# 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 $\sigma(E)$?

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?

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 $$\sigma(K(Y))=K(\sqrt{Y})$$.
• Why $N$ is not splitting field for $S'$ over $\sigma(K(Y))$ ? Commented May 2 at 19:52
• 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. Commented May 5 at 11:04
• 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 $\sigma(K(Y))$. Commented May 7 at 11:53
• 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 $\sqrt{Y}$? Commented May 7 at 11:55