# 3 Galois Theory. 3.1 Definitions and Examples

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1 3 Galois Theory 3.1 Definitions and Examples This section of notes roughly follows Section 14.1 in Dummit and Foote. Let F be a field and let f (x) 2 F[x]. In the previous chapter, we proved that there always exists a finite extension K of F that contains the roots of f (x). The big idea of Galois Theory (named after Évariste Galois, ) is to consider the relationship between the group of permutations of the roots of f (x) and the algebraic structure of its splitting field. The explicit connection is given by the Fundamental Theorem of Galois Theory, which we will prove in the next section. In this section, we introduce all of the necessary terminology. Definition 3.1. Let K be a field. The collection of all automorphisms of K is denoted Aut(K). An automorphism 2 Aut(K) is said to fix 2 K if ( )=. If S K, then is said to fix S if it fixes all the elements of S (i.e., ( )= for all 2 S). Note that Aut(K), ; since the identity map is an automorphism, called the trivial automorphism. Recall that the prime subfield of K is generated by 1. Moreover, every 2 Aut(K) satisfies (0) = 0 and (1) = 1. It follows that fixes the prime subfield of K. In particular, Aut(Q) and Aut(Z p ) only contain the trivial automorphism. Definition 3.2. Let K/F be an extension of fields and let Aut(K/F) be the collection of automorphisms of K that fix F. Note that if F is the prime subfield of K, then Aut(K/F) = Aut(K) since every automorphism of K fixes its prime subfield. Theorem 3.3. For every field K, the set Aut(K) is a group under function composition. If K/F is an extension of fields, then Aut(K/F) is a subgroup of Aut(K). Theorem 3.4. Let K/F be an extension of fields and let 2 K be algebraic over F. Then for any 2 Aut(K/F), ( ) is a root of the minimal polynomial for over F (i.e., Aut(K/F) permutes the roots of irreducible polynomials). Equivalently, any polynomial with coe cients in F having as a root has ( ) as a root. Example 3.5. Here are two examples. (1) Consider Q( p 2). If 2 Aut(Q( p 2)) = Aut(Q( p 2)/Q), then ( p 2) is either p p 2 or 2 since these are the only two roots of the minimal polynomial for p 2. It follows that for a,2 Q, (a+b p 2) is equal to either a+b p 2 or a b p 2 (since fixes Q). The map determined by p 2 7! p 2 is the identity automorphism. The map determined by p 2 7! p 2 is the only other map in Aut(Q( p 2)) = Aut(Q( p 2)/Q). This implies that Aut(Q( p 2)) = Aut(Q( p 2)/Q) is a cyclic group of order. 2 (2) Now, consider K = Q( 3p 2). Let 2 Aut(K/Q). Then is completely determined by its action on 3p 2: (a + b 3p 2+c( 3p 2) 2 )=a + b ( 3p 2) + c ( 3p 2) 2. 38

2 Recall that the other two roots of x 3 2 are not elements of K. However, ( 3p 2) 2 K and must be a root of x 3 2. It follows that ( 3p 2) = 3p 2. Therefore, must be the identity map, and so Aut(K/Q) is the trivial group. In general, if K is generated over F by some collection of elements, then any automorphism 2 Aut(K/F) is uniquely determined by what it does to the generators. If K/F is finite, then K is finitely generated over F by algebraic elements, so by the previous theorem, the number of automorphisms of K that fix F is finite. That is, Aut(K/F) is a finite group. In particular, the automorphisms of a finite extension can be considered as permutations of the roots of a finite number of equations, but not every permutation of the roots gives rise to an automorphism (as in the previous example). We can also associate to each group of automorphisms a field extension. Theorem 3.6. Let H apple Aut(K), where K is a field. Then the collection F of elements of K fixed by all the elements of H is a subfield of K. Remark 3.7. In the previous theorem, H need not be a subgroup. Definition 3.8. If H apple Aut(K), then the subfield fixed by H is called the fixed field of H. Theorem 3.9. The association of groups to fields and fields to groups defined above is inclusion reversing, namely (1) If F 1 F 2 K are two subfields of K, then Aut(K/F 2 ) apple Aut(K/F 1 ), and (2) If H 1 apple H 2 apple Aut(K) are two subgroups of automorphisms with associated fixed fields F 1 and F 2, respectively, then F 2 F 1. Example Let s return to the previous examples. (1) The fixed field of Aut(Q( p 2)/Q) is just Q. (2) Since Aut(Q( 3p 2)/Q) is the trivial group, every element of Q( 3p 2) is fixed by Aut(Q( 3p 2)/Q), and so the fixed field of Aut(Q( 3p 2)/Q) is Q( 3p 2). Here s the current summary of the big picture. Given a subfield F of K, the associated group is the collection of automorphisms of K that fix F. On the other hand, given a group of automorphisms of K, the associated extension is defined by taking F to be the fixed field of the automorphisms. In Example 3.5(1), starting with the subfield Q of Q( p 2) one obtains the group {1, } (where : p 2 7! p 2) and starting with the group {1, } one obtains the subfield Q. In this case, we get a duality between the two concepts. In Example 3.5(2), starting with the subfield Q of Q( 3p 2), we only get the trivial group. Starting with the trivial group, we can t drop down from Q( 3p 2) to obtain Q. The shortcoming in this example is that there are not enough automorphisms to force the fixed field to be Q. The next theorem provides us with a bound on the size of the automorphism groups associated to a splitting field of a single polynomial. Theorem Let E be the splitting field of the polynomial f (x) 2 F[x]. Then with equality if f (x) is separable over F. Aut(K/F) apple[e : F] More generally, it is true that Aut(K/F) apple[e : F] for any finite extension. We will prove this later. 39

