This is a report paper which provides a brief summary of introduction to linear algebra by John R. Durbin. The paper shall provide a brief summary of the three chapters of the book; chapter 11, 12 and thirteen.
Introduction
Modern algebra has been causing much difficulty to the learners and the professors as well. This is attributed to the complexity of the unit and elaborative theorems and formula used in the unit .therefore in this papers I have tried to simplify the theorems and formulas to easen the understanding of the topics such as Galois theorems permutation and many more. This report will provide a comprehensive summary of the three algebraic topics. The summary has been made as comprehensible as possible. Just like the taenia solium depends on the already digested food in the small intestine, this report provides an eleventh hour recipe for excelling candidates especially in the area of algebra. Therefore, any student who depends on this summary is rest assured of passing algebra because it has been made as understandable as possible.
CHAPTER 12.
The main three major problems discussed in this chapter are.
- The duplication of the cube.
- The trisection of an arbitrary angle.
- The quadrature of the circle.
DUPLICATION
If the side of a given cube is taken as a unit of length and by X, then the volume of the cube is x3 cubic units and this can be illustrated as shown below.
Given a segment of length 1, construct a segment of length X where X=3√2.
- Definitely we can illustrate that the trisection of an arbitrary angle. Is impossible in general. This can be done by illustrating that an angle of 600 cannot be trisected. However, it is easy to illustrate that any angle can be trisected if and only if a segment of the length of its cosine can be constructed from a segment of a unit as demonstrated below.
If A is any angle then. Cos A= 4 cos3(A/3)-3cos (A/3).
Assume A= 60O and X to be=cos(A/3). This therefore cancels to 8x3 – 6x- 1=0.
Therefore trisecting a 60o can be summarised as shown below.
Assuming that we are given a segment of unit length, construct a segment of length X where X=8X2 – 6X -1=0.
Take the radius of particular circle as a unit length. The area of the circle is π square units. Hence the problem turns to be that f constructing a segment of length X such that x2=π. Hence the quadrature of the circle has been explained.
Note: a necessary and sufficient condition that a segment of a given length say X can be constructed using a straight ruler and a pair of compass starting with a segment of length 1, is that X can be obtained from 1 by a finite number of rational operations and square roots.
Theorem 1. states that the set K of constructible numbers is a field.
Proof. Assume that agiven unit of length has been provided. Therefore, it is prudent to show that if a€K then a + b is also a subset of K, and a-b is also€ of K, ab € K,. However, we can also proof that if b ≠ 0 then a/b € K.
We can also proof that a/b € K as follows.
Given that a and b are constructible numbers with b≠0. It is easy to prove that a/b € K for all a and b> 0.
CONSTRUCTIBLE NUMBERS.
Before we explore deep into this sub-topic, it is prudent to define what a constructible number is. The term constructible numbers refers to a numbers which can be written as a finite number of additions, subtractions, multiplications, divisions and as square roots of the integers. Therefore, a complex number x can be said to be constructible if x=0 or x=1 or else x is an intersection point of a pair of lines, a line and a circle or instead it is a pair of circles which can be drawn with a straight ruler and a compass.
A constructible line is any line which is on the x and y-plane passing through any given two
Constructible points. Secondly a constructible circle is a circle which is on the x and y-plane with its centre at constructible point with it’s and radius equal to the distance between two constructible points. Finally, a constructible number is a number that can either be described as the distance between any given two constructible points or the negative of such a distance. In a nutshell, a number is said to be constructible if it can be represented by a finite number of addition, subtractions, multiplications, divisions, and even countable square root extractions of the numbers or even integers.
A necessary and sufficient condition that a segment of a given length says x can be constructed using a straight ruler and a pair of compass staring with a segment of length 1, is that x can be obtained form 1 by a finite number of rational operations and square roots. For example, when a given unit length of a line has been given, we can work on assumption that this length is fixed in all the operations.
2.
