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NOTES on GALOIS THEORY Alfonso Gracia-Saz, MAT 347

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NOTES on GALOIS THEORY Alfonso Gracia-Saz, MAT 347 Galois theory MAT 347 NOTES ON GALOIS THEORY Alfonso Gracia-Saz, MAT 347 Go to the roots of these calculations! Group the operations. Classify them according to their com- plexities rather than their appearances! This, I believe, is the mission of future mathematicians. This is the road on which I am embarking in this work. { From the preface to Galois' final manuscript, written the night before he died. About these notes In the fourth and final part of the course we are going to study field extensions and Galois theory. This corresponds to Chapters 13 and 14 in the book. However, the book goes into more depth than we need (and I am also following a different order) so I am providing these notes to help you as a skeleton. These notes complement, but do not replace, lectures. I am not including proofs and motivation, which we will take care of in class. It is a good exercise to start with these notes and to try and reconstruct all the proofs. You are welcome to follow the book instead (or as well) if you prefer so. In particular, the book contains many extra examples that can be helpful. If you find any typos or mistakes, please let me know. Parts of these notes are based on a similar handout written by Mira Bernstein (Wellesley College, 2005), with her permission. Last updated on February 10, 2020 (by Florian Herzig, for the 2018/19 course). 1 Galois theory MAT 347 1 Prime subfields and characteristic of a field This appears in section 13.1 of the book. Definition 1.1. Let K be a field. The prime subfield of K is the smallest subfield of K. Equivalently, it is the intersection of all the subfields of F . Lemma 1.2. Let K be a field. Let 1K be its multiplicative identity. For every n 2 Z, we can define n · 1K 2 K because (K; +) is an abelian group. Consider the map : n 2 Z ! n · 1K 2 K. Then is a ring homomorphism. Moreover, ker ¢ Z is a prime ideal. Definition 1.3. Under the conditions of Lemma 1.2, let ker = (n) = Zn where n 2 Z is either a positive prime number or zero. Then we say that the characteristic of K is n. We write char K = n. Proposition 1.4. Let K be a field. If char K = p is a prime, then the prime subfield of K is Fp. If char K = 0, then the prime subfield of K is Q. Hint: Apply the first isomorphism theorem to . If char K = 0, remember that Q is the field of fractions of Z. Lemma 1.5. If char K = p is prime, then p · a = 0 for every a 2 K. Examples 1.6. What is the characteristic of C? Find an example of an infinite field with characteristic prime. 2 Galois theory MAT 347 2 Field extensions This appears in sections 13.1{13.2 of the book. 2.1 Definitions Definition 2.1. A field extension is a field homomorphism i : F ! K, where F and K are fields. Notice that every such homomorphism is necessarily injective. (The kernel is an ideal, and we require i(1F ) = 1K , so i cannot be the zero map.) Examples 2.2. The inclusion maps i1 : Q ! C and i2 : F ! F (X) (where F is any field and X is a formal variable) are field extensions. Remark 2.3. If i : F ! K is a field extension, we often identify F with its image i(F ). Thus we think of F as a subfield of K and i as the inclusion (identity) map. In this case we say that K is an extension of F , and use the notation K=F for the extension. Whenever there is no ambiguity, we will use these two points of view (field homomorphism vs subfield) interchangeably. For the rest of this section, let K=F be a field extension. Definition 2.4. Suppose a1; : : : ; an 2 K. The smallest subring of K containing both F and all the ai's is denoted F [a1; : : : ; an]. It is called the subring of K generated by a1; : : : ; an over F . The smallest subfield of K containing both F and all the ai's is denoted F (a1; : : : ; an). It is called the subfield of K generated by a1; : : : ; an over F . Proposition 2.5. F (a1; : : : ; an) is the field of fractions of F [a1; : : : ; an]. Examples 