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$\DeclareMathOperator{\Gal}{Gal}$I've been self-learning Galois theory (and so some field theory background as well), and came across the following exercise in Artin's Algebra, Chapter 16, Exercise 3.1.

Let $f$ be a polynomial of degree $n$ with coefficients in $F$, and let $K$ be a splitting field for $f$ over $F$. Prove that $[K:F]$ divides $n!$.

If $f$ is irreducible and separable, then I think (correct me if I'm wrong) this can be proven by noticing that the Galois group $\Gal(K/F)$ is a subgroup of $S_n$. Since, for Galois extensions, $[K:F]=|\Gal(K/F)|$, the result follows in this case. For $f$ not necessarily irreducible or separable, it can nevertheless be split into irreducible and separable factors, and then the result should follow.

But I'm trying to get a proof without Galois theory, just by thinking about the fields in an elementary way. I think I have a complete argument, but I'm not sure about its correctness, since I am new to these sorts of arguments and my intuition feels a bit shaky. I would appreciate the help if someone could check it for me!


Attempted Proof. Assume that $f$ is irreducible over $F$. We can describe $K$ explicitly as $K=F(\alpha_1,\dots,\alpha_k)$ for some algebraic numbers $\alpha_i$ over $F$, corresponding to a subset of the roots of $f$. This field can be constructed as a series of extensions $F\subseteq F(\alpha_1)\subseteq F(\alpha_1,\alpha_2)\subseteq\dots\subseteq F(\alpha_1,\dots,\alpha_k)=K$. Let $K_i=F(\alpha_1,\dots,\alpha_i)$ for $1\leq i\leq k$. Then, notice that the extension $K_{i}/K_{i-1}$ is isomorphic to $K_{i}\cong K_{i-1}[x]/(f_{i-1}(x))$ where $f_i(x)$ is defined by $f_0=f$ and $f_{i}=f_{i-1}/m_{i-1}$, where $m_{i-1}$ is the minimal polynomial of $\alpha_{i-1}$ over $K_{i-1}$. Thus, the degree of the extension $[K_i:K_{i-1}]$ is the degree of $m_{i-1}$. Since $f$ can be written as $f=\prod m_i$, then $n=\sum\deg m_i=\sum[{K_i}:{K_{i-1}}]$. It follows that $[K:F] = \prod[K_i:K_{i-1}] \mid n!$.

Note that we have used the claim that for all sequences of positive integers $a_i$ such that $a_1+\dots+a_k = n$, we have $a_1\dotsm a_k\mid n!$. To see this, consider the multinomial coefficient $n!/\prod(a_k)!$ which is an integer.

If $f$ is reducible, we can nevertheless factor it into the product of irreducible factors, and consider constructing the splitting field of $f$ as a series of extensions corresponding to the splitting fields of each irreducible factor. In this way, we get $n$ as the sum of degrees of extensions corresponding to each splitting field extension. By the same multinomial coefficient argument, the result holds. $\square$


By the way, if the proof is correct but there is a much easier way to see that it is true that I'm missing (which I think is highly likely), I would greatly appreciate pointing this out as well.