Construct perpendicular lines using straightedge and compasses in the x and y axes. In order to determine the coordinates of x and y, follow the normal way as shown in the diagram below.
The point o,p and q are called constructible, we need to thoroughly understand the real meanings of constructible circle, number and even the constructible line as already been discussed above
3.
Proof of the lemma
.Normally the equation for the line through two points say, (x1, y1) and (x2, y2) is
y – y1 = y2 – y1
x –x1=x2-x1 if and only if x1≠x2.therefore this equation can be written in the form shown below
ax + by + c = 0 with each of the coefficients in F since every rational combination of x1, y1, x2, and y2. If x1 = x2, the line through (x1 y1) and
(X2, Y2) is X - XI = 0, which can also be written in the form has the ax + by + c = 0 where a, b, c are all subsets of F. II.'
4.The main relationship between statement two and ruler and compass is that we can use these two instrument to construct a particular segment of unit length say 1, then we can construct a segment of length x with 8x3 - 6x - 1 = O. This can critically be used to determine the radius which could also be determined by the mathematical manipulation of the real numbers. In this case the real number r is the radius of the segment of a unit length.
Lemma 1. Let F be a sub-field of R
- Assuming that a line has two points of whose coordinates are in F, then we can take or assume that the equation of the line can be written as ax + by + c=0 for a, b, c € F.
- Assuming that the radius of the circle in F and its centre at a point whose coordinates are in F, then the circle can be represented by the equation x2 + y2 + dx + ey + f= 0
Proof (a). Suppose that x1≠x2, then the equation for the line passing through (x1,y1) and (x2,y2) will be equal to (y-y1)(x2-x1)= (x-x1)(y2-y1).
This equation can further be simplified in the form of ax + by + c=0 with each of the coefficients in F. This is because each coefficient is taken to be a rational combination of x1, y1, x2 and y2. However, in the case where x1=x2, the line through (x1,y2) and (x2,y2) is x-x1 = 0. This can also be written in the form of ax + by + c= 0 with a, b, c € F.
Proof (b). Suppose that x1≠x2, then the equation for the line passing through (x1,y1) and (x2,y2) will be in the form of (x – x1)2 + (y-y1)2=r2.
Similarly the equation of a circle can be given by (x – x1) 2+ (y-y1)2=r2
Theorem 1. A real number R is constructible if and only if a definite sequence Q=F0 combination F1 combination Fk-1 combination FK of subfields of R where r € FK then Fj = Fj-1(aj-1)1/2 and aj-1 is appositive real number in Fj-1 for 1 ≤ j ≤ k.
CHAPTER TEN
Galois Theory.
SIMPLE EXTENSIONS.DEGREE
Under this subtopic, we shall begin by looking at how to construct fiel extensions which can be used to solve aparticular kind of problem such as that of providing roots for polynomials, the extension of R and C so as to obtain root for 1+ X2
Theorem
Assume f(α) and F(β) are simple algebraic extensions of the field F,and α and β are roots of the same polynomials P(x) irriduceble over F. therefore F(α) and F(β) are isomorphic under an isomorphism such that θ(α) = and θ(α)=a for each a €F.
example+
if θ is an asomorphosm of a field F onto a field E and P(x)=a0+a1x+anxn € F(x),then θ p(x) is the polynomial defined byθp(x)=θ(a0) +θ(a1)x++θ(an)xn
Let E be an extension field of the field F. Assume that F≤E. Also let S bea subset of E. This implies that there is at least one subfield of E which contains both F and S, namely E itself, then the intersection of all the subfields of E that contain both F and S is a subfield of E. Therefore, it shall be denoted that F(S). And if S≤F, then F(S)=F. And then if S=(a1, a2an ) then F(S) will be denoted F((a1, a2an ). For instance, if R(i)=C. The field F(S) consistes of all the elements of E which can be obtained from F and S through repeated applications of the operations of E- addition, multiplication and also the taking of additive and multiplicative inverses.