2.6. (i) In C, R[i] = R(i) = C. p p p (ii) In R, Q[ 5] = Q( 5) = fa + b 5 : a; b 2 Qg. p p p 3 3 3 (iii) In R, Q[ 5] 6= fa + b 5 : a; b 2 Qg. Any idea what it might be instead? How about Q( 5)? p p (iv) In R, Q(i + 5) = Q(i; 5). Remark 2.7. The ring of polynomials in n variables x1; : : : ; xn with coefficients in F is denoted F [x1; : : : ; xn]. Its field of fractions (rational expressions in x1; : : : ; xn) is denoted F (x1; : : : ; xn). The similarity of this notation to Definition 2.4 is intentional and consistent, since F [a1; : : : ; an] = ff(a1; : : : ; an): f 2 F [x1; : : : ; xn]g; and F (a1; a2;:::) is the field of fractions of F [a1; : : : ; an]. In other words, F [a1; : : : ; an] consists precisely of all polynomial combinations of the ai's, while F (a1; : : : ; an) consists of quotients of such polynomial combinations. However, as we saw in Example 2.6, there are often simpler ways of characterizing the elements of F (a1; : : : ; an) and F [a1; : : : ; an], since different polynomials and rational expressions in a1; : : : an can simplify and evaluate to the same element in K. 3 Galois theory MAT 347 Definition 2.8. If there exists α 2 K such that K = F (α), then K=F is called a simple extension. Definition 2.9. An element α 2 K is said to be algebraic over F if α is the root of some nonzero polynomial in F [X]; otherwise, α is said to be transcendental over F . Real or complex numbers that are algebraic (resp. transcendental) over Q are called simply algebraic (resp. transcendental) numbers. The field extension K=F is called algebraic when α is algebraic over F for every α 2 K. Examples 2.10. p n (i) For all k; n 2 N, k is algebraic, since it is the root of Xn − k. (ii) The field extension C=R is algebraic. Given α = a + bi 2 C, we see that α is a root of (X − a)2 + b2. Remark 2.11. The set of algebraic numbers (over Q) is countable. (Why?) Since we know that R is uncountable, it follows that \most" real numbers are transcendental. However, explicit examples of numbers transcendental over Q are very hard to find, since it is difficult to prove that a given number is not the root of any polynomial. It was shown in the 19th century that both e and π are transcendental over Q. 2.2 Simple extensions We continue assuming that K=F is a field extension. Proposition 2.12. Let α 2 K. Consider the evaluation map Eα : f(X) 2 F [X] ! f(α) 2 K. Then Eα is a ring homomorphism with image F [α]. Moreover, ker Eα ¢ F [X] is a prime ideal. Theorem 2.13. Continue with the notation of Proposition 2.12. The map Eα is injective iff α 2 K is transcendental over F . In that case, F [α] =∼ F [X] and F (α) =∼ F (X). In particular, F [α] 6= F (α). Definition 2.14. Let α 2 K be algebraic over F . Let I = ker Eα as in Proposition 2.12. Since F [X] is a PID, we know that I is principal. Since I 6= 0, we know that I = (f(X)) for a unique monic polynomial f(X) 2 F (X). Moreover, since I is prime, we know that f(X) is irreducible. We say that f(X) is the minimal polynomial of α over F and we denote it by mα,F (X) := f(X). Lemma 2.15. Let α 2 K be algebraic. Let g(X) 2 F [X]. Then α is a root of g(X) if and only if mα,F (X) divides g(X). Lemma 2.16. Let α 2 K be algebraic. Let g(X) 2 F [X]. Assume that α is a root of g(X). TFAE: • g(X) is the minimal polynomial of α over F . • g(X) is monic and irreducible. • g(X) is monic and it has the smallest degree among non-zero polynomials that have α as a root. p Examples 2.17. Both i and 2 are algebraic over and over . Over they both have degree 2, with Q R Q p minimal polynomials x2 + 1 and x2 − 2 respectively. Over , i still has degree 2, but 2 has degree 1, with p R minimal polynomial x − 2. Proposition 2.18. An element α 2 K has degree 1 over F if and only if α 2 F . 4 Galois theory MAT 347 Exercise 2.19. What complex numbers have degree ≤ 2 over Q? over R? Theorem 2.20. Suppose α 2 K is algebraic, with minimal polynomial f(X) over F . Then F (α) = F [α] =∼ F [X]=(f(X)): Example 2.21. R[X]=(X2 + 1) =∼ R[i] = R(i) = C. p p p 3 3 3 Example 2.22.
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