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Let $N/E$ be a (possibly normal) field extension. Let $\sigma : E \to N$ be a embedding (monomorphism) of fields. Assume that $N=E(X)$ ( the intersection of all subfields of $N$ that contain $E \cup X$ ) is a splitting field over $E$ of the set $S:=\{ f_i\}\subseteq E[x]$ of polynomials ; $X$ is the set of roots of all the polynomials in $S$. Then my question is, $N$ is also a splitting field of $S':=\{\sigma f_i \}$ over $\sigma(E)$? Here $\sigma f_i$ means the image of $f_i$ under the induced isomorphism $\bar{\sigma} : E[x] \to \sigma(E)[x]$. I think I understand that every polynomial in $S'$ splits in $N[x]$ ( $\because$ Basic question about isomorphisms and irreducible polynomials ) and stuck at showing that $N= \sigma(E)(X')$, where $X'$ is the set of roots of all the polynomials in $S'$. Perhaps, $X'= \sigma(X)$ and $\sigma(E)(\sigma(X))=E(X)$?

This question originates from trial to show next lemma ( C.f. Hungerford, Algebra, proof of the Lemma 6.11, p.286 )

Lemma : Let $K \subseteq E \subseteq N$ be a field extension with $N/K$ normal. Let $\sigma : E \to N$ be a $K$-monomorphism. Then $\sigma$ extends to $K$-automorphism of $N$.

( First trial to proof ) By the Hungerford, Theorem 3.14 ( p. 264 ), $N$ is splitting field over $K$ of some set $S\subseteq K[x]$ of polynomials. By hungerford's book Exercise 3.2 ( p.267 ), $N$ is splitting field of $S$ over $E$. Note next theorem ( Hungerford's book Theorem 3.8 ( p.260 ) ) :

Theorem 3.8. Let $\sigma : K \to L$ be an isomorphism of fields, $S:=\{f_i\}$ a set of polynoimals (of positive degree) in $K[x]$, and $S':=\{ \sigma f_i\}$ the corresponding set of polynomials in $L[x]$. If $F$ is a splitting field of $S$ over $K$ and $M$ is a splitting field of $S'$ over $L$, then $\sigma$ is extendible to an isomorphism $F\cong M$.

If our question above is true, then by substituting $\sigma|^{\sigma(E)} : E \to \sigma(E)$ to the $\sigma :K \to L$ in the theorem and by substituting $N$ to $F$ and $M$, we obtain an isomorphism $\bar{\sigma} : N \to N$ which extends $\sigma|^{\sigma(E)}$. This $\bar{\sigma}$ is a desired $K$-automorphism in the above Lemma. And our first question is true?

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I was struggling understanding a definition with another question that can be related with your question.

Take $K=\mathbb{Q}$ and the extension $K(Y)$ with $Y$ trascendental over $K$. Then you can consider the embedding $\sigma: K(Y) \rightarrow K(\sqrt{Y})$ that sends $\sigma(Y) = \sqrt{Y}$, that is an isomorphism. Then we take $S = \{X^2-Y\} \subset K(Y)[X]$. Then the splitting field for $S$ over $K(Y)$ is $N = K(Y)(\sqrt{Y}) = K(\sqrt{Y}) $.

For this embedding $S’=\{X^2-\sqrt{Y}\}\subset K(\sqrt{Y})[X]$ and $N$ is not the splitting field for $S’$ over $\sigma(K(Y))=K(\sqrt{Y})$.

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    $\begingroup$ Why $N$ is not splitting field for $S'$ over $\sigma(K(Y))$ ? $\endgroup$
    – Plantation
    Commented May 2 at 19:52
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    $\begingroup$ We can see that the root is $\sqrt[4]{Y}$. But that’s not a demostration. Maybe we can do the argument by dimensions. $[N:\mathbb{Q}(\sqrt{Y})] = 1$. Meanwhile $[N:\mathbb{Q}(Y)] = 2$. Theres no automorfism $\hat{\sigma}: N/\mathbb{Q}(Y) \rightarrow N/\mathbb{Q}(\sqrt{Y}) $ over $\sigma$. If the root was in N we could extend the homomorphism. $\endgroup$
    – user_sion
    Commented May 5 at 11:04
  • $\begingroup$ I think you are right. If $\sqrt[4]{Y}$ is in $N$, then we may apply the theorem 3.8 in my question. But since the existence of automorphism $\hat{\sigma} : N \to N$ over $\sigma$ is impossible, $Y^{1/4} \notin N$ which means that $N$ is not splitting field for $S'$ over $\sigma(K(Y))$. $\endgroup$
    – Plantation
    Commented May 7 at 11:53
  • $\begingroup$ For making certainly sure, the $\sigma : K(Y) \to K(\sqrt{Y})$ is really field isomorphism? Where the transcendental element $Y$ come from? What is the definition of the square root $\sqrt{Y}$? $\endgroup$
    – Plantation
    Commented May 7 at 11:55
  • $\begingroup$ And, perhaps, do you have the Hungerford's Algebra book? In fact, for the Lemma in my question ( Hungerford's book, p.286, proof of the Lemma 6.11 ), he suggested to use the Theorem 3.14 ( p. 264 ), Exercise 3.2 ( p.267 ), Theorem 3.8 ( p.260 ), as I wrote in question. Can we follow his suggestion for the lemma? If so, how? $\endgroup$
    – Plantation
    Commented May 7 at 11:56

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