy of $\mathbb{C}$ for each irreducible quadratic factor of $f(x)$ over $\mathbb{R}$.
We can even be more explicit.  Linear factors of $f(x)$ over $\mathbb{R}$ just correspond to roots of $f(x)$ in $\mathbb{R}$, and irreducible quadratic factors correspond to complex conjugate pairs of roots of $f(x)$ in $\mathbb{C}\setminus\mathbb{R}$.  So $K\otimes\mathbb{R}\cong\mathbb{R}^a\times\mathbb{C}^{b/2}$, where $a$ is the number of real roots of $f(x)$ and $b$ is the number of nonreal complex roots of $f(x)$.  If you want to describe these directly in terms of $K$ without mentioning $f(x)$, you can say $a$ is the number of field embeddings $K\to\mathbb{R}$ and $b$ is the number of field embeddings $K\to\mathbb{C}$ whose image is not contained in $\mathbb{R}$ (since to embed $K=\mathbb{Q}(\alpha)$ in another extension $L$ of $\mathbb{Q}$, you just have to pick a root of $f(x)$ which is where you will send $\alpha$).
(Note that in all of this, I am describing the tensor product $K\otimes\mathbb{R}$ not just as a vector space but as a ring, or in fact as an $\mathbb{R}$-algebra.)
