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Let $f ∈ F[x]$ have degree $n > 0$, and let $L$ be the splitting field of $f$ over $F$. I wish to show if $[L : F] = n!$ then $f$ is irreducible over $F$.

Also, the converse of this is false with some counterexample?

I know $[L:F]$ divides $n!$ from this result but i can't make the same method work to prove this~ Let $K$ be a field and $f(x)\in K[X]$ be a polynomial of degree $n$. And let $F$ be its splitting field. Show that $[F:K]$ divides $n!$.

user26857
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Homaniac
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  • If $f$ is reducible, write $f = gh$ with the degrees $m, k$ of $g, h$. With your cited question you can bound the degree of the splitting field by $m! k!$ which is strictly smaller then $n!$ if $m, k > 0$. –  Sep 16 '18 at 16:39
  • The only way to get n! is if splitting off a single root needs a degree $n$ extension (and so on). But the degree $n$ extension for a root means $f$ is irreducible. The converse goes wrong in the "(and so on)" part. – user10354138 Sep 16 '18 at 16:48
  • Ah I see, let me try to figure that out, I am also trying to find a counterexample to prove the converse is false. Would $x^4+1$ over Q work hmm – Homaniac Sep 16 '18 at 16:49
  • Yes that would do. – user10354138 Sep 16 '18 at 16:50
  • Ah I just don't quite understand how the degree n extension for a root means f is irreducible though – Homaniac Sep 16 '18 at 16:53
  • For the counterexample, any cyclotomic polynomial of degree greater than 2 will work. – C Monsour Sep 16 '18 at 17:50
  • Could you explain why cyclotomic polynomial works? – Homaniac Sep 16 '18 at 17:53

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Let $f = gh$ with $\deg(g) = m$ and $\deg(h) = k$ be a reducible polynomial. Let $F(a) = K$ and $K(b) = L$ where $a$ is a root of $g$ and $b$ is a root of $h$. We know that $[L :F] = [L : K][K : F] \mid m! k!$, but also that $m!k! < (m + k)!$ (if $m, k > 0$). Hence, $[L : K] < (m + k)! = n!$; which is a contradiction, therefore our original assumption must have been false, meaning that $f$ is irreducible.

Jonas Hardt
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Tamas Kanti Garai
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