What are some good books for learning Galois Theory?

[Mamta25]

May anyone/someone please suggest/recommend some books on learning Galois Theory? Before learning this pure mathematics subject, is the knowledge of group theory required in order to study Galois Theory? I have the e-textbook of Galois Theory by Ian Stewart, 4th edition but was wondering if there are other Galois Theory books for practice.

Hey, here are some of the best simple books for learning Galois Theory.
1. Galois Theory by Ian Stewart
2. Galois Theory by Steven H. Weintraub
3. Galois Theory by Emil Artin
4. Galois Theory Through Exercises by Brzezinski
5. Field and Galois Theory by Patrick Morandi.
I would suggest start with Galois Theory by Ian Steward because this book is beginner friend and has easy explanation.
I hope this will help.

 

[Math100]

Thank you very much for all the important and useful sources on Galois Theory, everyone!

 

[mathwonk]

I have spent some time pondering the first part of the translation of Galois' "first memoir" by Neumann, roughly pages 107-117 in Neumann's, The mathematical writings of Evariste Galois, from the European Mathematical Union, and comparing it to H. Edwards AMS Notices exposition. I confess I found Galois himself much more clear and helpful than the article by Edwards which attempted to explain Galois to a modern reader. I am grateful however for that link, which illustrated the value of Edwards' own advice.

I.e. to quote Edwards himself, in that article:

" The introduction to my 1974 book Riemann’s Zeta Function was an exhortation to “Read the classics!” A few years later I discovered Niels Henrik Abel’s remark that “It appears to me that if one wants to make progress in mathematics one should study the masters and not the pupils,” which then became my constant refrain. Abel’s contrast of “masters” to “pupils” is an important addition to
the message. “Read the classics” doesn’t just mean read good texts; it means read the texts that gave birth to the subjects or gave them their most vivid statements—those written by the “masters”—not the ones written by later “pupils”, who themselves learned the ideas from the masters and are trying to make presentations that are more accessible or that conform to newer styles."

I.e. Edwards recommends reading Galois, as do I.

Honestly, I admit I found Galois puzzling, but it was easier to understand it by thinking about it, and finding my own explanation, rather than by reading what had puzzled Edwards, and how he resolved his own confusion. E.g. some things that Edwards thought his readers needed explaining, or correcting, were not a problem for me.

Not to dissuade anyone who is helped by Edwards, (or me), but I think many of us, I at least, are reluctant to tackle original works, and I just wanted to put in a plug for actually looking at Galois, and then spending some time wrestling with it.

Now of course I face the same (AI editor changed it to "insane") conundrum as Edwards: to what extent should I continue to attempt here to explain how I understand what Galois meant, and have it be even less clear than Galois, (or Edwards)? The answer is that although it may not help anyone else, trying to explain does help me. That's one main reason people write articles and books and give lectures.

 

[mathwonk]

Ok, going down same rabbit hole as Edwards, trying to explain Galois in modern terms, in particular, why the table at the bottom of page 916 of the linked AMS Notices article, which is on page 116-117, of Neumann, does display the Galois group of the equation g whose roots are a,b,c….

Permutation groups as “substitutions”
We will describe the Galois group of an equation as a certain group of permutations, or substitutions, of the roots. To clarify, a permutation of say the first 5 integers, is a list of those integers, in some order, say: (2, 1, 4, 5, 3). One may consider this as defining a substitution, by taking as the original permutation the usual ordering (1, 2, 3, 4, 5).
Then the permutation (2, 1, 4, 5, 3), defines the substitution taking (1, 2, 3, 4, 5) to (2, 1, 4, 5, 3), i.e. taking 1—>2, 2—>1, 3—>4, 4—>5, 5—>3.

If we had begun with a different permutation as original, say (2, 1, 4, 5, 3) rather than the usual one, then this same permutation (2, 1, 4, 5, 3) would define a different substitution, namely the identity substitution taking (2, 1, 4, 5, 3) to itself.

But once an initial arrangement has been chosen, then each arrangement, i.e. each permutation, defines a unique substitution. Thus when an initial permutation is chosen, we use the terms permutation and substitution interchangeably. Thus if we are given the permutations (1, 2, 3), (2, 3, 1), (3, 1, 2), we think of the first one as the initial arrangement, and then the three given permutations define three substitutions, namely those taking the first to itself, the first to the second, and the first to the third. These three substitutions define a cyclic group of order three. The sequence (2, 3, 1), (3, 1, 2), (1, 2, 3) define exactly the same substitutions, where now (2, 3, 1) is the initial arrangement. It is the group of substitutions that matter; it can be described using any initial ordering of the elements.

