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Let $f(x) \in F[x]$ have degree $n>0$ and let $L$ be the splitting field of $f$ over $F$. Show that if $[L:F]=n!$ then $f(x)$ is irreducible over $F$.

My approach: I attempted to prove the contrapositive. Suppose $f(x)$ is reducible. Then $f=gh$ where $deg(g)=m > 0$ and $deg(h)=k >0$ such that $m+k=n$. If we let $K$ be a splitting field of $g$ over $F$ then $L$ is a splitting field of $h$ over $K$ and we know that $[E:F] \mid m!$ and $[L:E] \mid k!$. Thus, it holds that $[L:F] \mid k!m!$. We know want to show that $[L:F] \neq n!$ to prove the contrapositive that is, $k!m!|n! \notin \mathbb{Z}$. Is this possible? Or is there another approach.

Any and all help will be greatly appreciated!

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You can end this way: $[L:K]\le m!\,k!$. But we know $\,m!\,k!<(m+k)!$ if $m,k>0$. Indeed $m!\,k!=(1\,2\,\dots\,m)\cdot(1\,2\cdots k)<(1\,2\,\dots\,m)\cdot(m+1)\,(m+2)\cdots(m+k)$. Hence $\,[L:K]<(m+k)!=n!$

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You can also argument like this: $Gal(L/F)$ is a subgroup of $S_n$ with $[L:F]=n!$ elements, so we deduce that $Gal(L/F) = S_n$, hence it acts transitively on the roots of $f$, which shows the irreducibility.

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