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Let $K$ be a field and $f \in K[x]$ be a non zero polynomial of degree $n$. Let $L$ be the splitting field of $f$ over $K$.

Prove that $[L:K]$ divides $n!$ - I already proved this.

Now I am stuck at this:

Prove that if $n!=[L:K]$, the polynomial is irreducible.

I have no idea how to proceed, although I think I should go by induction, not sure how to actually carry it out.

user26857
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darkside
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1 Answers1

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Suppose not, $f=gh$ with $\deg(g)=a$ and $\deg(h)=b=n-a$ where $a,b\ge1$.

Now let $K_1$ be the splitting field of $g$ over $K$. So $[K_1:K] \le a!$ and $[L:K_1]\le b!$. So $[L:K]\le a!b!$ which is strictly less than $n!$ if any of $a,b>1$.

Edit 1: $[K_1:K] \mid a! \implies [K_1:K]\le a!$ and ${n\choose k}>1 \iff k\ne0,n$.

user26857
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π-quark
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  • Why does $[K_1:K]\le a!$? amd how do you prove $a!b! <n! , for a>0 $ or $b>0$ ? – darkside Sep 20 '23 at 07:21
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    $\frac{n!} {a!b!} = {n \choose a}$ – π-quark Sep 20 '23 at 07:43
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    One should point out that $L$ is the splitting field of $h$ over $K_1$, and this is why $[L:K_1]\le b!$. – user26857 Oct 21 '23 at 08:05
  • +1, but this answer should be deleted, as a duplicate of that one. – Anne Bauval Oct 21 '23 at 08:08
  • @user26857 Both should be done. It is recommended not to duplicate answers, but this post needs to be reopened in order to be re-closed as a duplicate. – Anne Bauval Oct 21 '23 at 08:14
  • Hi there. I have one unanswered question. Would you like to see this question: https://math.stackexchange.com/questions/2531008/subextension-of-a-field-with-galois-series-of-subextensions-of-prime-degree Thank you very much for your help! – Hermi Nov 15 '23 at 03:35