Tensor product of Galois extension and algebraic closure 
Let $K$ be a field and $L$ is a finite Galois extension of $K$ such that $[L:K] = n$. $\bar{K}$ is the algebraic closure for $K$.
I need a proof for the following facts
  
  
*
  
*$L \otimes_K \bar{K} \cong \prod_{i=1}^n \bar{K}$.
  
*The isomorphism from (1) has the following form:  $\phi(l \otimes k)
        = (k \phi_1(l), k \phi_2(l), \dots, k \phi_n(l))$, where $l \in L, k \in \bar{K}$ and $\{\phi_i: L \to \bar{K}\}$ - are distinct homomorphisms.
  
*Does $(1_{\bar{K}}, 0, \dots, 0) \in \bar{K}^n$? Which $l \in L, k \in \bar{K}$ correspond to $(1_{\bar{K}}, 0, \dots, 0) \in \bar{K}^n$ if does.
  

My idea was the following:
$L=K\left(\alpha\right)$ (via primitive element theorem). There exists an irreducible polynomial $P = P_{min}\left(\alpha, K\right) \in K[X]$ such that $P$ splits in $L$ i.e.  $P(X) = (X - \alpha_1)(X - \alpha_2) \dots (X - \alpha_n)$, where $\alpha_1, \alpha_2, \dots, \alpha_n \in L$ are distinct roots of $P$. I also have the following one $L \cong K[X]/P(X)$ as soon as $P$ is the minimal polynomial for the primitive element $\alpha$.
Therefore 
$L \otimes_K \bar{K} \cong K[X]/P(X) \otimes_K \bar{K}$ 
or 
$L \otimes_K \bar{K}\cong \bar{K}[X]/(X - \alpha_1)(X - \alpha_2) \dots (X - \alpha_n)$. 
Finally with Chinese remainder theorem: 
$L \otimes_K \bar{K}\cong \prod_{i=1}^n \bar{K}[X]/(X - \alpha_i) \cong \prod_{i=1}^n \bar{K}$. (1)
For the second one, note that $\{1, \alpha_i, \dots, \alpha_i^{n-1}\}$ is a $K$-basis for $L$ and we have $n$ different basises constructed in such they. Thus $\phi_i(\alpha^k)$ forms a basis, as soon as, $\alpha_i^k = \phi_i(\alpha^k)$. The basis can be written as columns in a matrix. The linear independency for columns gives the same result for rows i.e. the set $\{v_j = (\phi_1(\alpha^j), \phi_2(\alpha^j), \dots, \phi_n(\alpha^j))\} \subset \bar{K}^n$ consists of $n$ linearly independent elements i.e. it is a basis for $\bar{K}^n$. As result for any $l \otimes k = (\sum_{i=0}^{n-1}k_i \alpha^i) \otimes k \in L \otimes_K \bar{K}$ we have the following element from $\bar{K}^n$: $k (\phi_1(\sum_{i=0}^{n-1}k_i\alpha^i), \phi_2(\sum_{i=0}^{n-1}k_i\alpha^i), \dots, \phi_n(\sum_{i=0}^{n-1}k_i\alpha^i)) = k (\phi_1(l), \phi_2(l), \dots, \phi_n(l))$.
For the last one the main problem for me is that $\phi_i$ is a homomorphism between fields i.e. injection: $\ker \phi_i = \{0\}$. 
Any info about errors in the first and second parts of the proof or hint, solution for the last one is really appreciated.

Updates:
Original statement was the following: $L$ is a normal extension and not necessary be separable. The statement makes the proof invalid (thanks @Starfall for the hint).
Additional 3d question was added
Proof for the second part was added

 A: Your proof for (1) is incorrect - $ L $ need not be isomorphic to $ K[X]/(P(X)) $. For instance, take $ K = \mathbf Q $, $ L = \mathbf Q(2^{1/3}, \zeta_3) $ and $ P(X) = X^3 - 2 $. Then, $ K[X]/(P(X)) \cong \mathbf Q(2^{1/3}) $, which is not isomorphic to $ L $. (This argument can be salvaged in the case when $ L/K $ admits a primitive element, but this need not be true in general.)
Instead, we argue using linear independence of characters. Pick a basis $ b_1, b_2, \ldots, b_n $ for $ L/K $. Then, the elements $ v_i = (\phi_1(b_i), \phi_2(b_i), \ldots, \phi_n(b_i)) $ are linearly independent in $ \bar K^n $: indeed, if they were linearly dependent, the matrix $ A $ whose $ i $th row is $ v_i $ would have linearly dependent rows, thus it would also have linearly dependent columns. A linear dependence relation among the columns of $ A $ would look like
$$ \sum_{k=1}^n c_k \phi_k(b_i) = 0 $$
for all $ i $, and thus, upon taking $ K $-linear combinations, this would give a linear dependence relation among distinct characters $ L^{\times} \to \bar K $, which is impossible. Thus, the $ v_i $ are linearly independent, and by linear algebra, they form a basis of the $ \bar K $-vector space $ \bar K^n $. This proves that the map $ \phi $ is both injective and surjective, thus it is an isomorphism of $ \bar K $-vector spaces. Since it is also an algebra homomorphism, it follows that it is an isomorphism of $ \bar K $-algebras.
