Neukirch’s Number Theory – why is $ℂ \otimes_ℚ K → K_ℂ,~z \otimes a ↦ (j(z)a)_τ$ an isomorphism?

Question

In §5 of chapter 1 of Neukirch’s Algebraic Number Theory on Minkowski theory, it is claimed that $$ℂ \otimes_ℚ K → K_ℂ,~z \otimes a ↦ zj(a)$$ yields an isomorphism. Here, $K$ is an algebraic number field, $K_ℂ = \prod_τ ℂ$ (where $τ$ runs through all field embeddings $K → ℂ$) and the map $j$ is given by $$j \colon K → K_ℂ,~a ↦ (τa)_τ.$$

I’m having a hard time seeing this isomorphism. What’s the inverse map?

Answer 1

Let $P(x) \in \mathbb{Q}[X]$ be an irreducible polynomial such that $K= \mathbb{Q}[X]/P(X)$. There is a natural isomorphism : $$\mathbb{C} \otimes_{\mathbb{Q}} K \cong \mathbb{C}[X]/P(X).$$ Let $S$ be the set of roots of $P$ in $\mathbb{C}$. On the one hand, by the chinese remainder theorem, there is an isomorphism $$\mathbb{C}[X]/P(X) \rightarrow \prod_{S} \mathbb{C}, \quad Q(X) \mod P(X) \mapsto (Q(s))_{s \in S}.$$ On the other hand, the embeddings of $K$ in $\mathbb{C}$ are the $$\mathbb{Q}[X]/P(X) \rightarrow \mathbb{C}, \quad Q(X) \mod P(X) \mapsto Q(s).$$ where $s$ runs through $S$.

This is why $\mathbb{C} \otimes_{\mathbb{Q}} K \rightarrow \prod_{S} \mathbb{C}$ is an isomorphism.