The explicit isomorphism $K \otimes_k k_v \cong \prod_{w / v} K_w$ (linked to Adele ring)

Question

I post here cause I think I don't understand something about the Adele ring. I've just begun to learn about it, then sorry if my question seems trivial. Actually, if $K/k$ is a field extension, we have proved that we have the following isomorphism (of topological group), for $v$ a place of $k$ : $$K \otimes_k k_v \cong \prod_{w / v} K_w \quad (1)$$

To prove it, we say that : $K \cong k[X]/(P)$ where $P$ is irreducible over $k$, and we find that $K_w = k_v[X] /(P_i)$, where $P_i$ is an irreducible factor of $P$ on $k_v[X]$. So we have the natural embedding :

$$ i : k[X]/(P) \longrightarrow k_v[X]/(P_i)$$

which sends $\overline{Q(X)}$ on $[Q(X)]_{P_i}$. Now, we have $K=k(x)$ where $x$ is a root of $P$ on $\overline{k}$, and the natural map : $k(x) \longrightarrow k[X]/(P)$, and in the same way, $K_w = k_v(\alpha)$, where $\alpha$ is a root of $P_i$ on $\overline{k_v}$.

Now, I want to describe explicitly the isomorphism $(1)$. We know that we can deduce it by the isomorphism :

$$ \begin{array}{ll} \phi : k[X]/(P) \otimes_k k_v &\longrightarrow \prod k_v[X]/(P_i) \\ \overline{Q(X)} \otimes a &\longmapsto ([aQ(X)]_{P_i}) \end{array}$$

But actually, I'm stuck here. If $x \notin k_v(\alpha)$, then we have the natural embedding :

$i : k(x) = K \longrightarrow k_v(\alpha) = K_w$ which sends $R(x) \in k(x)$ on $R(\alpha)$ on $k_v(\alpha)$. Then I imagine that the isomorphism in $(1)$ will be :

$$ \begin{array}{ll} \phi : K \otimes_k k_v &\longrightarrow \prod K_w \\ Q(x) \otimes a &\longmapsto (aQ(\alpha) = (a.i(Q(x))) \end{array}$$

But if $x \in k_v(\alpha)$, which is what we would want in order to have $K \subset K_w$, then we have the previous embedding, and also the embedding :

$$K \hookrightarrow K_w$$

which sends $x$ to $x$ (i.e the "identity" on $K$).

But about the isomorphism on $(1)$, is it :

$$ \quad (1') \quad \begin{array}{ll} \phi : K \otimes_k k_v &\longrightarrow \prod K_w \\ Q(x) \otimes a &\longmapsto (aQ(\alpha) = (a.i(Q(x))) \end{array}$$

or

$$ \quad (2') \quad \begin{array}{ll} \phi : K \otimes_k k_v &\longrightarrow \prod K_w \\ Q(x) \otimes a &\longmapsto (aQ(x) = (a.(Q(x))) \end{array}$$

It's seems to be the $(1')$ cause if we consider : $k[X]/(P) \longrightarrow k_v[X]/(P_i)$, then $\overline{Q(X)}$ (which is the image of $Q(x)$ is sent to $[Q(X)]_{P_i}$ which is the image of $Q(\alpha)$ (cause $\overline{X}$ is a root of $P$, and sent to a root of $P_i$, i.e $[X]_{P_i}$ which is the image of $\alpha$ (cause after all, $x$ might not be a root of $P_i$).

But it's seems to be weird, cause we would like the isomorphism to be the canonical embedding, no ?

I'm afraid I confuse things... So if someone could help me to understand it, I would thanks him very much !

Answer 1

Why not try with $k=\Bbb{Q},K=\Bbb{Q}(i), k_v=\Bbb{Q}_5$ ? $$\Bbb{Z}[x]/(x^2+1)\otimes_{\Bbb{Z}} \Bbb{Z}_5\overset{(a+bx)\otimes c\to ac+bcx}\longrightarrow \Bbb{Z}_5[x]/(x^2+1)$$ $$\overset{ax+b \to \{ax+b\bmod 5^n\}_n}\longrightarrow\varprojlim \Bbb{Z}[x]/(x^2+1)/(5^n)$$

$$\overset{\{ax+b\bmod 5^n\}_n\to (\{ax+b\bmod (2+i)^n\}_n,\{ax+b\bmod (2-i)^n\}_n)}\longrightarrow\varprojlim \Bbb{Z}[i]/(2+i)^n\times \varprojlim \Bbb{Z}[i]/(2-i)^n$$

In general given $O_{k_v} = \varprojlim O_k/\mathfrak{q}^n$ and $\mathfrak{q}O_K=\prod_j \mathfrak{Q_j}^{e_j}$ the isomorphism is $$O_K \otimes_{O_k} O_{k_v} \to \varprojlim O_K/ (\mathfrak{q}O_K)^n \to\varprojlim \prod_j O_K / \mathfrak{Q_j}^{e_j n} \to \prod_j \varprojlim O_K / \mathfrak{Q_j}^{e_j n}=\prod_j O_{K_{w_j}}$$

Answer 2

It is also easy to give a general explanation of what the map is. Namely, for $k$ a ground field, for $K$ a finite separable extension of $K$ (so it has a primitive element $\alpha$), and for an arbitrary field extension $E$ of $k$, let $P$ be the monic irreducible for $\alpha$ over $k$, and let $P$ factor as $P=\prod_i P_i$ in $E[x]$. Then $$ K\otimes_k E \approx k[x]/P \otimes_k E \approx E[x]/P \approx \bigoplus_i E[x]/P_i $$ where the last isomorphism is given by Sun-Ze's theorem (a.k.a. "Chinese Remainder Theorem").