General Letters in Mathematics Vol. 3, No. 3, Dec 2017, pp.132-153 e-ISSN 2519-9277, p-ISSN 2519-9269 Available online at http:// www.refaad.com Field Extension by Galois Theory Md Taufiq Nasseef University of Kent, UK [email protected] Abstract. Galois Theory, a wonderful part of mathematics with historical roots date back to the solution of cubic and quantic equations in the sixteenth century. However, beside understanding the roots of polynomials, Galois Theory also gave birth to many of the central concepts of modern algebra, including groups and fields. In particular, this theory is further great due to primarily for two factors: first, its surprising link between the group theory and the roots of polynomials and second,the elegance of its presentation. This theory is often descried as one of the most beautiful parts of mathematics. Here I have specially worked on field extensions. To understand the basic concept behind fundamental theory, some necessary Theorems, Lammas and Corollaries are added with suitable examples containing Lattice Diagrams and Tables. In principle, I have presented and solved a number of complex algebraic problems with the help of Galois theory which are designed in the context of various rational and complex numbers. Keywords: Field Extension, Splitting fields, Separability, Galois theory. MSC2010 12E30. 1 Introduction and Preliminaries Evariste Galois (French pronunciation: [evarist galwa]) (October 25, 1811-?May 31, 1832) was a French mathemati- cian born in Bourg-la-Reine. While still in his teens, he was able to determine a necessary and sufficient condition for a polynomial to be solvable by radicals, thereby solving a long-standing problem. His work laid the foundations for Galois theory and group theory, two major branches of abstract algebra, and the subfield of Galois connections. He was the first to use the word "group" (French: groupe) as a technical term in mathematics to represent a group of permutations [1]. Galois' most significant contribution to mathematics by far is his development of Galois theory. He realized that the algebraic solution to a polynomial equation is related to the structure of a group of permutations associated with the roots of the polynomial, the Galois group of the polynomial. He found that an equation could be solved in radicals if one can find a series of subgroups of its Galois group, each one normal in its successor with abelian quotient, or its Galois group is solvable [2]. This proved to be a fertile approach, which later mathematicians adapted to many other fields of mathematics besides the theory of equations to which Galois originally applied it. Field extension is the focal ambition to work. So it would be a very good idea to start with the definition of field extensions. 1.1 Field extensions A field L is an extension of another field K if K is a subfield of L. Definition 1.1. A field extension is a monomorphism i : K ! L where K and L are fields. K is the small field, L the large field. Field Extension by Galois Theory 133 1.2 Simple extensions A simple extension is an extension L:K having the property that L = K(α) for some α 2 L: Polynomials are known to all.It is important to know about the specific group of polynomials and properties which are needed in my field extension. A polynomial n f(t) = a0 + a1 + ::: + ant over a field K is monic if an = 1: Definition 1.2 Let K(α): K be a simple extention.If there exists a non-zero polynomial p over K such that p(α) = 0 then α is an algebraic element over K and the extension is simple algebraic extension. Definition 1.3 Let L : K be the field extension, and suppose that α 2 L is algebraic over K.Then the minimum polynomial of over K is the unique monic polynomial m over K of the smallest degree such that m(α) = 0: An example could be the best thing to clear the definitions. For example, i 2 C is algebraic over R. If we let m (t) = t2 + 1 then m(i) = 0.Clearly m is monic. The only monic polynomials over R of smaller degree are those of the form t + r(r 2 R) or 1.But i cannot be zero of any of these. Hence the minimum polynomial of i over R is t2 + 1: Two theorems are going to present with no proofs that are closely related to this project. Theorem.1.1 If K is any field and m is any irreducible monic polynomial over K, then there exists an extension K(α): K such that α has minimum polynomial m over K. Theorem.1.2 Suppose K and L