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  • $\begingroup$ @reuns, thanks for your response; why is that not true? We pick any root $a_1$ of $f$, then its minimal polynomial must divide $f$, so we factor it out and repeat the process. In the end we have $f=\prod m_i$, right? What am I missing? Also, I did think about just using $\deg{m_i}\leq n-i+1$ and you're probably right that it suffices, but I thought it was a bit annoying to justify the case where two of the $m_i$ are equal. In particular e.g. $3\times 3\nmid 5!$ even though it satisfies the inequality. $\endgroup$
    – YiFan Tey
    Oct 31, 2022 at 0:00
  • $\begingroup$ $f_1$ does not make sense since you do not consider $\alpha_0$. Do you mean by $m_{i-1}$ the minimal polynomial of $\alpha_{i}$ over $K_{i-1}$? If so, consider the irreducible polynomial $x^3-2$ over $\mathbb{Q}$. Let $\alpha_1=\sqrt[3]{2}, \alpha_2=\sqrt[3]{2}\frac{-1+\sqrt{-3}}{2}$. $f_0=m_0=x^3-2$ so that $f_1=1$ but $K_1[x]/(1)=0$. $\endgroup$
    – Acrobatic
    Nov 1, 2022 at 3:27
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    $\begingroup$ @Acrobatic Great observation, thanks! What I was trying to accomplish was to adjoin each root of the polynomial one at a time by taking quotients by the minimal polynomial of each, and at each step getting rid of "the part of $f$ that had already been done" and repeating the process. I realise now I didn't quite do it correctly, perhaps the quotient should be by $m_{i-1}$ instead of $f_{i-1}/m_{i-1}$? And I think I forgot to account for issues of existence of roots of unity perhaps, since $x^3-2$ doesn't split over $\mathbb{Q}[x]/(x^3-2)$. $\endgroup$
    – YiFan Tey
    Nov 1, 2022 at 3:48
  • $\begingroup$ @Acrobatic Does it sound like the idea should work if I correct the problems in this way or do you see an easy remedy? Would greatly appreciate the help. (If not, I'd gladly accept it if you expand into an answer explaining the problems with this proof, as well.) $\endgroup$
    – YiFan Tey
    Nov 1, 2022 at 3:50
  • $\begingroup$ The point is that you do not have to construct the minimal polynomials. You can use the fact that if a polynomial $P$ has a root in an extension field, then its minimal polynomial divides $P$. Assume $\operatorname{deg}f=n$, allowing $f$ to be reducible. Adjoining its $n$ roots $\alpha_1 \dots \alpha_n$, which may not be distinct, to $F$ we obtain its splitting field. Let $K_i=F(\alpha_1 \dots \alpha_i), K_0=F$, $g_i(x)=f(x)/(x-\alpha_1)\dots (x-\alpha_i)$. Then, the minimal polynomial of $\alpha_{i+1}$ over $K_i$ divides $g_i$ which is of degree $\leq n-i$. $\endgroup$
    – Acrobatic
    Nov 1, 2022 at 4:14

2 Answers 2

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That possibility did not occur to me. I was wrong.

Alternatively, we can use induction on $n$. Suppose that for polynomials of degree $\leq n-1$ the assertion holds. Let $f=f_1\dots f_r$ the irreducible decomposition of $f$. If $\operatorname{deg}f_1=m\lt n$, then we can use the induction hypothesis to show $[K:F]$ divides $m!(n-m)! $, which also divides $n!$ and it suffices. Thus, we can reduce to the case $f= f_1$ i.e. $f$ is irreducible. Adjoining a root of $f$ to $F$ we obtain $K_1=F(\alpha_1)$ which satisfies $[K_1:F]=n$. The splitting field $K$ of $f$ over $F$ is also the splitting field of $f/(x–\alpha_1)$ over $K_1$. By hypothesis, $[K:K_1]$ divides $(n-1)!$, so that $[K:F]=[K:K_1][K_1:F]$ divides $(n-1)!n=n!$.

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  • $\begingroup$ Wow, that's really clean! Not sure why induction in this way did not occur to me. Many thanks for your time :) $\endgroup$
    – YiFan Tey
    Nov 1, 2022 at 5:07
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For the sake of completion, let me expand the observations by @Acrobatic in the comments into an answer, explaining why my original attempt was wrong.

The most jarring fundamental mistake in the way the proof is written is probably the fact that the quotient $K_{i}[x]/(f_i(x))$ that I was trying to do to adjoin new roots is sort of the opposite of the correct quotient, because we should be quotient-ing by the minimal polynomial and not by something like $f_{i-1}/m_{i-1}$. We can fix this by just writing $m_i$ instead. But even here the proof isn't complete, because the construction was supposed to make $f$ split completely in the end, but this isn't even true, because the quotient $K[x]/(f(x))$ does not give the splitting field of $f$ over $K$ in general, even if $f$ is separable and irreducible --- just take $f(x)=x^3-2$ and $K=\mathbb Q$.

To fix these problems, it makes more sense to instead just divide $f$ out by $(x-\alpha_i)$ for each root $\alpha_i$ in the field with $\alpha_i$ adjoined to that obtained in the previous step; there's no need to think about the minimal polynomial. I was worried about doing this because $\alpha_i$ did not belong to the original base field so this quotient is not well-defined, but there's nothing to worry about because we can just work in the field with $\alpha_i$ adjoined instead (duh). There is also no need to suppose that the $\alpha_i$ are distinct. Even if we make these amendments, though, I'm not sure if there is a way to remedy my original proof based on the step-by-step adjoining of roots, because it doesn't seem obvious to show that the product of degrees of extensions divides $n!$. It feels like there should be a way, but even now, I don't see how.

The value of my failed proof, at least, is that it motivates the inductive solution in the answer by @Acrobatic. :)

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