The structure of $\mathbb Q[t]/P\otimes_{\mathbb Q}\mathbb R$ and the real and complex roots of the polynomial $P$ Let $P\in \mathbb Q[t]$ be a polynomial with $r$ different real roots and $2s$ different complex but not real roots. How can we prove that $\mathbb Q[t]/P\otimes_{\mathbb Q}\mathbb R = \bigoplus\limits_{r}\mathbb R\oplus\bigoplus\limits_{s}\mathbb C$?
 A: Hint: Assume that $P$ is a monic polynomial. Degree of $P$ is $r + 2s.$ Then the $\mathbb{Q}$ algebra $\mathbb{Q}[t]/(P)$ is a free $\mathbb{Q}$ module of rank $r + 2s.$ (this follows from Cayley-Hamilton theorem.) So $\mathbb{Q}[t]/(P) \otimes_{\mathbb{Q}} \mathbb{R}$ is a free $\mathbb{R}$ module of rank $r + 2s.$
A: First note that $\mathbf Q[X]/(P) \otimes \mathbf R \cong \mathbf R[X]/(P)$: The map
$$ \Phi \colon \mathbf Q[X]/(P) \otimes \mathbf R \longrightarrow \mathbf R[X]/(P), \, \overline f \otimes r \longmapsto \overline{rf}$$
is $\mathbf R$-linear and also surjective ($\overline X^i \otimes 1$ is mapped to $\overline X^i$ and $(\overline X^i)_i$ is a generating set of $\mathbf R[X]/(P)$). Since both $\mathbf R$-modules have dimension $\deg(P)$, the map $\Phi$ is an isomorphism.
Thus we need to determine the structure of the quotient $\mathbf R[X]/(P)$. To do this we want to use the Chinese remainder theorem and the factorization of $P$. By assumption, the polynomial $P$ factorizes over $\mathbf C$ as
$$ P = (X-\alpha_1)\dotsm(X-\alpha_r)(X-\beta_1)(X-\overline {\beta_1})\dotsm (X-\beta_s)(X-\overline {\beta_s})$$
with $\alpha_1,\dotsc,\alpha_r \in \mathbf R$ and $\beta_1,\dotsc,\beta_s \in \mathbf C \setminus \mathbf R$. Write $g_i = (X-\alpha_i)$ and $h_j = (X-\beta_j)(X-\overline{\beta_j})$. I claim that $g_i$ as well as $h_j$ are elements of $\mathbf R[X]$. For the linear polynomials $g_i$ this is obvious. For $h_j$ this follows from
$$ h_j = (X-\beta_j)(X-\overline{\beta_j}) = X^2 - 2 \operatorname{Re}(\beta_j) X + \lvert \beta_j \rvert ^2.$$
Moreover, $g_i$ and $h_j$ are irreducible elements of $\mathbf R[X]$ (the $g_i$ are linear and the $h_j$ do not have a root in $\mathbf R$).
Thus
$$ P = g_1 \dotsm g_r \cdot h_1\dotsm h_s $$
is the factorization of $P \in \mathbf R[X]$ into irreducible polynomials. By Chinese remainder theorem we obtain
$$ \mathbf R[X] / (P) \cong \bigoplus_{i=1}^r \mathbf R[X]/(h_i) \oplus \bigoplus_{j=1}^s \mathbf R[X]/(g_j).$$
For each $i$ the field $\mathbf R[X]/(g_i)$ is an extension of $\mathbf R$ of degree $\deg(g_i) = 1$, that is, $\mathbf R[X]/(g_i) \cong \mathbf R$.
For each $j$ the field $\mathbf R[X]/(h_j)$ is an extension of $\mathbf R$ of degree $\deg(h_j) = 2$, that is, $\mathbf R[X]/(h_j) \cong \mathbf C$.
