TITLE: Neukirch’s Number Theory – why is $ℂ \otimes_ℚ K → K_ℂ,~z \otimes a ↦ (j(z)a)_τ$ an isomorphism?
QUESTION [4 upvotes]: 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?

REPLY [2 votes]: 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.
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