Other What are some good books for learning Galois Theory?

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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.
 
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Galois theory sits at the intersection of group theory, field theory, and linear algebra. Learning more about groups, rings, and fields will prepare you for Galois theory. Linear Algebra concepts are good to have under your belt as well.

Some folks recommend:

- Introduction to Galois Theory by Hernandez & Laszlo -- a modern and concise undergrad text

or

- Galois Theory by David Cox — many students and self-learners like how Cox presents multiple perspectives and emphasizes clarity.

In any event, you need to remember that self-study means you'll be using many books to develop your understanding of Galois theory.
 
jedishrfu said:
Galois theory sits at the intersection of group theory, field theory, and linear algebra. Learning more about groups, rings, and fields will prepare you for Galois theory. Linear Algebra concepts are good to have under your belt as well.

Some folks recommend:

- Introduction to Galois Theory by Hernandez & Laszlo -- a modern and concise undergrad text

or

- Galois Theory by David Cox — many students and self-learners like how Cox presents multiple perspectives and emphasizes clarity.

In any event, you need to remember that self-study means you'll be using many books to develop your understanding of Galois theory.
I've taken Linear Algebra before but haven't taken classes/courses for group theory nor field theory. I think I have to buy/purchase these books then. Thank you for the suggestion/recommendation.
 
You need to know some group theory and field theory first. Why do you want to study galois theory? I think that most books on the subject are fine, as long as the style of presentation suits you. You can take a look at jmilne.org in the course notes section. He has galois theory notes as well as group/field theory.
 
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I do not think group or field theory is essential in the sense that most books on Galois theory will introduce the important concepts. I think I used Postnikov back in the time.
 
Van der Waerden Modern Algebra
 
Thank you very much for the suggestion/recommendation, guys!
 
I liked the one by Ian Stewart. It was pretty much self-contained.
 
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I may be wrong. Its been a while since I learned the basics.

I believe a member on this forum, Mathwonk, wrote some notes explaining Galois Theory first and filling in the blanks.

I can be wrong.

In anycase, check out his algebra notes.

If you want something easier but well written, to get into group/rings. Have a look at Galian: Contompary Algebra. Artin Algebra can also be a good book, but it requires more mathematical maturity.

In any case,

Not knowing what a group/field is may make understanding something simple like a Galois Group imposible imo.
 
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I have learned it from van der Waerden's Algebra book
https://www.amazon.de/Algebra-German-B-van-Waerden/dp/3642855288/
and I also read Artin's book specifically about Galois Theory
https://www.amazon.de/Galois-Theory-Delivered-University-Mathematical/dp/0486623424/

Both are rather old-fashioned and written in a pre-Bourbaki style. Hence, it is more of a question about the style of representation than it is a question about sources. Here is a more modern treatment by Milne
https://www.jmilne.org/math/CourseNotes/FT.pdf

I found Milne while searching for "galois theory pdf".
 
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Those notes 844-1, 844-2, of mine linked in post #12, have as prerecquisite the earlier notes 843-1, 843-2, on the same webpage:
https://www.math.uga.edu/directory/people/roy-smith

The notes 843-1 are on groups, the notes 843-2 use groups and specifically Galois groups to give a necessary condition for solvability of equations, and to identify some polynomials which do not have solution formulas in terms of radicals. Then the notes 844-1 discuss rings and fields, and the notes 844-2 use all of this to give a sufficient condition for solvability, and to actually give solution formulas.

I have recently expanded the notes 843-1, hopefully making them more readable, but have not posted them anywhere.

I think Milne's writing is excellent.

Tomorrow I will try to post a summary of Galois theory.
 
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Harder than I thought. A beginning:
Sketch of Galois theory
Galois theory translates a problem about solutions of polynomials over a field, into a problem in group theory, and the analysis uses the theory of dimension of abstract vector spaces, quotients of polynomial rings, and quotients of groups. Hence one must know something of all those subjects to even understand the statements.

Groups and fields
A (concrete) group G s a collection of permutations of the elements of some set S, together with their inverses, and which is closed under composition; an abstract group is a set G with an associative operation with an identity s.t. all elements have inverses. [An abstract group can be viewed concretely as a group of permutations of its own elements via the given operation.]. G is called abelian if the operation is commutative.