We are used to discussing permutations as reorderings of the usual ordering of integers, i.e. we always take the usual order as initial, so to us a permutation is the same as a substitution, but roots of a polynomial do not have a usual ordering, and Galois was trying to emphasize that his group does not depend on the choice of ordering.

Automorphisms of a simple field extension:
Let F(V) = K be a simple algebraic field extension of degree n of F. Then V satisfies a unique irreducible monic polynomial h of degree n over F, and we have an isomorphism of fields K = F(V) ≈ F[X]/(h), i.e. V corresponds to X, mod h. Assume K contains all n roots of h, say V, V’, V’’,……,V^(n-1).

Then K is generated over F also by every root of h, i.e. K = F(V) = F(V’) = F(V’’) = ……. Thus the isomorphism of K with F[X]/(h) can be given so that Xmodh, corresponds to any one of the n roots V, V’, V’’,…. In particular, by composing these isomorphisms K—>F[X]/(h)—>K, we get exactly n automorphisms of K over F, and these take an element p(V) to itself, or to p(V’), or to p(V’’),…..

The Galois group of an equation
Now let F be the coefficient field of g, and let a,b,c…. be all roots of g in a suitable extension field, e.g. an algebraic closure of F. Then let K = F(a,b,c…) be the field generated over F by the full set of roots of g, i.e. K = the splitting field of g over F. Suppose m = degree(g), so there are m roots a,b,c…., and let n = the degree of K over F, i.e. the dimension as vector space.

Then by the theorem of the “primitive element”, proved earlier by Galois, the splitting field K is actually a simple extension of F, i.e. there is some element V such that K = F(a,b,c…) = F(V). This means that every element of the splitting field, in particular every root a,b,c…, is expressible as a rational function, in fact actually as a polynomial, in V, with coefficients in F. E.g. let a = ƒ(V), b = ƒ1(V), c = ƒ2(V),…, ƒm-1(V). (don't know why these alternate f's look different)

Now since the degree of K over F is n, the degree of the irreducible polynomial of V over F is n, so let V, V’, V’’,…,V^(n-1) be the full set of roots of this polynomial.

Then the splitting field K has exactly n automorphisms over F defined by sending an element p(V) to itself, or to p(V’), or to p(V’’),…., p(V^(n-1)). Moreover these automorphisms form a group. Since the roots a,b,c,…of g, generate K as a field over F, each automorphism of K is determined by its restriction to the set of roots. Moreover, each root of g is mapped by any F-automorphism of K to another root of g. Hence the group of automorphisms of K over F restricts isomorphically to a group of “permutations” (substitutions) of the set of roots of g.

Choosing some initial ordering of these m roots, we write them as polynomials in V, i.e. choose as above a = ƒ(V), b = ƒ1(V), c = ƒ2(V),…, ƒm-1(V). then we get the group of the equation g from the following sequence of n permutations of these m roots:

ƒ(V), ƒ1(V), ƒ2(V),…, ƒm-1(V).
ƒ(V’), ƒ1(V’), ƒ2(V’),…,ƒm-1(V’).
…….
ƒ(V^(n-1)), ƒ1(V^(n-1)),…….,ƒm-1(V^(n-1)).

Compare this to the table at bottom of page 916 of the AMS article of Edwards, and to that at top of page 117, of Neumann. Galois explains this in about two pages, (115,117, Neumann), which I recommend.

Remark: Edwards objects, on page 917, to how Galois describes the actions of the permutations on the roots in this table, but in my mind, his revised version of Galois' Prop.1 is no different from what I understood Galois to mean originally. I.e. I think the objection is due to a different interpretation of the term "rational function of the roots". It is true the action on such a function, of a permutation of the roots, is not well defined if you mean by that term "the value of the function", but it is well defined if you mean by that term "the form of the function", i.e. if you mean a rational expression in the roots, rather than the numerical value of that expression. I believe Galois made it clear he considered both meanings when he emphasized that "invariance" of a function, under a substitution, referred not just to invariance of the form, but also of the value of the function.

I.e. let p(x1,...,xm) be a rational function of m variables, with coefficients in the base field F, and given a permutation s of the variables, let (sp)(x1,...,xm) = p(s(x1),...,s(xm)).

If a1,...,am are the m roots of a polynomial g, in a splitting field K, there is no reason for the value p(a1,...,am), an element of K, to equal the value (sp)(a1,...,am) = p(sa1,...,sam), usually another element of K. If they are equal, i.e. if p(a1,...,am) = (sp)(a1,...am), Galois says "the rational function of the roots" p(a1,...,am), is invariant under the permutation s.

