For $\sigma : E \to N$, $N$ is splitting field of $S \subseteq E[x]$ over $E$ $\Rightarrow$ $N$ is splitting field of $\sigma(S)$ over $\sigma(E)$?

Question

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?

Answer 1

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 $\sigma(K(Y))=K(\sqrt{Y})$.