Isomorphism between $K_\mathbb{R}$ and $K\otimes_{\mathbb{Q}}\mathbb{R}$ I am reading Neukirch's Algebraic Number theory and I am a little bit confused about the part where he mentioned that $K_\mathbb{R}$ and $K\otimes_{\mathbb{Q}}\mathbb{R}$ are isomorphic via the map $\varphi: a\otimes x\mapsto(ja)x$, where $j:K\to K_\mathbb{R}$ is the map given by $a\mapsto (\tau a)$, where $\tau\in \operatorname{Hom}(K,\mathbb{C})$.

My questions are:
1) What type of isomorphism do we have here? Is it a $\mathbb{Q}$ or $\mathbb{R}$-vector space isomorphism?
2) And can someone show me how $\varphi$ is an isomorphism?

Thanks in advance!
Your question is a bit unclear. 


  Neukrich probably said that with $\tau_1(a),\ldots,\tau_{r_1+r_2}(a)$ all the (pairs of) complex embeddings then we have an isomorphism $$K\otimes_{\Bbb{Q}}\Bbb{R}\to \Bbb{R}^{r_1}\times \Bbb{C}^{r_2}, a\otimes b\to (\tau_1(a)b,\ldots,\tau_{r_1+r_2}(a)b) \tag{1}$$


Given a complex embedding $\tau$ there is a corresponding absolute value $|a|_v=|\tau(a)|$.

$K_v$ is the completion of $K$ for $|.|_v$.

$v$ is the same for the complex conjugate embedding $\overline{\tau}$. 

Either $\tau$ is a real embedding and $K_v\cong \Bbb{R}$, or it is a non-real embedding and $K_v \cong \Bbb{C}$. 

On the other hand $K\cong \Bbb{Q}[x]/(f(x))$, $K\otimes_{\Bbb{Q}}\Bbb{R}\cong \Bbb{R}[x]/(f(x))$ and the latter is a field iff $f$ is irreducible over $\Bbb{R}$.

Conversely if  $\dim_\Bbb{R}(K\otimes_{\Bbb{Q}}\Bbb{R})=\dim_\Bbb{R}(K_v)$ then $\dim_\Bbb{R}(K\otimes_{\Bbb{Q}}\Bbb{R})=\deg(f)$ is $1$ or $2$ and if it is $2$ then $K_v=\Bbb{C}$ thus $K=\Bbb{Q}(\sqrt{-d})$ and $f$ is irreducible.

For any $\tau$ we have a surjective homomorphism $K\otimes_{\Bbb{Q}}\Bbb{R}\to K_v, a\otimes b\to \tau(a)b$.

$(1)$ follows from
$$\Bbb{R}[x]/(f(x)))=\Bbb{R}[x]/(\prod_j f_j(x))\cong \prod_j\Bbb{R}[x]/(f_j(x))$$
Your question is a bit unclear. 


  Neukrich probably said that with $\tau_1(a),\ldots,\tau_{r_1+r_2}(a)$ all the (pairs of) complex embeddings then we have an isomorphism $$K\otimes_{\Bbb{Q}}\Bbb{R}\to \Bbb{R}^{r_1}\times \Bbb{C}^{r_2}, a\otimes b\to (\tau_1(a)b,\ldots,\tau_{r_1+r_2}(a)b) \tag{1}$$


Given a complex embedding $\tau$ there is a corresponding absolute value $|a|_v=|\tau(a)|$.

$K_v$ is the completion of $K$ for $|.|_v$.

$v$ is the same for the complex conjugate embedding $\overline{\tau}$. 

Either $\tau$ is a real embedding and $K_v\cong \Bbb{R}$, or it is a non-real embedding and $K_v \cong \Bbb{C}$. 

On the other hand $K\cong \Bbb{Q}[x]/(f(x))$, $K\otimes_{\Bbb{Q}}\Bbb{R}\cong \Bbb{R}[x]/(f(x))$ and the latter is a field iff $f$ is irreducible over $\Bbb{R}$.

Conversely if  $\dim_\Bbb{R}(K\otimes_{\Bbb{Q}}\Bbb{R})=\dim_\Bbb{R}(K_v)$ then $\dim_\Bbb{R}(K\otimes_{\Bbb{Q}}\Bbb{R})=\deg(f)$ is $1$ or $2$ and if it is $2$ then $K_v=\Bbb{C}$ thus $K=\Bbb{Q}(\sqrt{-d})$ and $f$ is irreducible.

For any $\tau$ we have a surjective homomorphism $K\otimes_{\Bbb{Q}}\Bbb{R}\to K_v, a\otimes b\to \tau(a)b$.

$(1)$ follows from
$$\Bbb{R}[x]/(f(x)))=\Bbb{R}[x]/(\prod_j f_j(x))\cong \prod_j\Bbb{R}[x]/(f_j(x))$$