If the values are equal for all p such that p(a1,...,am) belongs to F, then Galois says s belongs to the group G of the polynomial g. Moreover, when G is so defined, he then says that conversely, if (the values) p(a1,...,am) = (sp)(a1,...,am) are equal for all s in G, then (the value) p(a1,...,am) belongs to F. This is Galois' definition of the Galois group of g.

It follows that if p(x1,...,xm) is any polynomial relation among the aj, i.e. such that p(a1,...,am) = 0, then also (sp)(a1,...,am)= 0, for any s in G, (since 0 is in F), which implies that each s in G acts, not just on polynomials p(x1,...,xm), but on the field K of their values p(a1,.,..,am). I.e. since the field K is a quotient of the polynomial ring F[x1,...,xm], by the maximal ideal of all such relations, and since s induces a ring automorphism on the polynomial ring, it then also induces a field automorphism of K, fixing F.

Thus to me, this rephrasing of Galois' Prop.1 by Edwards, although perhaps more precise, is not essentially different, nor is Galois' version "flawed". But these are matters of individual taste. For me, I never realized until reading Galois, that one can actually write down the Galois group in such an explicit elementary way. Of course "explicit" is generous, since upon examination, although one can prove that, in Galois' table, the entry ƒ(V') say, is one of the roots, one does not know which one it is!
It also follows easily from this approach that the only elements of K fixed by all permutations in G are those of the base field F. I.e. an element of K is expressible as a polynomial f(V) over F, where f(X) is a polynomial over F of degree < n. If it is fixed by all n permutations of G, then all n elements f(V), f(V'), f(V''),...,f(V^(n-1)), are equal in K. But a polynomial of degree <n over a field, cannot have the same value at n distinct arguments unless it is constant. Hence the polynomial f(X), over F, is a constant, hence f(V) belongs to F. I did not realize before how easy this is.

 

[mathwonk]

In terms of quality of exposition versus price, there can perhaps be nothing to match the great, but succinct and theoretical, treatment from 1942 by Emil Artin, (Mike's father), (ed. and supplem. by Arthur Milgram) , available for $7.39 at amazon! Artin throws in a complete elementary introduction to linear algebra in the first 20 pages. (My copy cost $2.50 in 1963.)
https://www.amazon.com/Galois-Theor...-Mathematical/dp/0486623424?tag=pfamazon01-20

Even clearer, to me, is the exposition provided by Emil Artin's son Mike, in his book: Algebra, by Michael Artin, more expensive, but a great bargain as well, especially since its covers a great deal more. See chapter 14, and pages 552-560 in particular for this topic.

 

[WWGD]

There's an element of subjectivity in the quality of the book. I aways suggest, if you have a chance, drop by a bookstore* , or school library and browse through several books to see which feel right to you. As a rule of thumb, if the book is well-organized, has a few solved problems, the author has made efforts to write a good book.
*RIP Borders Books.

 

[mathwonk]

Here is a link to a beautiful and eloquent talk given recently by a friend of mine, Michael Comenetz, containing the gentlest and clearest possible introduction to Galois theory, amazingly aimed at "anyone", i.e. essentially no background at all assumed by the reader, (well, maybe high school algebra). Dr. Comenetz, (PhD in algebraic topology), is a tutor at the St. Johns College, Annapolis, famous for its high level undergraduate teaching of authentic scholastic texts.
https://michaelcomenetz.wordpress.com/2025/10/29/modern-mathematics/#more-1878

 

[mathwonk]

Trying to answer this question made me aware that I did not have much of a grasp on Galois theory, and has sent me off thinking about it for months. I think I understand it better now, and offer these comments:

Galois theory in characteristic zero (any field containing the rational numbers):

Any given finite field extension E of degree n of a field F, can be represented as isomorphic to a quotient of F[X] by many different irreducible polynomials of degree n over F; and if galois, the galois group is thus represented as a transitive permutation group of the n roots of any one of them. I.e. given a finite extension E of F, we can choose, in many ways, an irreducible polynomial g over F, such that E is isomorphic to F[X]/(g), where X corresponds to a distinguished root of g in E, and if all roots of g also lie in E, the field automorphisms of E fixing F, are determined by what they do to the roots of g, indeed even by what they do to the one distinguished root corresponding to X.

Moreover, all the elements of the field are represented as polynomials of degree < n in the root corresponding to X under the isomorphism, and that root can be taken as any one of the roots of g in E. It follows that the number of field automorphisms of E fixing F, equals exactly the number of roots of g contained in F, and thus is at most equal to the degree n of the extension; moreover if all roots of g are in E, i.e. if the extension is Galois, that group has order equal to the degree n.

