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Let $p\in F[x]$ be an irreducible polynomial in a perfect field $F$. Let $\Sigma_p$ be the splitting field of $p$ over $F$ and $G_p$ its Galois group. If $\deg(p)=n$, then $$G_p\cong H≤S_n$$ But this implies $$|G_p|≤|S_n|=n!$$ In my last question, i proved that, if $\alpha\in \Sigma_p$, then its minimal polynomial $m_\alpha$ shows: $$m_\alpha=\prod_{\sigma\in S}(x-\sigma(\alpha))$$ For some $S\subseteq G_p$. This leads to the obvious conclusion that $\deg(m_\alpha)≤n!$. My question is: is this upper bound tight? If it is tight, can you give an example for $\deg(m_\alpha)>n$?

More formally, let $\tau:\Sigma_p\rightarrow \mathbb{N}$ be the function defined by: $$\tau(x)=\deg(m_x)$$ What's the least upper bound for $\tau$?

Simón Flavio Ibañez
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    By the primitive element theorem, there must be an element $\alpha\in\Sigma_p$ whose minimal polynomial is of degree precisely $[\Sigma_p: F]$. – Mark Dec 17 '24 at 19:39
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    What Mark said! I remembered an example I once did for the heck of it. I figured out a primitive element of the splitting field of a quintic with full Galois group $S_5$. And proceeded to give its minimal polynomial, so of degree $120$. I would never try this without a computer algebra system at hand :-). A commenter said that it looks like a Christmas tree! – Jyrki Lahtonen Dec 20 '24 at 05:10
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    Hi @JyrkiLahtonen it is, indeed, a good-looking polynomial for this time of the year. I just have one question: in wikipedia says the notion of "primitive polynomial" belongs to finite field theory, but you found a primitive polynomial for the splitting field of $x^5-3x^2+1$. Does this mean that primitive polynomials exist for every algebraic field extension of a perfect field? – Simón Flavio Ibañez Dec 22 '24 at 23:42
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    That is a very good question, Simòn! Unfortunately in field theory, the meaning of a primitive element varies a bit depending on the context. In the case of extension of number fields $L/K$, an element $\rho\in L$ is called primitive, if $L=K(\rho)$. That is, $L$ is the smallest field containinf $K$ and $\rho$. OTOH in the context of a finite field $K$, an element $g$ is called a primitive element only when it is a generator of the (cyclic) multiplicative group $K^*$. This is a bit unfortunate. – Jyrki Lahtonen Dec 23 '24 at 05:00
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    (cont'd= My best guess as to the origin of this namespace duplication is here. The meaning of a primitive polynomial then also varies. In the finite field side it is simply the minimal polynomial of a primitive element over the prime field $\Bbb{F}_p$. However, in $\Bbb{Z}[x]$ a polynomial is called primitive, if the gcd of its coefficients is $\sim 1$. – Jyrki Lahtonen Dec 23 '24 at 05:02
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    (cont'd) But, you correctly observed that as every finite algebraic extension of a perfect field is simple, it has a primitive element in the first sense (irrespective of whether the base field is finite). – Jyrki Lahtonen Dec 23 '24 at 05:06

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