Particular Tensor Product I've found while reading and article the following thing:
Let $\mathbb{K}=\mathbb{Q}(\alpha)$ be an algebraic number field. What do we mean by $$\mathbb{K}_{\mathbb{R}}=\mathbb{K}\otimes_{\mathbb{Q}} \mathbb{R} $$
What are the elements inside looks like? Is it isomorphic to the smallest field containing $\mathbb{K}$ and $\mathbb{R}$ (as a vector space over $\mathbb{Q})$?
 A: $\mathbb{K}=\mathbb{Q}(\alpha)$ is a simple extension, so it is either isomorphic to $\Bbb Q(X)$ or to $\Bbb Q[X]/(f)$ with $f \in \Bbb Q[X]$. In the first case, $\Bbb K_\Bbb R \cong \Bbb R(X)$; in the second case, $\Bbb K_\Bbb R \cong \Bbb R[X]/(f)$. If $\deg f > 2$, then $f$ would be reducible (by Fundamental Theorem of Algebra), so $\Bbb K_\Bbb R$ would not be an integral domain.
However, the significance of $\Bbb K \otimes_\Bbb Q \Bbb R$ is that if you quotient out by a maximal ideal $\mathfrak m$ (which exists by Zorn's lemma), then you would obtain a field $F := (\Bbb K \otimes_\Bbb Q \Bbb R)/\mathfrak m$ with maps $\Bbb K \to F$ and $\Bbb R \to F$ such that the following diagram commutes:
$$\begin{array}{rcl}
\Bbb Q & \longrightarrow & \Bbb K \\
\downarrow && \downarrow \\
\Bbb R & \longrightarrow & F
\end{array}$$
So $F$ can be thought of as a larger field containing both $\Bbb K$ and $\Bbb R$.

In this case, $\alpha$ is the primitive $m$th root of unity, so $\Bbb K \cong \Bbb Q[X]/(\Phi_m(X))$ where $\Phi_m(X)$ is the $m$th cyclotomic polynomial. Then, $\Bbb K_\Bbb R \cong \Bbb R[X]/(\Phi_m(X))$.
Let's look at a few examples.


*

*$m=3$: $\Phi_3(X) = X^2+X+1$, so $\Bbb K_\Bbb R = \Bbb R[X]/(X^2+X+1)$. This is already a field. It is non-canonically isomorphic to $\Bbb C$, which is indeed a field containing $\Bbb R$ as well as the $3$rd primitive root of unity.

*$m=4$: $\Phi_4(X) = X^2 + 1$, so $\Bbb K_\Bbb R = \Bbb R[X]/(X^2+1)$. This is still a field. It is isomorphic to $\Bbb C$, still non-canonically, but it's way more canonical than the last example (considering how $\Bbb R[X]/(X^2+1)$ is a way of defining $\Bbb C$).

*$m=5$: $\Phi_5(X) = X^4 + X^3 + X^2 + X + 1$, so $\Bbb K_\Bbb R = \Bbb R[X]/(X^4 + X^3 + X^2 + X + 1)$. Over $\Bbb R$, $X^4 + X^3 + X^2 + X + 1 = \left( X^2 + \frac{1+\sqrt5}2 X + 1 \right) \left( X^2 + \frac{1-\sqrt5}2 X + 1 \right)$. By the Chinese remainder theorem, $\Bbb K_\Bbb R = \Bbb R[X]/\left( X^2 + \frac{1+\sqrt5}2 X + 1 \right) \times \Bbb R[X]/\left( X^2 + \frac{1-\sqrt5}2 X + 1 \right) \cong \Bbb C \times \Bbb C$ where the last isomorphism is non-canonical. Quotienting out by a maximal ideal would get you a non-canonical copy of $\Bbb C$.

*$m=6$: $\Phi_6(X) = X^2 - X + 1$, so $\Bbb K_\Bbb R = \Bbb R[X]/(X^2 - X + 1)$, which is a field non-canonically isomorphic to $\Bbb C$.

*$m=7$: $\Phi_7(X) = X^6 + X^5 + X^4 + X^3 + X^2 + X + 1$, $\Bbb K_\Bbb R \cong \Bbb R[X]/(X^2-2\cos(2\pi/7)+1) \times \Bbb R[X]/(X^2-2\cos(4\pi/7)+1) \times \Bbb R[X]/(X^2-2\cos(6\pi/7)+1) \cong \Bbb C \times \Bbb C \times \Bbb C$

*$m=8$: $\Phi_8(X) = X^4 + 1$, $\Bbb K_\Bbb R \cong \Bbb R[X]/(X^2+(2+\sqrt2)X+1) \times \Bbb R[X]/(X^2+(2-\sqrt2)X+1) \cong \Bbb C \times \Bbb C$

*$m=9$: $\Phi_9(X) = X^6 + X^3 + 1$, $\Bbb K_\Bbb R \cong \Bbb R[X]/(X^2-2\cos(2\pi/9)+1) \times \Bbb R[X]/(X^2-2\cos(4\pi/9)+1) \times \Bbb R[X]/(X^2-2\cos(8\pi/9)+1) \cong \Bbb C \times \Bbb C \times \Bbb C$

*$m=10$: $\Phi_{10}(X) = X^4 - X^3 + X^2 - X + 1$, $\Bbb K_\Bbb R \cong \Bbb R[X]/(X^2-2\cos(2\pi/10)+1) \times \Bbb R[X]/(X^2-2\cos(6\pi/10)+1) \cong \Bbb C \times \Bbb C$

Let's generalize the examples. We know that the primitive $m$th roots of unity are $\{ \exp(2ki\pi/m) \mid k \in (\Bbb Z/m\Bbb Z)^\times \}$.
Unless $m=1$ or $m=2$, the roots of unity would not be real, so the corresponding term in the factorization of $\Phi_m(X)$ would include its conjugate $\exp(-2ki\pi/m)$ as a complex root. So the corresponding term is $X^2 - 2\cos(2k\pi/m)X + 1$, since the sum of the roots is $2\cos(2k\pi/m)$ and the product of the roots is $1$. It pairs $k \in (\Bbb Z/m\Bbb Z)^\times$ with $-k$, so we can choose the terms with $0 \le k \le m/2$ as a set of representatives:
$$\Phi_m(X) = \prod_{k \in (\Bbb Z/m\Bbb Z)^\times \\ 0 \le k \le m/2} \left( X^2 - 2 \cos(2k\pi/m)X + 1 \right)$$
Applying the Chinese Remainder Theorem:
$$\Bbb K_\Bbb R = \prod_{k \in (\Bbb Z/m\Bbb Z)^\times \\ 0 \le k \le m/2} \Bbb R[X]/\left( X^2 - 2 \cos(2k\pi/m)X + 1 \right)$$
which is non-canonically isomorphic to $\Bbb C^{\varphi(m)/2}$, where $\varphi$ is the Euler totient function.