If a1,…,ar are the roots of g in E, then there are exactly r automorphisms of E over F, defined by composing the isomorphisms ƒj:F[X]/(g)—->E taking X to aj. I.e. every F-automorphism of E has form ƒj o ƒ1^-1, taking a1 to one of the aj.

E.g. all irreducible polynomials over the real field R are either linear or quadratic, and adjoining the roots of any one of the quadratic ones gives the same finite extension of R, namely the complex numbers. Hence the complex number field is isomorphic to any quotient R[X]/(q(X)) with q an irreducible quadratic real polynomial. We usually use the simplest such polynomial X^2+1, and represent the complex numbers as R[X]/(X^2+1), with X corresponding to i. Then every complex number is represented as a linear real polynomial in i. The multiplication in the field is defined by multiplying two such polynomials and then setting all occurrences of X^2, or i^2, equal to -1.

We could also use any other irreducible real quadratic, such as X^2-2X+2, which we will write as t^2-2t+2, since the variable t will represent a different complex number now than before, namely a root of t^2-2t+2. Then the complexes, isomorphic to R[t]/(t^2-2t+2), are represented as linear polynomials in t, where multiplication is defined as polynomial multiplication, then setting all occurrences of t^2 equal to 2t-2.

Both these fields R[X]/(X^2+1), and R[t]/(t^2-2t+2), are isomorphic to the complex numbers, and in particular to each other, since sending t to X+1, sends t^2-2t+2 to (X+1)^2-2(X+1) + 2 = X^2+1. Since X corresponds to i, and t corresponds to X+1, this means that in the representation R[t]/(t^2-2t+2), i corresponds to t-1. That is, t represents a distinguished root of t^2-2t+2 rather than a root of X^2+1.

Thus each complex number could be represented by a real linear polynomial in t, or i+1, just as well as by a real linear polynomial in X, or i.

Key results:

F(a) ≈ F[X]/(g) if g is the irred polynomial for a over F.
This is the main result for defining homomorphisms, hence isomorphisms and automorphisms, of fields. If F(a) is the splitting field of g, this implies the galois group of F(a) over F has order equal to degree g, and more generally that the galois group of any finite extension E/F has order ≤ degreeE/F.

If E is generated over F by one root a of the polynomial g, and if E also contains all the other roots of g, a=a1,a2,…an, then E is the splitting field of g, and is equal to F(aj) for every j, hence is isomorphic over F to F[X]/(g) via sending X to any aj. Hence there are n isomorphisms of E over F, by sending a to any aj, i.e. by composing the isomorphisms E = F(a) —> F[X]/(g)—>F(aj) = E. Explicitly, if b = f(a) is any element of F(a) = E, where f is any F polynomial of degree < n, then sending b = f(a), to f(aj), defines an automorphism of E over F.

I.e. to define automorphisms of E, we must know how to define F- algebra maps, and this is done most easily using the universal property of the polynomial ring F[X], and its quotients F[X]/(g). Namely, if E is any F algebra, and a is any element of E, there is a unique F algebra map F[X]—>E sending X to a, which sends each polynomial f(X) to f(a). This descends to an F algebra map F[X]/(g)—>E if and only if g(a)=0.

It follows also that if E is the splitting field of the irreducible polynomial g, (i.e. the "root field" generated by all roots of g), these properties of E do not depend only on g. I.e. if K is an algebraically closed field containing E, then it follows that any F -homomorphism of E into K, must permute the roots of g, and since these roots generate E, any such map is an automorphism of E. Hence if h is another irreducible F-polynomial with one root b in E, and such that that root generates E, i.e. if E = F(b), there is an F isomorphism from E to F(bj) for all roots bj of h; since this map must be an automorphism of E, all these roots bj lie in E.

In fact it is not necessary for the root b to generate E, i.e. if E is the splitting field of some irreducible polynomial g, and if h is any other F irreducible polynomial with one root in E, then all roots of h lie in E. This follows from the basic extension principle for homomorphisms, which also follows from the fundamental rule above for defining F algebra homomorphisms. We state it next:

Lemma: If E/F is any finite extension, and H is a subfield of E over F, and K is algebraically closed, containing E, then any F homomorphism of H into K extends to an F homomorphism of E into K.
proof: Write E as a simple extension of H, i.e. E = H(a) and let H’ be the image of H in K under the given homomorphism. Then H[X] ≈ H’[X], and the irreducible H polynomial h of a corresponds to an irreducible H’ polynomial h’, and we let a’ be a root of this polynomial in K. We can thus define an F isomorphism E = H(a) ≈ H[X]/(h)—>H’[X]/(h’) = H’(a’). qed

Now in the situation above, where E is the splitting field of g over F, if h is any irred F polynomial with root b in E, and another root b’ in K, then the F isomorphism from the subfield F(b) of E to F(b’) in K extends to an F homomorphism of E into K, which must be an automorphism, hence b’ also lies in E. qed.