Therefore, if E=F(a) for some a € E., then E can be said to be a simple extension of F. We can therefore classify the simple extension of F by making use of F(x) , the ring of the polynomial in the indeterminate X over F and
F (a)=a0 + a1 a ++: an an : a0 a1an€ F
Theorem
The ring of the polynomials in a. The difference between F(x) and F(a) is that the two polynomials in F(x) are equal only in the case where the coefficients on like powers of x are equal, whereas if a is algebraic over F, then two polynomials in F(a) can be equal without the coefficients on like powers of a being equal. The following is a typical example of what has been discussed above.
1 + 3√2=-1 + 3√2 +√22 in Q(√2).
But 1+3x#-1+3x+x2 in Q[x].
Therefore if E is a simple extension of F with E= F(a) and a algebraic over F, then
E≈F(x)/(p(x)) where p(x) is irreducible over F and (p(x) is the ideal consisting of all F(x) € F(x) such that F(a)=0
The following equation can be proved as shown below
theorem 42.1 Define θ: F[x] F[a] by
Θ (a0 + a1x ++ an xn)=a0 + a1a + anan,
Theorem 4.2 If a is transcendental over F, then F ( a) is isomorphic to the field of quotients of F(x).
Proof. If a is transcendental over F, then Ker θ= (0) in the proof of theorem 42.1 and F(x) ≈F(a). The field of quotients of F(x) can be thought of as being a set of all quotients f(x)/g(x)≠0, where f(x) an g(x) have no common factor of positive degree. Also, F(a) consists of all quotients f(a)g(a)-1 with f(a), g(a) € F(a) and g(a) ≠0. It follows that F(a) is isomorphic to the field of quotients of F(x).
Theorem 42.3. Assume that F is a field and that p(x) € F(x) is irreducible over F. Then F(x) is a field extension of F and p(x) has a root in F(x)/(p(x)).
Solution(proof).
Let I=(p(x)) and let p(x) =a0 + a1x + an xn, and also let a denote the element I + x € F(x)/ I. Then p(a) = a0 + a1(I + X) ++ an( I + x)n
=I + ( a0 + a1x + anxn)
=I + p(x)
=I
But I is the zero of F(x)/I. Thus a is a root of p(x) in F(x)/ I.
Roots of polynomials
An element c of a field F is a root of a polynomial f(x) € F(x) if f(c)=0.
This section will prove that a polynomial degree n has at most n roots. We shall also see that any polynomial of degree n over the field.
Theorem a polynomial f(x) of degree n≥ 1 over a field F has at most n roots in F.
Proof. The proof of this theorem will be done by the induction on n. If n=1, then f(x)=a0 + a1x with a1≠ 0 and the only root is –a1-1a0. Thus assume that n is greater than 1 and assume that the theorem true for polynomials of degree less than n. If f(x) has no root, we are through. On the other hand if c is a root then by the factor theorem f(x) = (x-c) f1(x) for some f1(x) € F(x) and degree f1(x)=n-1. By induction hypothesis, f1(x) has at most n-1 roots in F. it will follow that f(x) has at most n roots in F if f(x) has no roots in F except c and the roots of f1(x). But this is so because if a € F, then f(a)=(a-c), so that f(a)=0 only if a-c=0.
Section 44 fundamental theorem: introduction.
This section outlines the connection between roots of polynomials, fields and automorphism groups.
Example
If E is an extension of Q, and ό is an automorphism of E, then ό(a)=a for all a€Q.
Proof. First ό(1) = 1 therefore, ό(2) = ό(1 + 1) = ό(1) + ό(1) =2. We can extend this mathematical induction to prove that ό(n) =n for every positive integer n. But then for -m €Z and m is less than zero, we have m greater than zero and ό(-m)= -m, so ό(m)= ό(-(-m))=- ό(-m)=-(-m)=m. Thus ό(r)=r for all r € Z. Also if s € Z and s≠0, then ό(1/s) =1/ ό(s). Thus if r,s€Z with s≠0, then ό(r/s)= ό(r) ό(1/s)=r/s. This therefore completes the proof.