are fields and i : K ! L is an isomorphism.Let K(α), L(β) be simple algebraic extensions of K and L respectively, such that α has minimum polynomial mα(t) over K and β has minimum polynomial mβ(t) over L.Suppose further thatmβ = i(mα(t)).Then there exists an isomorphism j : K(α) ! L(β) such that jjK = iandj(α) = β: Definition 1.4. The degree [L : K] of a field extension L : K is the dimension of L considered as a vector space over K. Example 1.1. The complex numbers C are 2-dimensional over the real numbers R; 1; i being a basis.Hence [C : R] = 2: Definition 1.5. If K,L,M are fields and K ⊆ L ⊆ M, then L is the Intermediate Field. Theorem.1.3 If K,L,M are fields and K ⊆ L ⊆ M, then [M : K] = [M : L][L : K] p p p p Example 1.2Suppose we wish top find [Q( 2; 3) : Q].p Itp is easy to seep that f1; 2g isp ap basis for Q( 2)p overp Q. It isp a littlep harder to see that f1; 3g is a basis for Q( 2; 3) over Q( 2).Hence [Q( 2; 3) : Q] = [Q( 2; 3) : Q( 2)][Q 2 : Q] = 2:2 = 4: [2] 1.3 Splitting fields Definition 1.6.If K is a field and f is a polynomial over K then f splits over K if it can be expressed as a product of linear factors f(t) = k(t − α1)::::(t − αn) where k, α1...αn 2 K: Example 1.3 Let f(t) = (t2 − 3)(t3 + 1) over Q. We can construct a splitting field for f as follows: inside C f splits into linear factors p p p p −1 + i 3 −1 − i 3 f(t) = (t + 3)(t − 3)(t + 1)(t − )(t − ) 2 2 so there exists a splitting field inside C, namely p p −1 + i 3 Q( 3; ) 2 p This is clearly the same as Q( 3; i): 134 Md Taufiq Nasseef 1.4 Discriminant The discriminant of X2 + aX + b = (X − r)(X − r0) is (r − r0)2 = r2 − 2rr0 + r02 = (r + r0)2 − 4rr0 = a2 − 4b; which is the usual discriminant of a monic quadratic polynomial.In low-degree cases, explicit formulas for discriminants of some trinomials are disc(X2 + aX + b) = a2 − 4b; disc(X3 + aX + b) = −4a3 − 27b2; disc(X4 + aX + b) = −27a4 + 256b3; disc(X5 + aX + b) = 256a5 + 3125b4 Example 1.4 The discriminant of X3 −X −1 is -23, the discriminant of X3 −3X −1 is 81, and the discriminant of X3 − 4X − 1 is 229. [2] 1.5 Normality Definition 1.7. An extension L:K is normal if every irreducible polynomial f over K which has at least one zero in L splits in L. 1.6 Separability Definition 1.8. An irreducible polynomial f over a field K is separable over K if it has no multiple zero in a splitting field. This means that in any splitting field f takes the form k(t − σ1):::(t − σn) where the σi are all different. Lemma 1.1. Let L:K be a separable algebraic extension and let M be an intermediate field. Then M:K and L:K are separable. Minimum polynomial is new term to us. To know about it we need to follow the Eisentein's Irreducibility Criterion. Eisentein's Irreducibility Criterion Let n f(t) = a0 + a1t + ::::::: + ant be a polynomial over Z. Suppose that there is a prime q such that (1)q . an (2)q j ai(i = 0; 1; ::::::::n − 1) 2 (3)q . a0: Then f is irreducible over Q 2 5 5 4 3 1 5 4 3 Example 1.5 Consider f(t) = 9 t + 3 t +t + 3 over Q. This is irreducible if and only if 9f(t) = 2t +15t +9t +3 is irreducible over Q.Eisentein's Criterion applies with q=3, showing that f is irreducible. [2] 2 Galois Theory Definition 2.1. Let K be a subfield of the field L.An automorphism α of K-automorphism of L if α(k) = k for all k 2 K: A simple but effective theorem would be the next one. Theorem.2.1 If L:K is a field extension then set of all K-automorphisms of L forms a group under composition of maps. The next example will give us a rough idea about our problem. Definition 2.2. The Galois group Γ(L : K) of the extension L : K is the group of all K-automorphisms of L under composition of maps. p Example 2.1 The extension C : R .Suppose that α is an R-automorphism of C. Let j = α(i) where i = −1: Then j2 = (α(i))2 = α(i2) = α(−1) = −1 Field Extension by Galois Theory 135 since α(r) = r for all r 2 R.Hence either j = i or j = −i: Now for any x; y 2 R we have α(x + iy) = α(x) + α(i)α(y) = x + jy Thus we have two candidates for R-automorphisms: α1 : x + iy ! x + iy α2 : x + iy ! x − iy Now α1 is the identity, and thus is an R-automorphism of C.The map α2 is known as complex conjugation and can be shown to be an R-automorphism as follows: α2((x + iy) + (u + iv)) = (x + iy) − (u + v)i = α2(x + iy) + α2(u + iv) α2((x + iy)(u + iv)) = α2(xu − yv + i(xv + yu)) = xu − yv − i(xv + yu) = (x − iy)(u − iv) = α2(x + iy)α2(u + iv) Thus α2 is an automorphism.