Let $G = \text {Gal}(L/K)$. How do I see that $\phi \in G$ is completely determined by $\phi\mathbb( \sqrt[4] {3})$ and $\phi(i)$? Let $K = \mathbb Q$ and $L = \mathbb( \sqrt[4] {3}, i)$. I see that $L \supset K$ is a an Galois extension. Also, after computations I've $[L : K] = 8$ (degree of $L$ over $K$ considered as a vector space).
Let $G = \text {Gal}(L/K)$. 
How do I see that $\phi \in G$ is completely determined by $\phi\mathbb( \sqrt[4] {3})$ and $\phi(i)$ ?
I see that if $\mathbb {\sqrt[4] {3}}, i, 1$ is a basis for $L$ over $K$ then it must be true, since $x = a_1 \mathbb { \sqrt[4] {3} } + a_2 i + a_3 1$ for every $x \in L$ with $a_i \in K$, but this is not the case since the degree of $L$ over $K$ is $8$.
 A: With $L=\Bbb Q(\sqrt[4]3)$ it's $[L:K]=4$ not $8$. But maybe you mean $L=\Bbb Q(\sqrt[4]3,i)$. In this last case the degree over $\Bbb Q$ is $8$ and this last one is a NORMAL extension of $\Bbb Q$ and this is necessary in order to consider Galois extensions: if $L$ is not normal (like $L=\Bbb Q(\sqrt[4]3)$) the Galois correspondance isn't guaranteed.
Said so, an automorphism $\phi\in G=\operatorname{Gal}(L|K)$ (with $L=\Bbb Q(\sqrt[4]3,i)$) is totally determined by knowing where $i$ and $\sqrt[4]3$ go because it's first of all an homomorphism which fixes pointwise $\Bbb Q$.
Explicitely, every element of $L$ is a linear combination of $1$ and powers of $i$ and $\sqrt[4]3$, with coefficients in $\Bbb Q$. To be more precise a base of $L$ over $K$ is given by
$$
\{1,\sqrt[4]3,(\sqrt[4]3)^2,(\sqrt[4]3)^3,i,i\sqrt[4]3,i(\sqrt[4]3)^2,i(\sqrt[4]3)^3\}
$$
hence by the properties of homomorphisms like $\phi(a_0i(\sqrt[4]3)^2)=
\phi(a_0)\phi(i)\phi((\sqrt[4]3)^2)=a_0\phi(i)(\phi(\sqrt[4]3))^2$ ($a_0\in\Bbb Q$) you can see that the knowledge of $\phi(i)$ and $\phi(\sqrt[4]3)$, the automorphism is totally determined.
A: Extending the comments to an answer.
Assume that a field $F$ is given as a subset of a bigger field $\Omega$. Let $S$ be a subset of $\Omega$. We denote by $F(S)$ the smallest subfield of $\Omega$ containing both $F$ and $S$. We can do that because such a thing exists. Namely $F(S)$ is the intersection of all those subfields of $\Omega$ that contain both $F$ and $S$. It follows easily from the subfield criterion that such an intersection is a subfield.
Lemma. If $S=\{a_1,a_2,\ldots,a_n\}$ then
$$
F(S)=\left\{\frac{P(a_1,a_2,\ldots,a_n)}{Q(a_1,a_2,\ldots,a_n)}\,\bigg\vert\,
P,Q\in F[x_1,x_2,\ldots,x_n], Q(a_1,a_2,\ldots,a_n)\neq0\right\}.
$$
Proof. Denote the set given above by $K$. Because $F(S)$ is a field, it is closed under multiplication. Therefore it contains all monomials $c a_1^{k_1}a_2^{k_2}\cdots a_n^{k_n}$, where $c\in F$ is arbitrary and $k_i,i=1,2,\ldots,n,$ are arbitrary non-negative integers.
Because $F(S)$ is closed under addition it must also contain all the elements
$P(a_1,a_2,\ldots,a_n)\in\Omega$ for all the polynomials $P\in F[x_1,x_2,\ldots,x_n]$.
