Embedding of Galois Group I am trying to prove the following:

Let $E/k$ be a splitting field of $f(x)\in k[x]$ with Galois group $G=\operatorname{Gal}(E/k)$. Prove that if $k^*/k$ is an extension field and $E^*$ is a splitting field of $f(x)$ over $k^*$, then $\sigma\mapsto\sigma|E$ is an injective homomorphism $\operatorname{Gal}(E^*/k^*)\to\operatorname{Gal}(E/k)$.

I have two questions:
1) Is it necessary that $E\subset E^*$? Since $E^*$ is a splitting field over $k^*$ and we don't know how $k^*$ is constructed from $k$, I think the answer is no, but I do believe that there should be some embedding from $E$ to $E^*$. Am I right?
2) Assume that somehow we can identify $E$ as a subfield of $E^*$. I can prove that $\sigma\mapsto\sigma|E$ is well-defined and is a group homomorphism. How should I prove that it is injective? I think it would be nice if we can write $E=k(z_1,\cdots,z_n)$ and $E^*=k^*(z_1,\cdots,z_n)$ where $z_1,\cdots,z_n$ are roots of $f(x)$, but can we do that? It is certainly true that $E=k(z_1,\cdots,z_n)$, but I am not sure if $E^*$ can be constructed in the same way as adjoining $z_1,\cdots, z_n$. If it is true, then $\sigma|E=1_E$ implies that $\sigma$ fixes $k^*$ and $z_1,\cdots,z_n$ and hence fixes $E^*$.
Any help?
 A: Firstly, from the definitions, if $E=k(z_1,\cdots,z_n),$ where these $z_i$ are all the roots of $f,$ then $E^*$ is isomorphic to $k^*(z_1,\cdots,z_n).$ So there is a natural embedding $E\hookrightarrow E^*.$ Hence we might view $E$ as a subfield of $E^*.$
Finally, if $\sigma, \sigma'\in\text{Gal}(E^*/k^*)$ with $\sigma\mid_E=\sigma'\mid_{E},$ then $\sigma$ and $\sigma'$ agree on all roots of $f(x).$ Since $E^*$ is generated over $k^*$ by the roots of $f(x),$ and as every element of $\text{Gal}(E^*/k^*)$ must fix $k^*,$ it follows that $\sigma$ and $\sigma'$ agree on each element of $E^*.$ This shows that $\sigma=\sigma'.$
Therefore we have shown that the homomorphism in question is injective.
Hope this helps.
P.S.
I.
We can also define the splitting field of a polynomial over a field as the smallest field extension in which the polynomial splits completely, up to isomorphisms. Then we find that, as $E^*$ is the splitting field of $f$ over $k^*,$ it is a field containing $k$ in which $f$ splits completely, thus it follows that $E\subset E^*.$
Notice that "the" splitting field of a polynomial over a field is unique up to isomorphisms.  

 To show that the splitting field is unique up to isomorphisms, we might use the isomorphism extension theorem to construct the required isomorphism.  

II.
As to what $k^*(z_1,\cdots,z_n)$ means, recall how we construct $k(z)$ for an algebraic element $z:$ let $g$ be the minimal polynomial of $z$ over $k,$ and then we shall have $k(z)\cong k[x]/(g(x)).$
Hope this clarifies some doubts.