E.g.the isometries of a square form a non abelian group of order 8, [a dihedral group]; the full collection of permutations of a set of n elements is a non abelian group of order n! called Sn = the symmetric group on n elements; the rotation group of a cube is the group S4 of order 24; the integers Z are an (infinite) abelian group under addition, the rational numbers Q are a group under addition, and the non zero rationals are an abelian group under multiplication. The real numbers and the complex numbers are also abelian groups under addition whose non zero elements are abelian groups under multiplication. The complex nth roots of 1, i.e. the n complex solutions of X^n-1, form an abelian multiplicative group of order n, consisting of the powers of any “primitive” root, and isomorphic to the additive group Z/nZ. Any number system F, with two operations, +, . , which is an abelian group under addition, and its non zero elements are an abelian group under multiplication, (s.t. multiplication distributes over addition), is called a field. We assume all our coefficient fields contain Q.

Splitting fields of polynomials
If F is any field, and f is a polynomial with coefficients in F, a root of f is an element ß of a field K containing F, and such that in K, f(ß) = 0. If f has degree n, and K contains F and n roots of f, and no smaller subfield of K contains F and all n roots, then K is called a root field, or solution field, or splitting field, of f over F. E.g. the complex number field C is the splitting field over the real field R, of the polynomial X^2+1. Hence the complex field can be viewed as the quotient R[X]/(X^2+1) = {real polynomials in X, but equated if they become equal when X^2 is set equal to -1} = { expressions of form a+bi with a,b real and i^2 = -1}.

Similarly, the splitting field over Q of X^2-2 is the set K = {a+bs, with a,b, rational and s^2 = 2, i.e. with s = sqrt(2)}. Then K ≈ Q[X]/(X^2-2) = {rational polynomials in X, but equated if they become equal when X^2 is set = 2}. These “quadratic” splitting fields K have vector dimension over their base field F, equal to 2, the degree of the irreducible polynomial over F from which K is obtained by adjoining one root of f. In general the field F[X]/(f) obtained by adjoining one root of the irreducible polynomial f, has vector dimension over F equal to the degree n of f, and the full root field K obtained by adjoining all n roots of f, has vector dimension d over F, where n divides d and d divides n!.

I.e. to obtain the full root field of f over F, one first adjoins one root, getting a field isomorphic to the quotient F[X]/(f) = F1 = F(ß), where ß = X(modf). Then in this extension field, the polynomial f has at least one root, so in F1, the polynomial f has at least one linear factor. Unfortunately, since in this quotient field, the letter X has been set equal to that root, we need a new letter for the variable of f thought of again as a polynomial. I.e. in F1, f(X(modf)) = (f(X))(modf) = 0. So let we could let f(Y) now be the polynomial with coefficients in F1, the same coefficients as f had, but now in F1. But that seems silly, so instead of thinking of our field F1 as the actual quotient F[X]/(f(X)), just pick a new letter ß1 for the root, and write the field F1 as F(ß1). Then think of f(X) as a polynomial in the variable X, with coefficients in this F1 = F(ß1). E.g. if F = Q, just pick ß1 as some complex root of f.

Anyway we have f(X) = (X-ß1)g(X) with g a polynomial with coefficients in F1. Now the question is whether g is irreducible over F1, or whether it factors in some way. If we are really lucky, maybe g already factors completely into linear factors over F1, and then F1 is already the full splitting field of f over F. Then the splitting field has vector dimension over F equal to n = degree(f). At the opposite extreme, if g is irreducible, and after adjoining a root ß2 of g, we get a field F2 where g factors as g = (X-ß2)h with h still irreducible, and suppose this keeps up, i.e. suppose every time we adjoin another root, we only get one linear factor and the other factor is always still irreducible. Then it takes n-1 steps to get a full splitting field of f, and the vector dimension of that splitting field over f is n!, where n= degree(f).

E.g. one extreme case happens for the polynomial f(X) = X^4+X^3+X^2+X+1, over Q, where (X-1)f(X) = X^5-1. If we adjoin one root ß of f, we get the field F(ß) = F1 of dimension 4 over Q, and we claim this field already contains all 4 roots of f. That is easy, since ß is a primitive 5th root of 1, (because 5 is prime), so its powers ß, ß^2, ß^3, ß^4, ß^5 = 1, are all 5 roots of X^5-1, and in particular the first 4 of those give all 4 roots of f. Thus the splitting field of f over Q has minimum vector dimension over Q, i.e. equal to n = degree(f).