Because $F(S)$ closed under division by non-zero elements it must contain all the fractions $P(a_1,a_2,\ldots,a_n)/Q(a_1,a_2,\ldots,a_n)$ listed above. So $K\subseteq F(S)$.
If we can show that $K$ is a subfield then the claim follows because clearly $K$ contains both $F$ and $S$. This is an immediate consequence of the subfield criterion and is left as an exercise. This implies that $F(S)\subseteq K$. Therefore $K=F(S)$. QED
Proposition. Assume that $F$, $S$ and $\Omega$ are as in the above lemma. Assume that we are given elements $b_1,b_2,\ldots,b_n\in\Omega$. Then there exists at most one $F$-homomorphism of fields $\phi:F(S)\to\Omega$ such that
$\phi(a_i)=b_i$ for all $i=1,2,\ldots,n$.
Proof. Assume that such a homomorphism exists (Note: this is not at all clear and for most choices of elements $b_i$ it does not exist). Because $\phi$ is an $F$-homomorphism, we have $\phi(c)=c$ for all $c\in F$. Because $\phi$ respects multiplication we have for all $c\in F$ is and all non-negative integers $k_i,i=1,2,\ldots,n,$ that
$$
\begin{aligned}
\phi(c a_1^{k_1}a_2^{k_2}\cdots a_n^{k_n})&=\phi(c)\phi(a_1)^{k_1}\cdots\phi(a_n)^{k_n}\\
&=cb_1^{k_1}b_2^{k_2}\cdots b_n^{k_n}.
\end{aligned}
$$
Because $\phi$ also respects sums we get from this that for all the polynomials
$P\in F[x_1,x_2,\ldots,x_n]$ we must have
$$
\phi(P(a_1,a_2,\ldots,a_n))=P(b_1,b_2,\ldots,b_n).
$$
We just proved this when $P$ is a monomial and any polynomial is a finite sum of monomials, so this is clear. 
Because a homomorphism of fields maps a non-zero element to a non-zero element we can conclude that if $Q(a_1,a_2,\ldots,a_n)\neq0$, then also
$$
0\neq\phi(Q(a_1,a_2,\ldots,a_n))=Q(b_1,b_2,\ldots,b_n).
$$
Because a homomorphism of fields maps a quotient of two elements to the quotient of the images of those elements we deduce that we must have
$$
\phi\left(\frac{P(a_1,a_2,\ldots,a_n)}{Q(a_1,a_2,\ldots,a_n)}\right)=
\frac{P(b_1,b_2,\ldots,b_n)}{Q(b_1,b_2,\ldots,b_n)}.
$$
In the preceding Lemma we saw that all the elements $z\in F(S)$ are of the form
$z=P(a_1,a_2,\ldots,a_n)/Q(a_1,a_2,\ldots,a_n)$ for some polynomials $P,Q$. We have shown that the image of $\phi(z)$ is fully determined. QED
A: $L$ need to be normal because otherwise it wouldn't make sense to consider automorphisms of $L$. Take $\Bbb Q(\sqrt[3]2)$: in order to have an automorphism $\sqrt[3]2$ must go in one of its conjugates, which are $\sqrt[3]2,\omega\sqrt[3]2,\omega^2\sqrt[3]2$ where $\omega=e^{\frac{2\pi i}{3}}$. But $\sqrt[3]2\mapsto\omega\sqrt[3]2$ is NOT an automorphism of $\Bbb Q(\sqrt[3]2)$.
Every element of $L$ is a linear combination of $1$ and powers of $i$ and $\sqrt[4]3$ with coefficients in $\Bbb Q$ by definition of $\Bbb K(\alpha)$ where $\Bbb K$ is a field and $\alpha$ is an algebraic element over $\Bbb K$.
This follows from: 


*

*$\Bbb Q[\sqrt[4]3,i]=\Bbb Q(\sqrt[4]3,i)$ (see showing $Q[\sqrt 2] = Q(\sqrt 2)$).

*$\Bbb Q(\sqrt[4]3,i)=\Bbb Q(\sqrt[4]3)(i)$

*if $\alpha$ is algebraic over $\Bbb K$ of degree $n$ then $\Bbb K[\alpha]=\{x_0+x_1\alpha+x_2\alpha^2+\dots+x_{n-1}\alpha^{n-1}\;\;|\;\;x_i\in\Bbb K\}$