3 Definition Let K/F be a finite extension. Then K is said to be Galois over F and K/F is called a Galois extension if Aut(K/F) =[K : F]. If K/F is Galois, Aut(K/F) is called the Galois group of K/F, denoted Gal(K/F). The next result follows immediately from Theorem 3.11 and the definition of Galois extension. In the next section, we will see that the converse is also true. Corollary If K is the splitting field over F of a separable polynomial f (x), then K/F is Galois. Note that the splitting field of any polynomial f (x) is the same as the splitting field of the product of irreducible factors of f (x) (i.e., remove any repeated factors of f (x)). The latter polynomial is separable by Corollary Corollary 3.13 implies that the splitting field of any polynomial over Q is Galois. Definition If f (x) is a separable polynomial over F, then the Galois group of f (x) over F is the Galois group of the splitting field of f (x) over F. Example Let s play with a few examples. (1) Previous calculations showed us that Aut(Q( p 2)/Q) =2=[Q( p 2) : Q]. Thus, Q( p 2)/Q is a Galois extension. In this case, the Galois group Gal(Q( p 2)/Q) is isomorphic to Z 2. Moreover, the Galois group of the separable polynomial f (x)=x 2 2 is Gal(Q( p 2)/Q). (2) Any quadratic extension K of any field F of characteristic di erent from 2 is Galois. (3) The extension Q( 3p 2)/Q is not Galois since its group of automorphisms does not have order equal to [Q( 3p 2) : Q] = 3. (4) Let s tinker with Q( p 2, p 3)/Q, where Q( p 2, p 3) is the splitting field of (x 2 2)(x 2 3). (5) Let s play the same game with the splitting field of x 3 2 over Q. (6) As in (3), the field Q( 4p 2) is not Galois over Q since any automorphism is determined by where it sends 4p 2 and of the four possibilities {± 4p 2,±i 4p 2}, only two are in the field Q( 4p 2) (the two real roots). Note that the degree 2 extensions Q( p 2)/Q and Q( 4p 2)/Q( p 2) are both Galois extensions (since both are quadratic extensions; see (2)). This shows that a Galois extension of a Galois extension need not be Galois. 40

