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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 $n!$.

Here $[F:K]$ denotes the dimension of $F$ over $K$ as a vector space.

I already know it is true for $K=F$, so the problem here is when $f$ is irreducible in $K[X]$.

user26857
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Guadalupe
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    For $f$ irreducible, adjoining one root corresponds to an extension of degree $\deg(f)$. After such an extension $f$ is not irreducible, but the worst case is when it factors into a linear term and a term of degree $\deg(f)-1$. – Michael Burr Apr 28 '16 at 20:00
  • @MichaelBurr this holds for non irreducibles factoring into many different factors, not just two. – Adam Hughes Apr 28 '16 at 20:03
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    @AdamHughes Perhaps I don't understand the point that you're trying to make. I was trying to point out where the $n!$ comes from because if $f$ is not irreducible, or it factors into more than $2$ pieces, those cases can be reduced to the irreducible case. – Michael Burr Apr 28 '16 at 20:57
  • Related: https://math.stackexchange.com/questions/62762 – Watson Jun 16 '16 at 12:00

1 Answers1

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Let $n_1,n_2,\ldots n_k$ be the degrees of the irreducible factors of $f$. Then $n_1+\ldots +n_k=n$. We know the result holds for irreducibles, hence $[F:K]|n_1!\cdot n_2!\cdot\ldots\cdot n_k!$. However the multinomial coefficient

$${n\choose n_1,n_2,\ldots, n_k}={n!\over n_1!n_2!\ldots n_k!}$$

is the number of ways to choose to separate $n$ things into $k$ groups of sizes $n_1,\ldots, n_k$ and as such is an integer. So

$$n! = j(n_1!\cdot n_2!\cdot\ldots \cdot n_k!)$$

for some $j\in\Bbb Z$ which is the definition of divisibility, hence a fortiori $[F:K]|n!$.

Edit (irreducible case lemma): The op seems to think that either $K=F$ or $f$ is irreducible, which is false eg $K=\Bbb Q$ and $f(x) = (x^2+1)(x^2-1)$, so I'm leaving the original answer and adding just the irreducible case.

When $f$ is irreducible, let $K=K_0$. We know that $K_1=K_0[x]/(f(x))$ has degree $n$ and has at least one root, so in $K_1[x], f(x) = (x-\alpha)g(x)$ for some $g(x)$ of degree $n-1$. Then perhaps there are more roots than just one, but when we factor $g(x)$ into irreducibles they are all of degree smaller than $n$, so when we write $g$ as a product of irreducibles we get things of degree adding up to something smaller than $n$. Then we induct as we form $K_2=K_1[x]/(g_1(x))$ where $g_1(x)$ is an irreducible factor of $g(x)$ and since each time the degree of the extension is lowered by at least $1$, we inductively see that $[K_i: K_{i-1}]$ for some $N$ we have $K_N=F$ and $[K_N:K_{N-1}]\cdot\ldots\cdot [K_1:K_0]=[F:K]$ and so since the product on the LHS divides $n!$ by the argument we are done.

Adam Hughes
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  • My problem is that $f$ is irreducible so it has no factors in $F$, aslo I doña know why it holds for irreducibles as you mentioned. – Guadalupe Apr 28 '16 at 20:11
  • @Guadalupe see the edit. – Adam Hughes Apr 28 '16 at 20:20
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    how to you know that $[F:K] \ |\ n_1!\cdot n_2!\cdot\ldots\cdot n_k!$ ? how do you construct the polynomial such that $F \simeq K[x] / P(x)$ ? – reuns Apr 28 '16 at 20:21
  • @user1952009 if you've got a separate question, please do feel free to ask on the forums, right now I'm trying to focus on the op, since his question is about divisibility not construction a primitive element. – Adam Hughes Apr 28 '16 at 20:24
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    lol... it is the same question since $[F:K] = deg(P)$ – reuns Apr 28 '16 at 20:26
  • Sorry, i mentioned it was the case left – Guadalupe Apr 28 '16 at 20:28
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    @user1952009 no it is not, one involves trying to construct a polynomial (hard to impossible) the other uses general facts about degrees (much easier). It is not required to actually find such a $P$, so that's a different question. – Adam Hughes Apr 28 '16 at 20:28
  • are you doing $K_{n+1} \simeq K_n[x] / Q_{n+1}(x)$ where $Q_n(x)$ are the irreducible factors of $P_i(x)$ over $K_n[x]$ with $P_i(x)$ the irreducible factors of $f(x)$ over $K[x]$ ? (you see what I mean, constructing by induction on the irreducible factors of $f(x)$ a field extension of the previous field extension) – reuns Apr 28 '16 at 20:29
  • @Guadalupe no problem it wasn't clear you had already handled the non-irreducible case where $F\ne K$, and I thought you might be assuming either $F=K$ or $f$ is irreducible, which is false since there is a third case. Hopefully the edit helps. Cheers!~ – Adam Hughes Apr 28 '16 at 20:30