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Some hours ago I was going through this post and I thought about the following argument to prove the result. For the sake of completeness I will be mentioning both the result and my attempt to prove it in what follows.

Theorem. Let $f(x)\in K[x]$ be a polynomial of degree $n$. Let $F$ be a splitting field of $f(x)$ over $K$. Then $[F:K]$ must divide $n!$.

My Proof. Let $\{a_1,a_2,…,a_r\}∈F$ denote the set of distinct roots of $f(x)\in K[x]$. Suppose also that $p_i(x)$ denotes the degree of the minimal polynomial of $a_i$ over $K(a_1,…,a_{i−1})$ for all $i∈\{2,…,r\}$. Let $\deg(p_i(x))=t_i$.

Then, \begin{align*}[F:K]&=[K(a_1,…,a_r):K]\\&=[K(a_1,…,a_r):K(a_1,…,a_{r−1})]⋯[K(a_1):K]\\&=∏_{i=1}^r \deg(p_i(x))\\&=∏_{i=1}^rt_i\end{align*}

Observe that, $\displaystyle\sum_{i=1}^r t_i\le n$. So, $$\dfrac{n!}{\left(\displaystyle\sum_{i=1}^r t_i\right)!}\in \mathbb{N}$$Furthermore since, $$\dfrac{\left(\displaystyle\sum_{i=1}^r t_i\right)!}{\displaystyle\prod_{i=1}^r (t_i!)}\in \mathbb{N}$$ is also an integer (being a multinomial coefficient) our conclusion follows immediately.

Question

Is my proof correct?

Community
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1 Answers1

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No. It is not correct. You have not justified the inequality $$\sum_{i=1}^rt_i\le n.$$ In fact, that inequality is often false. If $f(x)$ is irreducible, then more often than not $r=n$. And, in the generic case, it is possible that $\deg p_i=n+1-i$. Implying that $$\sum_{i=1}^nt_i=\sum_{i=1}^n(n+1-i)=n+(n-1)+\cdots+3+2+1=\frac12n(n+1).$$

Consider for example $f(x)=x^3-2$, when $a_1=\root3\of2$, $a_2=\omega a_1$, $a_3=\omega a_2$ where $\omega=e^{2\pi i/3}$. We have $p_1(x)=x^3-2$, $p_2(x)=x^2+\root3\of2x+\root3\of4$ and $p_3(x)=x-a_3$. So we have $t_1=3,t_2=2,t_3=1$ as promised. The same thing happens whenever the Galois group is isomorphic to the full symmetric group $S_n$.

Jyrki Lahtonen
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  • If $f(x)$ itself is irreducible then $[F:K]=n$ and we have nothing to prove. If $f(x)$ is not irreducible then it can be factored into irreducible polynomials. Consider $a_2$ Since $a_2$ is a root of the polynomial, there exists at least one irreducible polynomial $q_j(x)$ of which $a_2$ is a root. We can transform it to a minimal polynomial of $a_2$ over $K$, say to $s_j(x)$. Now if $p_j(x)$ be the minimal polynomial of $a_2$ over $K(a_1)$ then $p_j(x)\mid s_j(x)$ and hence $\deg (p_j(x))\le \deg(s_j(x))$. Summing over $j$ we get the inequality. –  Dec 22 '18 at 13:02
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    Wrong, @user170039. Look at the example of $f(x)=x^3-2$ again. It is irreducible over $\Bbb{Q}$ by Eisenstein's criterion. Its splitting field is $\Bbb{Q}(\root3\of2,\omega)$ a degree six extension. When $f(x)$ is irreducible over $K$, you can say that $[K(\alpha):K]=\deg f(x)$ for any root $\alpha$, but $K(\alpha)$ is usually not the splitting field. – Jyrki Lahtonen Dec 22 '18 at 13:06
  • Yes. I think you are right. Can you give a counterexample where $f(x)$ is not irreducible? –  Dec 22 '18 at 13:08
  • I meant a counterexample of my previous argument in the first comment. –  Dec 22 '18 at 13:10
  • In the worst case (when the splitting field is a degree $n!$ extension) it goes as follows. Let $\alpha_1,\alpha_2,\ldots,\alpha_n$ be the zeros of $f(x)$ in the splitting field. Then the worst case is that $$p_1(x)=f(x),$$ $$p_2(x)=f(x)/(x-\alpha_1)=(x-\alpha_2)(x-\alpha_3)\cdots(x-\alpha_n) ,$$ and for all $k$ $$p_k(x)=\prod_{j=k+1}^n(x-\alpha_j).$$ This is exactly what happens with the example $f(x)=x^3-2$ in my answer: $$x^2+\root3\of2x+\root3\of4=(x-\omega\root3\of2)(x-\omega^2\root3\of2)=f(x)/(x-\root3\of2).$$ – Jyrki Lahtonen Dec 22 '18 at 13:16
  • Also, in your comment you try to find the minimal polynomial of $a_2$ over $K$, but at that point what you want is its minimal polynomial over the BIGGER field $K(a_1)$. The same in the succeeding steps. The field get bigger at each step, meaning that we can split $f(x)$ into lower and lower degree factors. – Jyrki Lahtonen Dec 22 '18 at 13:18
  • Let us continue this discussion in chat. –  Dec 22 '18 at 13:21