A question on the relation of the Galois group as field automorphisms and the Galois group as permutations of roots Let $f\in K[X]$ be a monic separable polynomial and $L$ a splitting field of $f$. Let $M=\{l_1,\ldots,l_n\}$ be the set of roots of $f$ in $L$, i.e.
$$
f=(X-l_1)\cdots(X-l_n).
$$
The Galois group $Gal(L/K)$ is defined to be group of automorphisms $\delta:L\to L$ fixing $K$. There is a well defined injective group homomorphism
$$
\begin{array}{rcl}
h: Gal(L/K) & \to &\Sigma(M)\\
\delta &\mapsto & (l_i\mapsto \delta(l_i))
\end{array}
$$
with $\Sigma(M)$ the group of permutations of the elements of $M$.
I want to identify the image of $h$ in $\Sigma(M)$ and I have heard something like this: ''The Galois group consists of the permutations $\sigma$ of the roots of $f$ such that every algebraic equation in the roots with coefficients in the ground field $K$ remains true after applying $\sigma$.''
In interpreted this as the following statement:

The image of $h$ consists exactly of the permutations $\sigma\in\Sigma(M)$ such that for each polynomial $g\in K[X_1,\ldots,X_n]$ with the property that $g(l_1,\ldots,l_n)=0$ holds, the statement $g(\sigma(l_1),\ldots,\sigma(l_n))=0$ is also true.

My question is:

Does this really describe the image of $h$ and how can I see that?

Thank you very much.
Update: I found the statement on page 44 in Milne's book on fields and Galois theory but unfortunately no prove is given there. So the question on how to proof remains.
 A: Given any $\delta \in Gal(L/K)$, notice that $h(\delta)$ is exactly the restriction of $\delta$ to the set $M$.  The reason that $\delta \rvert_M$ maps $M$ to itself is that if we apply $\delta$ to both sides of the equation $f(l_i) = 0$, we see that $\delta$ will fix all the coefficients of $f$, as well as fixing the right hand side $0$, and the resulting equation is $f(\delta(l_i)) = 0$.  So if $l_i$ is a root, then $\delta(l_i)$ is also; in other words, $\delta$ sends roots to roots.  Now, since $\delta$ is an automorphism of $L$, it is one-to-one, so no two $l_i$'s are sent to the same root.  Thus $\delta$ must permute the roots $l_i$, which means that $\delta$ induces a permutation of $M$, which we may call $h(\delta)$.  I think you understand this part.
Now, we have $f(X) = (X-l_1)(X-l_2)\dots(X-l_n)$, where $l_i \in L$, and also $f(X) = a_nX^n + \ldots a_1X + a_0$ where $a_i \in K$. (By the way, you want $f$ to be monic in order to get that factorization, so we'll say $a_n =1$ also.)
So let $\delta \in Gal(L/K)$, and then let $g \in K[X_1, \ldots X_n]$ be such that $g(l_1, \ldots, l_n) = 0$.  If we apply $\delta$ to both sides of this equation, it will fix the coefficients of $g$ (since they were in $K$), and it will fix the right hand side of $0$, and it will permute the roots $l_i$.
So what we come up with is:
$$
g(\delta(l_1), \ldots, \delta(l_n)) = 0
$$
Since $h(\delta)$ is the restriction of $\delta$ to $M=\{l_1, \ldots, l_n\}$, it's equivalent to write:
$$
g\Big((h(\delta))(l_1), \ldots, (h(\delta))(l_n)\Big) = 0
$$
Or, writing $\sigma = h(\delta)$, we get
$$
g(\sigma(l_1), \ldots, (\sigma(l_n)) = 0
$$
All of this shows that $Im(h) \subseteq T$, where $T$ is the set of permutations $\sigma \in \Sigma(M)$ that satisfy the property you stated.
To show that $Im(h) = T$, you'd also want to show that $T \subseteq Im(h)$; in other words, given a permutation of the roots that satisfied that property, show that it extends to an automorphism $\delta \in Gal(L/K)$.  For this, you could use $L=K(l_1, \ldots, l_n)$, define $\delta(l_i) = \sigma(l_i)$ for all $l_i \in M$, and define $\delta(c) = c$ for all $c \in K$, and show that the result is an automorphism of $L$.