This illustrates that it is highly useful to be able to prove facts about a field extension using just one polynomial representation, and then deduce facts about other polynomials.

“Primitive element theorem”: every finite extension in char zero is simple, i.e. if E/F is an extension of finite degree (finite vector dimension as F-vector space), there is some element a in E such that E ≈ F(a) ≈ F[X]/(g) where g is the (monic) irreducible polynomial for a over F.
This is the key technical result allowing the basic technique for defining maps to be applied to all finite field extensions.

If G is finite group of automorphisms of E and F = fixed field of G, then for any element a of E, a is algebraic over F of degree ≤ #G, and the roots of the irred F polynomial of a, are the elements of the G orbit of a; in particular they all lie in E.
E.g. complex conjugation, together with the identity map, is a group of order two of automorphisms of the complex numbers, with the reals as fixed field. Then the roots of any irreducible real polynomial are conjugates of each other. This helps prove the FTGT, the 1-1 correspondence between subfields and subgroups, in particular that every subgroup occurs as Galois group of some subfield.

 

[mathwonk]

Summary:
The study of finite field extensions E/F by Galois theory is an analysis of the field automorphisms of E that leave F pointwise fixed. The logic proceeds in the following order:
Choose an element a of E-F;
Find the monic irred polynomial g for a over F;
Form the representation of the field extension generated by a as F(a) ≈ F[X]/(g); there always exists an a so that F(a) = E, by the primitive element theorem;
Form the galois group Gal(E/F) of F(a), or E, over F, i.e. Gal(E/F) = all automorphisms of E leaving F fixed. This group has order ≤ degreeE/F, or degree g in case of E = F(a) = F[X]/(g).
Determine the fixed field of Gal(E/F); it is an intermediate field between F and E and it equals F if and only if #Gal(E/F) = degree(E/F). In this case the extension is called galois. In this case, there is a 1-1 correspondence of intermediate fields between F and E, and subgroups of Gal(E/F). When E = F(a), the extension E/F is galois if and only if E contains not only the root a, but also all other roots of the irreducible polynomial g for a.

I.e. an element a algebraic over F determines an irreducible F- polynomial g, which determines a field extension E = F(a) ≈ F[X]/(g) over F, of degree = n = degree(g), and with a distinguished generator a. This field extension determines a galois group Gal(E/F) of order d | n. (i.e. d divides n).
This galois group has a fixed field F’ between F and E such that degree(E/F’) = d, and degree(F’/F) = n/d, where d = the total number of roots of g that lie in F(a)=E. The Galois group has useful information only about the extension E/F’, which is necessarily galois.

Galois group of a polynomial:
If F is a subfield of an algebraically closed field K, and g is an irreducible polynomial of degree n over F, let E be the field generated by all roots a1,…,an of g in K. Then any isomorphism of E into K must permute the roots of g, hence is an automorphism of E. As above, we can map the subfield F(a1) of E isomorphically to any subfield F(aj), and then extend to an isomorphism of all of E into K, which must be an automorphism of E. Thus the galois group of g, defined as the group Gal(E/F) of all automorphisms of E over F, is representable as a subgroup of Sn, and acts transitively on the roots of g.

Application to solvability of equations:
Galois used this theory to study solvability of polynomials by radicals as follows: a polynomial is solvable by radicals if and only if its root field is contained in an extension formed by repeatedly forming root fields of polynomials of form X^n-a. Then he showed that root fields of this sort have galois groups of special form, not shared by root fields of most polynomials.

In particular, the root field of X^n-1 over the rationals Q has abelian galois group, and the root field of any polynomial of form X^n-a, over the root field of the previous polynomial, also has abelian galois group. Hence the root field of any solvable polynomial has galois group composed of a tower of abelian groups, i.e. is a “solvable” group.

However the root field of the “general polynomial” of degree n, e.g., whose roots are variables, has galois group equal to the symmetric group on n elements, hence is not solvable for n ≥ 5. I.e. since the roots are variables, all n! permutations of them yield field automorphisms over say Q, by the universal principle for defining homomorphisms of polynomial rings. Then the fixed field of these permutations is the field of so called “symmetric functions”, which are by definition those preserved by all permutations. In particular the root field of the general polynomial of degree n is galois over the field of symmetric functions on n elements. But the elementary symmetric functions of the roots generate this field, and these are precisely the coefficients of the general polynomial. Thus the root field of the general polynomial as an extension of its coefficient field, is galois of degree n!, with galois group Sn, not a solvable group if n ≥ 5. Hence there is no general formula for solving all polynomial equations of degree n ≥ 5.