Chapter 11
Algebraic extensions
This section shall deal with basic ideas about automorphisms and degrees of field extensions
Theorem 45.1 assume that F (a) and F (ß) are simple algebraic extensions of a field F and a and ß are the roots of the same polynomials p(x) irreducible over F. Then F(a) and F (ß) are isomorphic under an isomorphism θ such that θ(a) =ß and θ(a)=a for each a € F.
Proof. Each element of F(a) can be expressed distinctively in the form (42.1) and each element of F(ß) can be expressed in the same form with a replaced by ß. The mapping θ which is determined by θ(a) = a for each a € F is an isomorphism because a and ß are roots of the same irreducible polynomial p(x).
Theorem 45.2 assume that θ is an isomorphism of a field F onto a field E and that F(a) is a simple algebraic extension of F with p(x) the minimum polynomial of a over F. Assume also that E(ß) is a simple algebraic extension of E with θp(x) the minimum polynomial of ß over E. Then θ can be extended to an isomorphism θ* of F(a) onto E (ß) such that θ*(a) = ß and θ*(a)= ß and θ*(a)= θ(a) for each a € F.
Theorem 45.3. Consider a simple extension F(a) of F. If we assume that a is atrancendential over F, then [F(a):F]=n, where n is the degree of the minimum polynomial of a over F and {1,a,, an-1 is a basis for F(a) as a vector space over F.
Proof. Assume that a transcendental over F. Then the elements 1,a,a2,are linearly independent over F, for otherwise a0, a1,an € F (not all zero), which would imply that a is algebraic over F thus[F(a) : F] is an infinite. If the minimum polynomial of a is of degree n, then the corollary shows that {1,a,an-1} is the basis for F(a) as a vector space over F. Thus [F(a) : F]=n.
Section 46. Splitting fields .Galois Groups.
Theorem.
If P(x) is a polynomial of degree greater than one over a field F,then there exist a splitting field K of P(x) over F such that[K :F] ≤ n!
Lemma 46.1. Assume that p(x) is a polynomial over a field F, and that θ is an isomorphism of F onto F1, then their exists an isomorphism θ* of K onto K1 extending θ, that is, such that θ*(a) = θ (a) for each a € F.
Proof.
Induction
Theorem
Since K is a splitting field of p(x) over F, K is a splitting field of q(x) over F(c). Also, [K: F(c)] < [K : F)]. Therefore the induction hypothesis implies there is an extension of isomorphism θ1: F (c) F1 (θ1(c)) to an isomorphism θ*: K K1.
Corollary. The Galois group G of a polynomial of degree n over a field F is isomorphic to a sub-group Sk, where k is the number of distinct roots in a splitting field of the polynomial over F. Thus, in particular, the order of G divides n!.
Proof. Any group of permutations of {c1ck} is isomorphic to a subgroup of Sk is isomorphic to a subgroup of Sn, since k≤n.
Separability and normality
Can be defined as follows: an irreducible polynomial f € [x] is separable if f hs no repeated roots in the splitting field. However, if f is not necessarily irreducible, then f is said to be separable only if each of its irreducible factors is separable.
Example: if f(x)= (x-1)2 (x-2) € Q is separable because its irreducible factors x-1 and x-2 are separable
Definition.
A polynomial p(x) of the degree n over a field F is separable over F if it has n distinct roots in a splitting field K over F. In a case where p(x) is not separable, then it implies that the polynomial p(x) is inseparable. Therefore, an algebraic element in an extension K of F is separable over F if its minimum polynomial is separable over F. On the other hand, an algebraic extension K of F is a separable extension if every element of K is separable over F.
Theorem
A polynomial p(x) over a field F is separable over F if and only if p(x) and its formal derivative p1(x) are relatively prime in K[x], where K is a splitting field of p(x), that is , if they lack common factor of positive degree in K[x).
