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I'm taking a graduate course in groups and fields. This is a theorem I came across in my professor's notes.

Lemma If $K$ is a splitting field for some polynomial $g$ over $F$. Say, $f(x) \in F[x]$ which is irreducible over $F$ and has zero in $K$. Then $f(x)$ splits in $K$.

The proof begins with the following statement.

If $\alpha_1, \alpha_2, \ldots$ are the roots of $f(x)$, then $[K(\alpha_i):K]$ is independent of $i$.

I'm not sure why this is true.

Can someone please explain why the degree of $K(\alpha_i)$ over $K$ should be independent of which $\alpha_i$ we choose among the roots?

Suppose, the root $\alpha_1 \in K$ then $\alpha_1$ satisfies a degree $1$ polynomial with coefficients in $K$, namely $x - \alpha_1$. However, we do not yet know whether each of the other roots $\alpha_2, \alpha_3, \ldots$ also satisfy a degree polynomial with coefficients in $K$.

This is the full proof in the notes:

enter image description here

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  • $\begingroup$ @Tuvasbien But $f$ is irreducible only over $F$, not over $K$. $f(x)$ does have a root in $K$ as per the lemma's statement. $\endgroup$
    – S.D.
    Oct 25, 2021 at 22:29
  • $\begingroup$ In $K$, that root $\alpha_1$ satisfies a degree 1 polynomial, namely $x - \alpha_1$. Whereas we can't say that about the other roots $\alpha_2, \alpha_3, \ldots$. $\endgroup$
    – S.D.
    Oct 25, 2021 at 22:31
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    $\begingroup$ Assuming that the degree of $K(\alpha _i)$ over $K$ is independent of $i$ is same as assuming the truth of the result to be proved. So this is a circular argument. $\endgroup$
    – Paramanand Singh ♦
    Oct 26, 2021 at 2:18
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    $\begingroup$ I agree with you. The beginning of this proof looks very strange, and this claim cannot be justified at that point. In the material I am using when teaching this stuff, the proof more or less begins with the sentence Since... two lines further down. In other words, the argument used in the posted answer. This proof is as delicate as it is clever, and easy to botch. I guess your teacher wanted to first write down the goal of the argument, in terms of the tower law, and was caught speeding. $\endgroup$
    – Jyrki Lahtonen
    Oct 26, 2021 at 3:45
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    $\begingroup$ The essence of the argument is that A) $[F(\alpha_i):F]$ is independent of $i$ (by irreducibility of $f$), and B) The unicity of the splitting field of $g$ (up to isomorphism) implies that there exists an isomorphism of the extensions $K(\alpha_i)/F(\alpha_i)$ and $K(\alpha_j)/F(\alpha_j)$, compatible with an isomorphism between $F(\alpha_i)$ and $F(\alpha_j)$. With the crucial fact that the isomorphism at the bottom maps $g$ to itself. The argument is sleek, but at least I always get the uneasy feeling that it goes too fast :-) $\endgroup$
    – Jyrki Lahtonen
    Oct 26, 2021 at 3:51

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Let $K_i$ be the spliting field of $g$ over $F(\alpha_i)$, then $[K_i: F(\alpha_i)]$ is independent of $i$, because $F(\alpha_i)\simeq F(\alpha_j)$ for all $i,j$.

But $K_i = K(\alpha_i)$, hence $[K(\alpha_i):F] = [K(\alpha_i):F(\alpha_i)][F(\alpha_i):F]$ is independent of $i$, and therefore $[K(\alpha_i):K] = [K(\alpha_i):F]/[K:F]$ is independent of $i$.

I guess the point here is this is true even if $f(x)$ has no zero in $K$, so it provides more info than the lemma itself.

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  • $\begingroup$ I can't agree with your first paragraph. Note that $f(x)$ is "irreducible" over $F$, not $K$ ($f(x)$ has a root in $K$). So all the $\alpha_i$'s may not share the same minimal polynomial over $K$. $\endgroup$
    – S.D.
    Oct 26, 2021 at 1:54
  • $\begingroup$ You're right. I completely misundersood the question. Now I have an updated answer. $\endgroup$
    – Just a user
    Oct 26, 2021 at 2:36
  • $\begingroup$ This is the way this Lemma is proved in the lecture notes I am using. $\endgroup$
    – Jyrki Lahtonen
    Oct 26, 2021 at 3:36
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    $\begingroup$ There are (at least) two routes to basic facts about Galois theory. One begins with groups of automorphisms, looks at fixed fields, and builds the theory from that direction. Another uses splitting fields. With a few facts about those in place you then get the correct number of automorphisms more or less free of charge. The latter is faster when you want to say something about the automorphisms of finite extensions of $\Bbb{Q}$. The former is more convenient when your starting point is a field that has some known automorphisms, but no preferred base field. Like the field of rational functions. $\endgroup$
    – Jyrki Lahtonen
    Oct 26, 2021 at 3:58
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    $\begingroup$ At least that's how I feel about this. I should probably put in the labor and write my own set of notes explaining both ways. Alas, I'm no Artin. Anyway, the two approaches each have their strengths. $\endgroup$
    – Jyrki Lahtonen
    Oct 26, 2021 at 3:59

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