One can also show that for most particular polynomials of degree n≥5, there is no formula in radicals solving even that one polynomial. As I recall, e.g. any irreducible polynomial of prime degree p over Q with exactly two imaginary roots has galois group Sp, and hence no such formula if p ≥ 5. E.g. X^5-80X+2 works.

 

[mathwonk]

Fundamental theorem of Galois theory in a nutshell:

Let K be a finite degree separable field extension of F (e.g. any extension of finite F vector space dimension, when F has characteristic zero), and define G(K/F) = Galois group of K over F = group of field automorphisms of K, fixing F. Thus F is by definition always contained in the fixed field of G(K/F), and one can show #G(K/F) = order of G(K/F) ≤ degree(K/F).

Then the following are equivalent:
1) F is precisely equal to the fixed field of G(K/F);

2) #G(K/F) = degree(K/F);

3) For every x in K, the minimal polynomial of x over F has all its roots in K;

4) Every embedding of K into an algebraic closure of K which is the identity on F, is an automorphism of K;

5) K is generated as an F algebra, by the roots of some polynomial with coefficients in F;

6) K is generated as an F algebra, by the roots of some irreducible polynomial with coefficients in F.

Defn: A finite, separable, field extension K/L with any of the previous 6 properties is called a Galois extension.

Cor: Fundamental theorem of Galois theory:

If K/F is Galois, e.g. if the fixed field of G(K/F) equals F, then there is a natural one-one correspondence between subgroups of G(K/F), and subfields of K containing F, defined as follows:

if H is a subgroup of G(K/F), H corresponds to its fixed field, i.e. to the subfield of K on which H acts as the identity; if L is a subfield of K containing F, then L corresponds to the subgroup G(K/L) of G(K/F) consisting of all elements of G(K/F) that fix all points of L.

These correspondences are inverse to each other, i.e. if L is the fixed field of the subgroup H, then by definition H is contained in G(K/L) but in fact H = G(K/L); and if L is any intermediate field and H = G(K/L) then by definition L is contained in the fixed field of H but in fact L equals that fixed field.

Moreover, if L is an intermediate field between F and K, the orbit of L under G(K/F) corresponds to the conjugacy class of G(K/L) in G(K/F). In particular, the field L is Galois over F if and only if the subgroup G(K/L) is normal in G(K/F); i.e. L is the only member of its orbit if and only if G(K/L) is the only element of its conjugacy class. In this case G(L/F) ≈ G(K/F)/G(K/L).

It follows that if K/F is Galois, and L is an intermediate field, then #G(K/L) = deg(K/L), and indexG(K/L) in G(K/F) = deg (L/F); and

If H is a subgroup of G(K/F), and L is the fixed field of H, then #(H) = deg(K/L), and index(H) in G(K/F) = deg(L/F).

Key lemmas:
1) If x is any element of K, algebraic over F, and m is the minimal polynomial of x over F, then F(x) ≈ F[X]/(m).
2) If L is an intermediate field of a finite extension K over F, then any F embedding of L into an algebraic closure of K, extends to an embedding of K.
3) If K is a finite extension of a field F of characteristic zero, there is an element x of K such that K = F(x).

Some indications of arguments, using the lemmas:
E.g. assuming K is generated by the roots a1,…an of some irreducible polynomial m over F, then by 1) each field F(aj) is a subfield of K isomorphic to F[X]/(m), hence all of them are isomorphic to each other. By 2), each isomorphism F(a1) ≈ F(aj) extends to an F isomorphism of K into an algebraic closure of K, which must permute the roots of m, hence must be an F automorphism of K. In particular K has at least n F-automorphisms. Using lemmas 3 and 1, property 3 above implies property 6.

I might visualize K as a circular candelbrum or chandelier, with F as the point at the bottom, the fields F(aj) as individual candle holder arms, and the elements of G as rotations of the candelabrum, permuting the positions of the candleholders.

 

[martinbn]

OT: @mathwonk , you often write long and good posts, which I enjoy reading even if I understand the topic well. I find them very insightfull. Have you considered formating them and posting them as insight blog posts here https://www.physicsforums.com/insights/

 

[mathwonk]

My friend, I thank you for the great compliment. I fear though that my weakest point is formatting. I seem to just keep generating poorly formatted content. (My apologies: I just posted even more.)

The following posts are my take on explaining the whole story, now that I have digested it all again, to anyone who knows the meaning of the key words.