To get an example where the splitting field has vector dimension n! over Q, where degree(f) = n, we can take f(X) = X^3-2. Then adjoining one root gives us a field isomorphic to the subfield of R generated by Q and the one real cube root ß of 2. I.e. all fields obtained by adjoining one root of f, are isomorphic to the quotient field Q[X}/(X^3-2), so we can take any one of them as a model. Thus we can regard F1 = F(ß) as a subfield of the reals, which makes it obvious that F(ß) does not contain any more roots of f, since those are not real. Hence to get the full splitting field of f over Q, we need to adjoin another (hence both) root of a quadratic polynomial over F1. This full splitting field F2 hence has dimension 2 over F1, so the dimension of F2 over Q is 6 = 3!, where degree(f) = 3.

For an intermediate example, take f(X) = X^4-2, over Q. Then adjoining a real 4th root ß of 2, gives also -ß as a second root, and thus f factors over Q(ß) into (X-ß)(X+ß).g(X) = (X^2-ß^2)(X^2+ß^2), where g = (X^2+ß^2) = (X^2 + sqrt(2)), is an irreducible quadratic with 2 imaginary roots. Thus the full splitting field of f over Q has vector dimension 8; i.e degree(f) = 4, so we have 4 < 8 < 4!.

The basic problem of solvability of equations by radicals
If f(X) is a polynomial with coefficients in a field F, we ask whether one can write a root the roots of f as an algebraic combination of elements of F and “radicals”. In the language of fields as given above, this asks whether a splitting field K for F, is contained in a field obtained from F by successively adjoining roots of equations of form X^r - c, where c is an element of a previously obtained field. I.e. a field extension is called radical if the larger field is obtained from the smaller field as the splitting field of a polynomial of form X^r - c, where c is an element of the smaller field, i.e. by adjoining rth roots of an element c. Then the question is whether K contained in a union of radical extensions, starting from F. Galois theory implies that if we can choose f irreducible over F, and so that the vector dimension of K over Q is n!, where n = degree f, then the answer is no for n ≥ 5. His proof depends on analyzing the “Galois group” of symmetries of K over F. Briefly, a union of radical extensions has a symmetry group that decomposes as a union of subgroups with abelian quotients, but this is not always the case for a general splitting extension.

More specifically, if n = degree(f), the Galois group G is a subgroup of Sn, with order(G) equal to the vector dimension over F of the splitting field K of f. Hence if that dimension equals n!, then G ≈ Sn, which does not have such an abelian quotient decomposition for n ≥ 5.
 
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I am noticing that to explain a topic simply, it helps to understand it, but I will persist. Be advised however to read an expert, I suggest e.g. Artin, Van der Waerden, Jacobson, Milne, Brauer, Lang, as well as Dummit and Foote, Stewart.

And check this article out, for a discussion of the history of Galois theory in textbooks, with many specific recommendations:
https://ncm.gu.se/pdf/normat/Juliusz_textbooks.pdf

A few words about symmetry
Solving a polynomial equation f with coefficients in F, means writing the roots of f in terms of the coefficients of f, and elements of F. This is inverse to the problem of writing the coefficients in terms of the roots, something easy to do, and also enlightening. Namely if r1,…,rn are the roots of a (monic) polynomial f, then f(X) factors completely into linear factors, as f(X) = (X-r1)(X-r2)….)(X-rn). Multiplying out thus gives the coefficients of f in terms of the roots:
f(X) = X^n -(r1+r2+…+rn)X^(n-1) + (r1r2+r1r3+…+rn-1rn)X^(n-2) -…±(r1r2r3…rn).

E.g. for a quadratic f with roots r,s, we have f(X) = (X-r)(X-s) = X^2 - (r+s)X + rs.

These coefficients are symmetric combination of the roots, in the sense that they do not change value under any reordering of the roots. In fact these particular symmetric functions of the roots, can be used to express any symmetric polynomial function of the roots. For this reason they are called the “elementary symmetric functions”. I.e. every symmetric polynomial function of the roots, that is one which does not change values under reordering the roots, can be written as a polynomial in the symmetric functions. This can be used to help solve polynomial equations.

