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.
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 $L$.