Proof.
Let K be a splitting field for p(x), then p(x) = c(x-c1) e1(x-ck)ek. With c, c1,ck € K, c≠0, and c1, .ck distinct. One term of p1(x) is ce1 (x-c1) e1-1 (x-c2) e2(x-ck) ek, and the other terms each have (x-c1) e1 as a factor. Therefore, if e1 > 1, then x-c1 is factor of both p(x) and p1(x). Similar statement will hold in a case where any ej is greater than 1. Conversely, if each of the ej=1, then p(x) and p1(x) do not have a common nonconstant factor. Meaning that p(x) has no repeated root in K if p(x) and p1(x) are relatively prime.
Theorem
If F is a field with no element, then every irreducible polynomial over F is separable over F.if F has an element which is greater than one then an irreducible polynomial p(x) over the field F if and only if P(x)=bo+b1xp++bkxp for bobk €F
Proof
Assume the polynomial p(x) with the degree n>1 which is in the form p(x) = a+ a x++an xn and irreducible over F.If q(x) £ then the either the greatest common divisor of p(x) and q(x) is e which is the unity of F. Or P(x)/ q(x).but due to the fact that deg p”(x) < deg p(x) we have p(x)/p’(x) if and only if p”(x) =0.therefore p(x)/ and p”(x) are relatively
Fundamentals Theorems of Galois Theorem
Describes a remarkable connection existing between the sub-fields F and K. Where K is any polynomial extension of F and the subgroups of G(K/F).
Theorem 1. Assuming that K is a polynomial extension of F and let G= G(K/F). Therefore let S be the set of the subgroups of G and T be the set of all the subfields of K which contain F. Then.
- H G(K/H). This is a 1-1 and onto map from the subfields of K which contains F to the set of the subgroups of G(K/H).
- The inverse of this map will in turn take each subgroup of G to its fixed field.
- H1≤ H2 if and only if G(K/H2) ≤ G(K/H1).
- L≤ K is a polynomial extension of F if and only if G(K/H) is a normal subgroup of G(K/F).
- L1:L2 = G(H(K/L2) / G(K/L1 whenever L2 ≤L1≤ K.
Theorem 2: Assume that K is apolynomial extension of F and H is an extension of F. Therefore, if Q:K H is an isomorphism such that FQ = F. Then K=H
Proof: suppose K=F[f(X) = 0]. Then KQ = FQ[fq(x) =0] = F[f(X) = 0].
Solvability by the Radicals.
A polynomial equation cab be said to be solvable by the radicals if f(x) = 0 where f (x) € F(x). However, this is only so if there is a sequence of fields. Such as F= F0 C F1 CCFk.
Finite Field
Given a prime p and an integer n, we can obtain only one finite field IF Pn with Pn elements. Therefore we can say that inside a fixed algebraic closure of Fp, then field Fpm lies inside Fpn if and only if m/n. In particular, FPn can be said to be the set of solutions of
Xpn – X = 0.
PROOF: in order to proof this, let E be an algebraic closure of FP. Let F(x) = Xp n –x in Fp [X]. Therefore, the algebraic derivative of F shall be -1, hence the GCD(F,F1) =1 and F has no repeated factors. If we let the digit K= Fp (α1,αpn) be the subfield of E which is generated over Fp by the roots of F(x)= 0. Therefore, we can conclude by saying that K is exactly the roots of F(x) =0.
Conclusion
Therefore for easy understanding of the most elaborative and complex topics in modern algebra, the professors should make an attempt of ensuring that all learners are involved in learning. The professors should also be aquatinted with the topics so as ensure full absorption of the content such as Galois theorem and others. This report has provided a comprehensive summary of the book mentioned above. The report provides a concise and easy to understand summary of the three topics in algebra which most of the student do find extremely difficult to understand.
Work Cited
Durbin, J.R. “Modern Algebra: an introduction”: U.S.A.Willey and Sons. 2009