 

[mathwonk]

Outline of Galois theory:
(key concepts: groups, normal subgroups, quotient groups, fields, rings, polynomial rings and their quotient rings, vector space dimension)

The problem is to decide whether all, or if not all, which polynomials can be solved by algebraic formulas in the coefficients involving only radicals. (All fields here have characteristic zero, i.e. contain the rational numbers, and all extensions considered are finitely generated algebraic, hence finite vector dimensional.)

Formulating the problem in terms of field theory:
If a polynomial is solvable as above, then the field extension of its coefficient field, generated by adjoining the solutions of the polynomial, i.e. the root field, is contained in an extension of the coefficient field, obtained by repeatedly adjoining nth roots for various n. Such extensions are called radical.
So the question becomes: which root fields are contained in radical extensions of the coefficient field? To answer this question, Galois invoked a powerful new invariant of field extensions.

Galois’ invariant: symmetry groups
Why should a root field be symmetric over its coefficient field? This follows from the fundamental theorem of Kronecker: if g(X) is an irreducible polynomial with coefficients in the field F, and if a is any root of this polynomial in an extension field, then the field F(a) generated by adjoining a to F, is F-isomorphic to the abstract field F[X]/(g), i.e. by an isomorphism fixing F point-wise.

If K is the root field of g, generated over F by the roots a1,…,an, of g, then K contains and is generated by, the subfields F(a1), …,F(an). By Kronecker’s theorem, these subfields of K are all isomorphic to F[X]/(g), hence also to each other. Since in fact each isomorphism F(ai) ≈ F(aj) extends to an automorphism of K, these isomorphisms of the fields F(aj) define symmetries of the root field K. Fundamentally, these symmetries of K form a finite group.

Basic definition: if K is any field extension of F, the Galois group G(K/F), of K over F, is the group of F-automorphisms of K, i.e. those leaving F point-wise fixed.

A root field over F, i.e. a field extension K generated over F by the full set of roots of a polynomial with coefficients in F, is called a (finite) Galois extension of F. Such a field necessarily has finite vector dimension over F.

Useful facts: Every finite dimensional extension E of F, in an algebraic closure F*, is F-isomorphic to only a finite number of other extensions of F in F*, and the smallest field K containing them all, their compositum, is a minimal Galois extension of F containing E.
The Galois group of a finite (dimensional) extension has order bounded above by the dimension, and the extension is Galois if and only if the order of the Galois group equals the vector dimension of the extension.

Galois realized that the group of a radical root field extension, i.e. of a radical Galois extension, has a special form, not shared by all Galois extensions; namely, as he put it: these “solvable” groups can be repeatedly “reduced by primes”, until the identity group is reached.

More precisely, a finite group G can be reduced by a prime p, iff (if and only if) it has a normal subgroup N of index p, i.e. such that G/N ≈ Z/pZ. G is solvable, iff it has a normal subgroup N of prime index, and N has a normal subgroup H of prime index, etc,.., until eventually the last normal subgroup itself has prime order.

Since this process is true of finite abelian groups, a finite group G is solvable iff it has a normal subgroup N such that G/N is abelian, and N has a normal subgroup H with N/H abelian, etc,…., until eventually the last normal subgroup is itself abelian.

Not all finite groups are solvable; e.g. since when n ≥ 5, the only non trivial normal subgroup of the symmetric group Sn is the alternating subgroup An, which has no non trivial normal subgroups, a Galois extension with group An or Sn cannot be solvable unless n≤4. This can be used to show that the general polynomial over Q of degree ≥ 5 is not solvable by radicals.

It follows from the Jordan Holder theorem of group theory that if G is a finite group and H is a normal subgroup, then G is solvable iff both H and G/H are solvable.

We explain next Galois’ criterion, that Galois group of a radical Galois extension field is a solvable group.

 

[mathwonk]

Galois groups of solvable polynomials, i.e. of polynomials with root fields contained in radical extensions.

Structure of radical extensions:
A radical extension of F has form F(a1,…an) where a1^r1 lies in F, a2^r2 lies in F(a1), …, an^rn lies in F(a1,…,an-1), where the powers rj are some non negative integers. Thus each new element is an rth root of some element from the previous field.

Note that if we want to adjoin say a 6th root of w, i.e. an element a such that a^6 = w, we could instead adjoin first a square root u of w, and then a cube root a of u. Hence we may assume that all powers rj occurring in a radical extension, are primes pj.

Remark: If F(a1,…,an) and F(b1,…,bm) are two radical extensions of F, then their compositum F(a1,…,an, b1,…,bm) is also radical.

Thus if q is the product of all the prime powers occurring in a radical extension F(a1,…,an), and if e is a primitive qth root of 1, then F(e,a1,…,an) is a radical extension.

