Galois theory basics, part 1

I was recently looking at algebraic proofs of FTA, and decided to review some Galois theory. So here’s some notes/an attempt to motivate Galois theory (perhaps up to or a bit before fundamental theorem of Galois theory), from the perspective of someone who much prefers field theory/polynomials to group theory/automorphisms (although maybe I should just stop being stubborn).

(Random side note: in the context of abstract algebra, polynomials are naturally motivated by “minimal (free) ring extensions”: if we have a commutative ring {R}, then {R[x]} is the smallest ring containing {R} with the additional element {x} satisfying no relations. On the other hand, any constraints/extra relations would be polynomials, so at least for monic polynomials {f} we get the quotient construction {R[x]/(f)}.)

Suppose {K = F(\alpha)} is a singly-generated field extension (by primitive element theorem this is broader than it seems). If {f} is the minimal polynomial of {\alpha} of degree {d}, then let’s look at how it splits/factors in {K[x]}. If {f} has some set of roots {r\le n} of roots lying in {K} (the “splitting part”/linear factors of {f} in {K[x]}), say {z_1 = \alpha, z_2, \ldots, z_r}.

Generally, the main object of Galois theory seems to be splitting fields (or if we’re lazy, algebraic closures), but I’m still fighting through the material myself so I don’t fully appreciate/communicate this myself. Perhaps the point is just that it’s much easier to work concretely with roots (in algebraic closures) than directly with irreducible polynomials. (For example, we can then work with symmetric sums of the roots, and generally draw lots of intuition from number fields.)

We’ll work in characteristic {0} for convenience, so e.g. the {z_i} are pairwise distinct.

1. “Symmetry” of the {z_i}: crux of Galois theory, and introducing Galois groups

The key is that {K = F(z_1) = F(z_2) = \cdots = F(z_r)} (recall {\alpha = z_1} by definition), since the {z_i} share minimal polynomials {f \in F[x]}.

To make this symmetry precise, we phrase things in terms of {F}-automorphisms of {K}; each {F}-automorphism fixes coefficients of {f}, hence is uniquely determined by sending {\alpha \rightarrow z_i}. Thus they form a Galois group {G := G(K/F)} of size {r}, since we easily check the automorphisms to be bijections of {K}.

2. Looking at the splitting/linear part more closely, and introducing fixed fields

A priori, we know (by definition) that {G} fixes (all elements of) {F}. But if {r} is small, then it’s reasonable that it might fix much more.

With or without this intuition, it’s natural to play around with these automorphisms. Of course, if {p_1p_2\in F[x]} for some {p_1,p_2\in K[x]}, then {p_1p_2 = g(p_1p_2) = g(p_1)g(p_2)} for all {g\in G} (which are {F}-automorphisms). So applying this to {g(f) = f}, we see that {g} “permutes the factors of {f}”. Focusing on the linear factors, we have {\{z_1,\ldots,z_r\} = \{g(z_1),\ldots,g(z_r)\}} (recall the {z_i} are distinct by characteristic {0}).

It follows that {\prod (x-z_i)} has coefficients fixed by {G}, hence in the fixed field {K^G\supseteq F} of {G} (easily check it’s a field). So {K} is a splitting field over {K^G} (it’s Galois over {K^G}). Furthermore, since {G(\alpha = z_1)} covers all the {z_i} ({G} acts transitively on the {z_i}), {G} fixes no proper subset of the {z_i}; in other words, by irreducibility, {\prod(x-z_i) \in K^G[x]} is the {K^G}-minimal polynomial of {\alpha = z_1}.

3. Computational/explicit perspective on fixed field

Certainly we have {J := F(\sum z_i, \sum z_iz_j, \ldots, z_1\cdots z_r) \subseteq K^G}. Does equality hold?

Well, certainly {\prod (x-z_i) \in J[x] \subseteq K^G[x]} is irreducible in {K^G[x]}, hence certainly in {J[x]} as well. So because {J(\alpha) = F(\alpha) = K = K^G(\alpha)}, we get {[K:J] = [K:K^G]}, and by {J\subseteq K^G} this forces {J = K^G}.