E.g. if r,s are the roots of a quadratic polynomial f(X) = X^2 -bX +c, where thus b = r+s and c=rs, then the expression D = (r-s)^2 does not change value after reordering the roots, hence D must be writable as a polynomial in b and c. It is well known that D, the “discriminant” of f, equals D = b^2-4c = (r+s)^2 - 4rs = (r-s)^2. Since we know that r+s = b, we can solve for r and s by taking the two square roots of D, to get r-s, and s-r, and then add to b to get 2r and 2s.
I.e. b ±sqrt(D) = (r+s) ± (r-s) = 2r, or 2s. hence r,s = (1/2)(b ± sqrt(b^2-4c)), the “quadratic formula.

The discriminant exists in general, and always tells us a little about how far away the solution field of a polynomial is from the coefficient field. I.e. since the roots are themselves in the root field, any polynomial in the roots also lies there, and any fully symmetric polynomial in the roots lies in the coefficient field. But if some only partially symmetric polynomial in the roots also lies in the coefficient field, then the root field is not as far away from the coefficient field as it might be.

E.g. if r,s,t are the roots of a cubic f(X), then ∆ = (r-s)(r-t)(s-t) is a square root of the discriminant, hence changes sign under any transposition of two roots, but remains unchanged under any even permutation, i.e. any composition of an even number of transpositions, (such even permutations form a normal subgroup An of Sn called the "alternating group"). If ∆ lies in the coefficient field, it implies that the solution field has only dimension 3 over the coefficient field, not dimension 6, but if not, the dimension is 6. For the cubic X^3 - 2, the discriminant D = ∆^2 = -27(4), is negative, so its square root ∆ does not lie in the coefficient field Q, so this gives another argument for why the solution field in this case has dimension 6 = 3! over Q.

Galois groups of splitting fields
Galois in his genius, after studying the works of LaGrange, realized that understanding the root field of a polynomial, and its relation to the coefficient field, rested on understanding which permutations of the roots can be realized by automorphisms of the root field that leave fixed all elements of the coefficient field. (To simplify matters, note that if the coefficient field is Q the rational numbers, then all automorphisms of the root field leave Q fixed.)

E.g in the case of a quadratic polynomial, the roots are already rational if and only if the only permutation of the roots that fixes the coefficient field is the identity, if and only if the discriminant has a rational square root. In general, for an irreducible polynomial over Q of degree n, the discriminant has a rational square root if and only if the root field has dimension at most (1/2)n! over Q, whereas in general it could be n!.

An automorphism of K over F, is a bijection *:K—>K that equals the identity on the subfield F, and preserves both addition and multiplication in K.

Definition: If F is a field and f(X) is a polynomial with coefficients in F, and if K is the field extension of F obtained by adjoining a full complement of roots of f, i.e. if K is a splitting field of f over F, then the set of field automorphisms of K which restrict to the identity on F, is a group called the Galois group of K over F, [and also the Galois group of f]. [K is uniquely determined by f and F, up to isomorphism.]

E.g. complex conjugation, sending a+bi to (a+bi)* = a-bi is an automorphism of C over R, hence an element of the Galois group of X^2+1 over R.

Basic theorem: The Galois group of f is naturally isomorphic to a permutation group of the roots of f, i.e. if degree f = n, to a subgroup of Sn.
proof: One only needs to show that an automorphisms of K over F carries a root of f to a (not necessarily different) root of f. Recall an automorphism of K over F, is a bijection *:K—>K that equals the identity on the subfield F, and preserves both addition and multiplication in K. Hence if ß in K is a root of f (with coefficients in F), then ß* is also a root of f, since * fixes the coefficients of f, including 0, so f=f*(= the polynomial obtained by applying * to the coefficients of f), hence 0 = f(ß) =(f(ß))* = f*(ß*) = f(ß*).
This shows every automorphism of K over F restricts to a permutation of the roots of f, and since K is generated over F by these roots, * is determined by what it does to the roots, so the restriction map is an injection from the Galois group to Sn. QED.

Remark: The Galois group of f, over F, is actually a subgroup of An if and only if the discriminant of f has a square root in F.

Galois’ insight was that this group has a special form in the case of root fields K obtained from polynomials that are solvable by radicals, i.e. that are contained in a union of field extensions of F, formed as root fields of polynomials of the special form X^r-c. so we look at those next, and their Galois groups.