Finally, if E is an algebraically closed extension of F, and if F(a1,…an) is any radical extension of F contained in E, then there are only a finite number of F-embeddings of F(a1,…,an) into E, and if K is their compositum, then K is a radical Galois extension of F. Moreover, if e is a primitive root of 1, every F- isomorphism of F(e) is an automorphism of F(e).

Thus every radical extension of F is contained in a radical Galois extension of form F(e,a1,…,an) where for each j, aj^pj lies in F(e,a1,…,aj-1) with pj prime, and F(e) contains a primitive pj-th root of 1 for every j. More precisely, we may assume that e is a primitive qth root of 1, where q is the product of all the primes pj, (or just of the distinct primes among them).

In particular, if g is a solvable polynomial over F, i.e. if every root of g is expressible algebraically in terms of radicals, then the root field K of g is contained in such a special radical Galois extension F(e,a1,…,an). Galois showed that the Galois group of F(e,a1,…,an) is a solvable group, and that the group of K/F is a quotient of that group by a normal subgroup, hence also solvable. We discuss this next.

The group of a radical Galois extension.
Every radical Galois extension is contained in one of the special form K = F(e,a1,…,an) described in the paragraph just above. We study the group of this extension in stages, the first being the Galois group of F(e)/F, the root field of the polynomial (X^q - 1). (This will turn out to be a quotient of the full Galois group G(K/F) by the normal subgroup G(K/F(e)).)

Since the roots of the polynomial (X^q - 1) form a multiplicative cyclic group ≈ Z/qZ, and generate F(e) as an extension of F, the group G(F(e)/F) restricts isomorphically to a subgroup of the automorphism group of Z/qZ. An automorphism of the cyclic group Z/qZ sends a generator to another generator, hence is determined by multiplying the generator by a unit in the ring Z/qZ, i.e. by an integer relatively prime to q. Hence G(F(e)/F) is isomorphic to the multiplicative group of units of the commutative ring Z/qZ; in particular, it is finite abelian.

The next case is:
Proposition: If E is a field of characteristic zero containing a splitting field for X^p - 1, with p prime, if a^p = c is in E, then E(a) is a splitting field over E for X^p - c, and there is an injective homomorphism G(E(a)/E) —> (Zp, +).
proof: In fact an automorphism of E(a) is defined by multiplying the element a by any root of X^p - 1, and these roots form a multiplicative group isomorphic to (Z/qZ, +). Thus the extension E(a) is Galois over E and the Galois group G(E(a)/E)) is finite cyclic, in particular abelian.

Now we can combine these results to describe the structure of the Galois group of a special radical Galois extension, using the famous:

Fundamental theorem of Galois theory (partial statement):
If E is a Galois extension of F, and K is an extension of E such that K is Galois over F, then K is also Galois over E, and G(K/E) is a normal subgroup in G(K/F), and G(E/F) is isomorphic to the quotient group G(K/F)/G(K/E).

Corollary: The Galois group of a special radical Galois extension K = F(e,a1,…,an), as described in the previous section, is a solvable group.
proof: K is Galois over F and each subfield F(e,a1,…,aj) is a splitting field, hence Galois, over the previous one F(e,a1…,aj-1), with abelian Galois group, so by the fundamental theorem, used n+1 times, G(K/F) is solvable.

Corollary: If a polynomial is solvable by radicals, the Galois group of its root field is a solvable group.
proof: The root field is contained in a special radical Galois extension, so by the fundamental theorem, its Galois group is a quotient of a solvable group, hence also solvable.

Remark; The converse is also true (in characteristic zero): i.e. if the root field of a polynomial has solvable Galois group, it is contained in a radical extension, hence the polynomial is solvable by radicals. The basic step is: if F contains all pth roots of 1, and if E is a finite extension of F with Galois group ≈ Z/pZ, then E = F(a) with a^p in F.

Advice for the perplexed: read the chapter on this in Mike Artin's book: Algebra. That is the clearest explanation I know of, and it is written by an expert.

 

[mathwonk]

Thus, a polynomial over F can be solved by radicals, i.e. its root field is contained in the union of an increasing sequence of fields F, F1,…,Fn, where each consecutive extension, Fj of Fj-1, is obtained by adding in the solutions of an equation of form X^r - c, if and only if the Galois group G of its root field is the union of an increasing sequence of subgroups {e}, H1, …,Hm =G, such that each subgroup Hj-1 is normal in the next one Hj, and their quotient Hj/Hj-1, is abelian. I.e. a polynomial is solvable by radicals, if and only if its group is solvable.

Roughly, since the group of the equation X^r - c is solvable, an equation whose root field is composed of the root fields of such equations, has group composed of solvable groups, hence is solvable.