Also, note that if two intermediate fields {L_1, L_2} give the same min poly (of {\alpha}), then this common set {S} of {F}-automorphisms is both {L_1} and {L_2}-automorphism. But you fix something in {L_1(\alpha)} only if it’s in {L_1} by a degree argument, so must have {L_2\subseteq L_1} and vice versa. Alternatively, note that the poly lies in {(L_1\cap L_2)[x]} and is certainly irreducible so by degree argument we have {L_1 = L_1\cap L_2}.

4. Summary of results

Cool, so because the (splitting parts in {F[x],K^G[x]} of the min polys of {\alpha}) are the same, the {F}– and {K^G}-automorphisms of {K} are the same. In other words, {G(K / K^G) = G = G(K / F)}. (Note that this is true for any field in between {F,K^G}—the splitting parts are the same.)

But furthermore, {K^G} is just large enough for for {\prod (x-z_i)} to lie in {K^G[x]} yet be irreducible. So finally,

\displaystyle [K:K^G] = r = \lvert G \rvert = \lvert{G(K/K^G)}\rvert = \lvert{G(K/F)}\rvert.

As a corollary, since {n = [K:F] = [K:K^G][K^G:F]}, we have {r\mid n} and {[K^G:F] = \frac{n}{r}}. I’d be interested in a direct derivation/interpretation of the {[K^G:F]}. Also note that {K/F} is a splitting field (of {f}) if and only if {r=n} if and only if {K^G = F}. I’ll probably have more to say later on, perhaps about intermediate fields.

(We can see some “competing goals” between these two paragraphs—in the former we want an intermediate field not too large, so the splitting part of the min poly stays the same; in the latter we want an intermediate field not too small, so the min poly completely splits. It’s quite nice that {K^G} is just right.)

5. More on the {r\mid n} thing

5.1. Good example to play with for the {r\mid n} thing

A good example to play with is {F = \mathbb{Q}}, {K = F(2^{1/6})}. Then {G} is simply between {\pm 2^{1/6}}, so order {2}, so {K^G = F(2^{1/3})}. With the above approach, it’s not clear where {\pm 2^{1/6}\rightarrow \pm 2^{1/6}\omega\rightarrow \pm 2^{1/6}\omega^2}, for instance, comes into play, where {\omega} is just a third root.

5.2. Another perspective on {r\mid n}

A different approach than the one above, that’s more satisfactory in a way: Take a primitive element {\alpha\in K} with min poly {f\in F[x]}, and again factor {f(x) = g(x) h(x) = g(x) \prod (x-\alpha_i)} for {h} of degree {r}, with {\alpha_i\in F(\alpha) = K} for {\alpha_1 = \alpha} a primitive element.

Extend {K} to some {L} (e.g. splitting field of {f} over {K}) so that {f} splits completely in {L}, with {n = \deg{f} = [K:F]}. Let {z_1,\ldots,z_n} be the roots. Now partition these based on their {F(z_i)}. Suppose {s\in S} give {F(s) = K_1}, and {t\in T} give {F(t) = K_2}. Then consider the {F}-isomorphism {\sigma: K_1\rightarrow K_2} taking {s_0\rightarrow t_0}. It must map {S} bijectively to {T}, because {F(p(s_0))} corresponds to {F(p(t_0))} (and {K_1=F(s_0)} corresponds to {K_2=F(t_0)}). So our partition splits guys into sets the same size {r} as {\{\alpha_1,\ldots,\alpha_r\}}, and we get {r\mid n} as desired.

5.3. A similar way to compare {S,T}

A similar way to compare {S,T}. Compare the factorizations of {f(x)} over {K_1} and {K_2}. Since the guys in {S,T} are primitive elements (essentially by definition), {\prod (x-s) = \prod (x-p(s_0))} maps to {\prod (x-p(t_0)) = \prod (x-t)}. In this proof it is natural to note that {S,T} are the roots of {f} lying in {K_1} and {K_2}, respectively, or equivalently that {F(s) = K_1} and {F(t) = K_2} for any {s,t \in S,T} respectively.

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