$K/\mathbb{Q}$ is a field extension of finite degree.Then $K\otimes_{\mathbb{Q}}K\cong K_1\oplus \cdots \oplus K_s$

Question

Let $K/\mathbb{Q}$ be a field extension of finite degree.

How to show the following ring isomorphism: $$K\otimes_{\mathbb{Q}}K\cong K_1\oplus \cdots \oplus K_s,$$ where each $K_i$ is a field over $\mathbb{Q}$ of finite degree.

Answer 1

By the primitive element theorem, one can write $K = \mathbb{Q}[X]/(f)$ for some irreducible polynomial $f \in \mathbb{Q}[X]$. Then

$K \otimes _{\mathbb{Q}} K = K \otimes \mathbb{Q}[X]/(f) = K[X]/(f \cdot K[X])$.

Write $f = p_1 \cdots p_k$ where $p_i$ are distinct irreducible polynomials over $K$ (distinct since $f$ was irreducible over $\mathbb{Q}$ and $\text{char}(\mathbb{Q}) = 0)$. Then by the Chinese remainder theorem,

$K[X]/(f \cdot K[X]) = K[X]/ (p_1 \cdots p_k) = \prod_{i=1}^k K[X]/(p_i),$

which is a product of finite extensions of $K$ (hence they are also finite over $\mathbb{Q}$).