Simplest case: K = root field of f(X) = X^n - 1. Then we claim the Galois group is abelian, in fact it is isomorphic to (Zn)* = the multiplicative group of units in the ring Zn = Z/nZ.
proof: The roots of X^n-1 themselves form a finite multiplicative subgroup of the complex numbers, the powers of e^(2πi/n) on the unit circle in the complex plane. Hence any field automorphism of the root field of f restricts to a multiplicative group automorphism of this group of roots, which is also injective since the elements of this group generate the root field. Each such automorphism sends a primitive root, such as e^(2πi/n), a generator of the group, to another primitive root, hence the map is an isomorphism from the Galois group to a subgroup of the group of units of (Z/nZ). The fact the map is onto that group follows from the (non trivial) algebraic result that the polynomial satisfied by the primitive roots of 1 is irreducible over Q, and the fact that the degree of this polynomial equals the order of the Galois group. In particular the Galois group in this case is abelian.

Next simplest case: K = root field of X^n - c.
Since the quotient of two nth roots of c is an nth root of 1, this root field contains the previous root field, so this one may be obtained in two stages, first adjoining all nth roots of 1, and then adjoining one root of X^n-c. Then an automorphism of this field, fixing the root field of X^n-1, sends each root a of X^n-c to some multiple ua, where u is a root of X^n-1, and u is independent of a. The map sending such an automorphism to u is an embedding from the Galois group of K over F = root field of X^-1, into the group of roots of X^-1, i.e. into the cyclic group Z/nZ. In particular, this group is abelian.

Remark: A key concept in this topic is that of a "normal" subgroup, which is a subgroup H of G such that one can form a quotient group G/H in a way such that the natural surjection G-->G/H is a homomorphism taking everything in H to the identity of G/H.

Corollary: If K is a field extension of Q, obtained in stages as the root fields of equations of form X^n-c, then the Galois group of K/Q is the union of a sequence of subgroups, each normal in the next larger one, with all successive quotient groups abelian.
proof: If F is contained in L contained in K, all root fields, then G(K/L) is a normal subgroup in G(K/F), and G(L/F) ≈ G(K/F)/G(K,L). QED.

Example: If F = Q, the rationals, and L is the root field of X^3-1, while K is the root field of X^3-2, then the Galois group G(K/Q) ≈ S3 has normal subgroup G(K/L) ≈ A3 ≈ Z/3Z. Both the subgroup G(K/L) and the quotient G(K/Q)/G(K/L) ≈ G(L/Q) ≈ Z/2Z, are abelian, (even though G(K/Q) ≈ S3 itself is not). Hence the Galois group of X^3-2 satisfies the necessary criterion for solvability over Q.

Here the sequence of subgroups in increasing order, is G(K/K) = {id}, G(K/L)≈Z/3Z, G(K/Q)≈S3, and the corresponding sequence of successive abelian quotients is G(K/L)/G(K/K) ≈ Z/3Z, and G(K/Q)/G(K/L) ≈ G(L/Q) ≈ Z/2Z.

In general, if the increasing succession of subfields of K, each formed by adding one more rjth root, is ....Fj, Fj+1,..., then the increasing sequence of subgroups of G(K/Q) is ....G(K/Fj+1), G(K/Fj),...., and the sequence of abelian quotient groups is......,G(K/Fj)/G(K/Fj+1) ≈ G(Fj+1/Fj),.....

Corollary: If f is a polynomial whose Galois group does not admit such a sequence of subgroups as described in the previous corollary, then f is not solvable by radicals.

Remark: If the splitting field H of f is merely contained in a field K of the form just described, rather than equaling it, then the Galois group G(H/Q) is a quotient of G(K/Q), i.e. then G(H/Q) ≈ G(K/Q)/G(K/H), and hence can be shown to have the same "abelian subquotient" property.

Corollary: The polynomial f(X) = X^5-80X+2 is not solvable by radicals.
proof: By calculus f has 3 real roots, hence 2 non real ones, and complex conjugation acts as a transposition on these 2 roots. Since f is irreducible, the Galois group G acts “transitively” on the 5 roots (every root can be sent to every other), hence G contains a 5-cycle and a transposition, hence equals all of S5. But S5 has A5 as index 2 subgroup; and A5 ≈ the Icosahedral group, which has no normal subgroups, i.e. A5 is “simple”. QED.
 
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