Continuing this post Why is this isomorphism $M \otimes_K L \stackrel{\simeq}{\longrightarrow} M^{[L:K]}$ an isomorphism of $M$ - algebras?
Suppose that $L/K$ is a finite separable extension of fields and let $M$ denote the Galois closure of $L$. Let $\textrm{Hom}_K(L,M)$ denote the set of all $K$ - algebra homomorphisms from $L$ to $M$. Since $L/K$ is separable we know that the number of elements in $\textrm{Hom}_K(L,M)$ is equal to $[L:K]$. Now I want to prove that we have an isomorphism of $M$ - algebras $$\varphi : M \otimes_K L \stackrel{\simeq}{\longrightarrow} M^{[L:K]}$$
Proof that they are isomorphic as $K$ - modules : Write $L = K(a)$ for some $ a\in L$ (we can do this via the primitive element theorem). Then
$$\begin{eqnarray*} M \otimes_K L &=& M \otimes_K K[a] \\ &\cong& M\otimes_K K[x]/(f) \hspace{3mm} \text{where $f$ is the minimal polynomial of $a$ over $K$} \\ &\cong& M[x]/(f)\\ &\cong& M[x]/(f_1\ldots f_{[L:K]}) \hspace{3mm} \text{where the $f_i$ are the distinct}\\ && \hspace{1.5in} \text{irreducible factors of $f$ since $M/L$ is Galois} \\ &\cong& M^{[L:K]}\end{eqnarray*}.$$
Lets suppose I have now $K=k( \sqrt[3]{2})$ where $K=Q(\omega)$ and $\omega$ is cube root of unity ; I know that $K \otimes_k K$ is isomorphic to $K^3$ which is just cartesian product (three copies of $K$) according to above result. If I proceed same way the proof given I reached upto last stage but I don't understand how to use chinese remainder theorem after that.
If $k=\Bbb Q(\omega)$ and $K=k(\sqrt[3]2)$ then the isomorphism $K\otimes_k K\to K^3$ goes as follows: $$(a+b\sqrt[3]2+c\sqrt[3]4)\otimes\xi\mapsto ((a+b\sqrt[3]2+c\sqrt[3]4)\xi, (a+b\omega\sqrt[3]2+c\omega^2\sqrt[3]4)\xi, (a+b\omega^2\sqrt[3]2+c\omega\sqrt[3]4)\xi)$$ where $a$, $b$, $c\in k$, $\xi\in K$. Basically its the $k$-algebra map that takes $$\sqrt[3]2\otimes\xi\mapsto(\sqrt[3]2\xi,\omega\sqrt[3]2\xi,\omega^2\sqrt[3]2\xi).$$
How does CRT come in? Given $\alpha$, $\beta$, $\gamma\in K$ then there is a common solution of \begin{align} f(T)&\equiv\alpha\pmod{T-\sqrt[3]2}\\ f(T)&\equiv\beta\pmod{T-\omega\sqrt[3]2}\\ f(T)&\equiv\gamma\pmod{T-\omega^2\sqrt[3]2}\\ \end{align} in the polynomial ring $K[T]$ which is unique modulo $T^3-2$.