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Let $E$ be a splitting field over $F$ of some $f(x)\in F[x]$, and let $K$ be a splitting field over $E$ of some $g(x)\in E[x]$.
If $\sigma \in Gal(K/F)$, then $\sigma |_E\in Gal(E/F)$.

Let $\lambda=\sigma|_E$.
Since $\sigma$ fixes $F$, we have $\lambda$ fixes $F$.
And also since $\sigma$ is an injection, so is $\lambda$.
So the things left are to prove that $\lambda(E)=E$.

Let $E=F(\alpha_1,\dots,\alpha_n)$ where $\alpha_i\in E$ are roots of $f(x)$.
Since $\lambda$ fixes $F$, $\lambda(\alpha_i)$ are also roots of $f(x)$.
Hence $\{\lambda(\alpha_i)\}\subseteq E$.

Let $x\in E$. Then $x=\sum_{i,j}k_{ij}\alpha_i^j$ where $k_{ij}\in F$.
Hence $\lambda(x)=\sum_{i,j}k_{ij}\lambda(\alpha_i^j)\in E$.
Hence $\lambda(E)\subseteq E$.

Is my last paragraph correct?
And I have no idea how to prove that it is surjection because $\lambda$ does not necessary permute $\alpha_i$. I still haven't use the fact that $K$ is a splitting field of $E$.

Wang Kah Lun
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  • Are you familiar with normal extensions? A finite extension $K/F$ is normal if $K$ is the splitting field of some polynomial $f \in F[X]$. – J. De Ro Jun 17 '19 at 13:24
  • Applying your argument to $\sigma^{-1}$, we have $\sigma^{-1}(E)\subseteq E$. Hence $\sigma(E)=E$. – 19021605 Jun 29 '25 at 07:17

1 Answers1

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Recall the definition of splitting field:

A splitting field of a polynomial with coefficients in a field is the smallest field extension of that field over which the polynomial splits or decomposes into linear factors.

In the second paragraph, you have shown that $\lambda(E)$ contains all the roots of $f(x)$. Also since $F$ is fixed by $\lambda$, $F\subseteq \lambda(E)$. And as you have shown above, $\lambda(E)\subseteq E$. Hence from the definition it follows that $\lambda(E)=E$.

Your last paragraph contains a minor error. An arbitrary element $x\in E$ should be written as a linear combination of arbitrary powers of all elements $\alpha_1,\ldots,\alpha_n$.

cqfd
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  • The OP wrote it as different powers? – J. De Ro Jun 17 '19 at 16:16
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    @EpsilonDelta Yes, it should be something like $\sum (\text{constant})\alpha_1^{i_1}\cdots\alpha_n^{i_n}$ where the sum varies over $i_1,i_2,\ldots,i_n$. – cqfd Jun 17 '19 at 16:21
  • One more question: How to know that $\lambda(E)$ is the smallest field extension? since $\lambda(E)$ maybe larger than $E$. – Wang Kah Lun Jun 18 '19 at 08:45