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$\textbf{Question}$: Would you say that normal extension $\Longleftrightarrow$ splitting field extension is intuitive or counterintuitive?

A splitting field extension $L:K$ is exactly large enough to allow a polynomial $f \in K[x]$ (or, more generally, a set $S$ of polynomials in $K[x]$) to split. It is constructed by introducing the roots $\alpha_1,...,\alpha_n$ of $f$ (or, more generally, the roots of the polynomials in $S$) to $K$, and so $L = K(\alpha_1,...,\alpha_n)$.

A normal extension $L:K$, on the other hand, requires that for each $\beta \in L$, the minimal polynomial $m_\beta \in K[x]$ splits.

($\Rightarrow$) The implication is straightforward to prove. Take $S$ = {$ m_\beta \in K[x]$ | $\beta \in L$}.

($\Leftarrow$) To me, the converse seems almost too much to ask. Specifically, pick a polynomial $f \in K[x]$, form the splitting field extension $L:K$, and take any $\beta \in L$, not necessarily a root of $f$. Then all the conjugate roots of $\beta$ are guaranteed to be in $L$ also? It seems reasonable to expect a counterexample. But, of course, the proof of the converse, which takes noticeably more work, makes the converse undeniable.

Apparently, a rational expression over $K$ in the roots of $f$ must have conjugate roots that are also rational expressions over $K$ in the roots of $f$. Is this intuitive?

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closed as primarily opinion-based by Lee David Chung Lin, Hans Lundmark, nmasanta, Daniele Tampieri, José Carlos Santos Aug 21 at 7:13

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    $\begingroup$ $L/K$ is the splitting field of $f $ with roots $a=a_1,\ldots,a_n$, let $c \in L$ ie. $c=P(a)$ for some polynomial $P \in K[x_1,\ldots,x_n]$, you need to show that any $K$-conjugate of $P(a)$ is of the form $P(b)$ for some $K$-conjugate $b$ of $a$, which follows from extending the isomorphism $\sigma:K(P(a)) \to K(P(b))$ to an isomorphism $\sigma:L \to \overline{L}$, the latter must send $a$ to themselves, thus $b= \sigma(a)$ is a permutation of $a$ and $\sigma(P(a)) = P(b) \in L$. $\endgroup$ – reuns Aug 14 at 19:54
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    $\begingroup$ Your notion of intuitive and mine are no more likely to agree than your taste and mine. Beyond that, the definition of normality that I was raised on is rather different, but whether this makes the equivalence more or less clear-cut I don’t recall. $\endgroup$ – Lubin Aug 14 at 20:06
  • $\begingroup$ By the way, mostly I’ve been following the book by Garling. The proof takes the isomorphism $\tau$ from $K(\beta)$ to $K(\gamma)$ fixing $K$ and sending $\beta$ to $\gamma$, where $\gamma$ is a conjugate root of $\beta$. Then extends it to an isomorphism between splitting field extensions of $f = \tau(f)$ over $K(\beta)$ and $K(\gamma)$, respectively. And then the tower law is applied. $\endgroup$ – Oscar Aug 15 at 0:14
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    $\begingroup$ I'd say it depends on how well-developed your intuition is. But I'd agree that at first sight it's surprising that being a splitting field for one polynomial makes it a splitting field for every irreducible polynomial that has a zero in it. $\endgroup$ – Gerry Myerson Aug 15 at 5:23
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    $\begingroup$ I recall being surprised by this fact the first two times I read the proof. After absorbing a bit more Galois theory it becomes intuitive. $\endgroup$ – Jyrki Lahtonen Aug 16 at 4:04