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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?

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  • $\begingroup$ Your suspicion 1 is well founded. It may be a tacit assumption that everything takes place inside a fixed extension $\Omega$ of $k$. In that case the splitting field of $f$ over $k$ is unique - not just unique up to isomorphism. Without such a $\Omega$ in the background you first need to do some identifications for it to make sense to restrict $\sigma$ to $E$. $\endgroup$ – Jyrki Lahtonen Jul 2 '14 at 6:01
  • $\begingroup$ @JyrkiLahtonen Should $\Omega$ be an algebraic closure of $E$? I am not sure because I have not learned it yet. $\endgroup$ – YYF Jul 2 '14 at 12:53
  • $\begingroup$ $\Omega$ should definitely be algebraically closed. This time an algebraic closure of $k$ would suffice. On another occasion you may want $\Omega$ to also have as many independent transcendental elements as you need (like infinitely many). Just to be sure that you have elbow room to do whatever you need :-) $\endgroup$ – Jyrki Lahtonen Jul 2 '14 at 14:17
  • $\begingroup$ @JyrkiLahtonen Thank you so much! I am just wondering why no one tell me about this or no textbooks say something about it until I get confused and you tell me this. ;) $\endgroup$ – YYF Jul 2 '14 at 14:19
  • $\begingroup$ Hard to say, really. It may well be that this is explained in an intro to this section. Some other text may be working under the assumption that everything takes place inside the field of complex numbers. It should be mentioned/explained at some point. A logical point is shortly after proving the uniqueness of the splitting field up to isomorphism. After that an exercise asking you to show that without loss of generality we can assume that when $k\subset k^*$ the splitting field over $k$ is also a subfield of the splitting field of $k^*$. $\endgroup$ – Jyrki Lahtonen Jul 2 '14 at 14:25
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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.

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  • $\begingroup$ Thanks, but I still have some confusion. Let us consider both $E/k$ and $E'/k$ are splitting fields of $f(x)$. Although $E$ is isomorphic to $E'$, they are literally not the same set. So, what if $k^*$ contains some roots of $f(x)$ in $E'$? In this case, $k^*$ does not contain all the roots of $f(x)$ in $E$. So, how can you simply say that $E^*$ contains all the roots of $f(x)$? I am not sure if I make my confusion easily understood. $\endgroup$ – YYF Jul 1 '14 at 14:26
  • $\begingroup$ Besides, we can write $E=k(z_1,\cdots,z_n)$ where $z_1,\cdots,z_n$ are roots and live in some splitting field of $f(x)$. But, $\sigma'$, in your notation, does not have to fix $z_1,\cdots, z_n$ and perhaps fixes some other $z_1',\cdots,z_n'$ of $f(x)$ in some other splitting field $E'$. If $E$ and $E'$ are splitting field of $f(x)$ over the same field $k$, I would say $E$ are $E'$ are isormophic, so I have nothing to worry about. But, $E^*$ is a splitting field over $k^*$. As I said in the post, we don't know how $k^*$ is constructed from $k$, so I am not sure if $E^*=k^*(z_1,\cdots,z_n)$. $\endgroup$ – YYF Jul 1 '14 at 14:30
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    $\begingroup$ Alright, but we have an isomorphism $E^*\cong k^*(z_1,\cdots,z_n),$ and then this gives us an embeding of $E$ into $E^*.$ In any case, we can choose $E$ and $E^*$ to lie in the same algebraic closure, and then everything is fine. In effect, an isomorphism does not affect the argument here, right? And thanks for reminding me of this issue. $\endgroup$ – awllower Jul 2 '14 at 4:07
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    $\begingroup$ Could you please show me how to construct an isomorphism $E^*\cong k^*(z_1,\cdots,z_n)$? If you can put it in your answer, I will accept it. BTW, your last comment helps! $\endgroup$ – YYF Jul 2 '14 at 12:52
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    $\begingroup$ BTW, how do you make sense of $k^*(z_1,\cdots,z_n)$ if $z_1,\cdots,z_n$ does not lie in some extension field of $k^*$? Actually, $z_1,\cdots,z_n\in E$. Now, I am wondering if an algebraic closure must be involved here to solve the issue? I have not learned algebraic closure yet, so I am not sure. $\endgroup$ – YYF Jul 2 '14 at 13:54

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