# Normal extension $\Longleftrightarrow$ splitting field extension, intuitive or counterintuitive? [closed]

$$\textbf{Question}$$: 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?

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

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• $L/K$ 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 $\sigma(P(a)) = P(b) \in L$. – reuns Aug 14 at 19:54
• 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. – Lubin Aug 14 at 20:06
• 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. – Oscar Aug 15 at 0:14
• 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. – Gerry Myerson Aug 15 at 5:23
• 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. – Jyrki Lahtonen Aug 16 at 4:04