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It is not hard to prove that for irreducible polynomial $f$ over field $\mathbb K$, the splitting field $\mathbb K_f$ has dimension $[\mathbb K_f:\mathbb K]\leq n!$. I need to find 2 examples $f,g$ that $[\mathbb K_f:\mathbb K]=n!$ and $[\mathbb K_g:\mathbb K]<n!$ for every $n>2$.

For example, for $\mathbb K=\mathbb Q$ and $f=x^3-2$, we have $[\mathbb K_f:\mathbb K]=6=3!$, because every time we extend the field, we only get one root $\alpha=\sqrt[3]2$ of $f$, and $\frac{f(x)}{x-\alpha}\in\mathbb Q(\alpha)[x]$ is still irreducible.

Could someone show me the general way to construct examples and counterexamples for $n>2$? It would be better answer for me that $\mathbb K=\mathbb Q$ and the answer shows me not only 2 examples but also the way to construct.

Dietrich Burde
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HXR
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    Why do you need to find (counter)examples for all $n$? Anyway $x^n-2$ over $\Bbb{Q}$ has a smaller splitting field for all $n>3$. Finding examples with the full extension degree $n!$ over $\Bbb{Q}$ is a bit harder, and more typically another base field is used. A recurring example is $\Bbb{K}$= the field of symmetric polynomials in $n$ unknowns. – Jyrki Lahtonen Apr 28 '24 at 04:47

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At least for a prime number $n=p$ we can find simple explicit examples over $\Bbb Q$. Indeed, let $p>2$ be a prime number. Then the splitting field $L$ of the irreducible polynomial $$X^p+pX^{p-1}+p$$ over $\Bbb Q$ has degree $[L:\Bbb Q]=p!$ with Galois group $Gal(L,\Bbb Q)\cong S_p$. The polynomial is irreducible by Eisenstein, and it has exactly one real root by Descartes sign rule. One can therefore show that the Galois group contains a transposition and a $p$-cycle (since it has an element of order $p$ by Cauchy's theorem). These generate the symmetric group $S_p$, so that the degree of $L$ over $\Bbb Q$ must be $p!$.

Secondly, as Jyrki pointed out, the polynomial $X^n-a$ has a splitting field of degree $\le n\cdot \phi(n)<n!$ for all $n>3$, see this post:

Computing the Galois group of polynomials $x^n-a \in \mathbb{Q}[x]$

Dietrich Burde
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