4 3.2 The Fundamental Theorem of Galois Theory This section of notes roughly follows Section 14.2 in Dummit and Foote. In the examples we ve looked at so far, every time we ve had a Galois extension, there was a nice 1-1 correspondence between the subgroups of the Galois group and the lattice of fixed subfields of the extension. Moreover, the correspondence has been inclusion reversing. It turns out that we always have this correspondence whenever we have a Galois extension. This is exactly the content of the Fundamental Theorem of Galois Theory. In order to prove this amazing theorem, we need to develop some more tools. Definition A (linear) character of a group G with values in the field L is a homomorphism from G to the multiplicative group of L: : G! L. That is, (g 1 g 2 )= (g 1 ) (g 2 ) for all g 1,g 2 2 G and (g), 0 for all g 2 G. A collection of characters 1,..., n of G are said to be linearly independent over L if they are linearly independent as functions on G, i.e., if there is no nontrivial relation with a 1,...,a n 2 L not all 0 as a function on G. a a n n =0 Theorem If 1,..., n are distinct characters of G with values in L, then they are linearly independent over L. Consider an injective homomorphism of a field K into a field L, called an embedding of K into L. Then is an injective group homomorphism from G = K into L, and so is a character of G = K with values in L. Since we know that (0) = 0, viewed as a character contains all of the information about as an injective field homomorphism on K. Corollary If 1,..., n are distinct embeddings of a field K into a field L, then they are linearly independent as functions on K. In particular, distinct automorphisms of K are linearly independent as functions on K. Theorem Let K be a field, G = { 1 =1, 2,..., n}appleaut(k), and F be the field fixed by G. Then [K : F]=n = G. Recall that Theorem 3.11 tells us that if K is the splitting field over F of the polynomial f (x) 2 F[x], then Aut(K/F) is bounded by [K : F] (with equality if f (x) is separable). The next result tells us that we have the same bound for any finite extension K/F. Corollary Let K/F be any finite extension. Then Aut(K/F) apple[k : F] with equality i F is the fixed field of Aut(K/F). That is, K/F is Galois i F is the fixed field of Aut(K/F). Corollary Let G be a finite subgroup of automorphisms of a field K and let F be the corresponding fixed field. Then every automorphism of K fixing F is contained in G, i.e., Aut(K/F)=G, so that K/F is Galois with Galois group G. Corollary If G 1, G 2 are distinct finite subgroups of automorphisms of a field K, then their fixed fields are also distinct. By the corollaries above, we see that taking the fixed fields for distinct finite subgroups of Aut(K) gives distinct subfields of K over which K is Galois. Moreover, the degrees of the extensions are given by the orders of the subgroups. A portion of the Fundamental Theorem of Galois Theory states that these are all of the subfields of K. 41

5 The next theorem provides the converse for Theorem Theorem The extension K/F is Galois i K is the splitting field of a separable polynomial over F. Furthermore, if this is the case, then every irreducible polynomial in F[x] that has a root in K is separable and has all of its roots in K (so in particular, K/F is a separable extension). Definition Let K/F be a Galois extension. If 2 K, the elements ( ) for 2 Gal(K/F) are called the conjugates (or Galois conjugates of 2 K. If E is a subfield of K containing F, the field (E) is called the conjugate field of E over F. The proof of Theorem 3.23 shows us that in a Galois extension K/F, the other roots of the minimal polynomial over F of any element 2 K are precisely the conjugates of under the Galois group of K/F. The second statement in Theorem 3.23 tells us that K is not Galois over F if we can find an irreducible polynomial in F[x] having a root in K but not all of its roots in K. Loosely speaking, Galois extensions are extensions with just enough distinct roots of irreducible polynomials. For convenience, let s put all of our characterizations of Galois extensions K/F in one place. (1) Fields with [K : F]= Aut(K/F) (Definition 3.12) (2) Splitting fields of separable polynomials over F (Corollary 3.13/Theorem 3.23) (3) Fields where F is precisely the set of elements fixed by Aut(K/F) (Corollary 3.20) (4) Finite, normal, and separable extensions (Theorem 3.23) We are finally ready to tackle the Fundamental Theorem of Galois Theory. Theorem 3.25 (Fundamental Theorem of Galois Theory). Let K/F be a Galois extension. Then there is a bijection Given by the correspondences {subfields E of K containing F}\${subgroups H of Gal(K/F)} {the fixed field of H} 7! H E 7! {the elements of G fixing E} which are inverses of each other. Under this correspondence, we have the following: (1) If E 1,E 2 correspond to H 1,H 2, respectively, then E 1 E 2 i H 2 apple H 1 (2) [K : E]= H and [E : F] = [Gal(K/F):H]. (3) K/E is always Galois with Galois group Gal(K/E)=H. (4) E is Galois over F i H is a normal subgroup in Gal(K/F). If this is the case, then the Galois group is isomorphic to the quotient group: Gal(E/F) Gal(K/F)/H More generally, even if H is not necessarily normal in Gal(K/F), the isomorphisms of E (into a fixed algebraic closure of F containing K) that fix F are in one-to-one correspondence with the cosets { (H)} of H in Gal(K/F). (5) If E 1,E 2 correspond to H 1,H 2, respectively, then the intersection E 1 \ E 2 corresponds to the group hh 1,H 2 i and the composite subfield E 1 E 2 corresponds to H 1 \ H 2. Hence the lattice of subfields of K containing F and the lattice of subgroups of Gal(K/F) are dual. 42

6 Example We conclude this section by tinkering with three examples. (1) Q( p 2, p 3)/Q (2) Q( 3p 2, 3 )/Q (3) Splitting field of x 8 2 over Q 43

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