Let A,B be sets. The set of all ordered pairs x,y. In symbols. The cartesian product R X R is then the set of cartesian coordinates of all points in the plane. Then R is called a relationfrom A to B. A relation from A to A is called a relation on A or in 4. Let R be a relation in the set X. If R is reflexive, antisymmetric, and transitive, then R is called a partial order on X. Examples 2. For x,y E Xlet xIIy mean that xis parallel toy. Let us further agree that every line is parallel to itself.
Then II is an equivalence relation on X. Similarly, congruence of triangles and similarity of triangles are equivalence relations. Consider the relation C determined by "set inclusion. Hence, set inclusion is reflexive, antisymmetric, and transitive; therefore it is a partial order on X. Let n be a fixed positive integer. For any x,y E Z, x is said to be congruent to y modulo n , written x y mod n , if n divides x — y.
II Relations This proves that congruence modulo n is an equivalence relation on Z. Let X be a set, and let be a partial order on X. The symbol here denotes an arbitrary partial order and does not necessarily have its usual meaning of less than or equal to" in real numbers. The set X together with the partial order is called a partially ordered set or, briefly, a poset. We refer to it as the poset X, or simply the poset X.
Let be then xis said to be a poset, and let x,yE X. Ifx y and x y, then xis said to be properlycontainedin y, written x 34 2t a 0 b c Let I, be a partially ordered set. Let S be a subset of X. A least upper bound l.
Show that the polynomial where a, E Z, has no rational root that is not integral. Also show that any integral root must be a divisor of 0 8.
Prove Wilson's theorem: For any prime p, p — 1! Prove Fermat's theorem: For any a,p E Z with p prime, a" a modp. Let n be a positive integer, and let 4 n denote the number of positive integers less than and prime to n.
Prove Euler's theorem: For any integer a relatively prime to n, I mod n. Furthermore, note that for a prime p. Thus, Euler's theorem generalizes Fermat's theorem. Constructing the rational numbers from integers is given in Chapter However, the construction of real numbers is not given because it is normally done in books on analysis see, for example, Landau A real number may be taken to be a terminating or nonterminating decimal.
Consider complex numbers of the form x,0. This allows us to identify every complex number x,0 with the real numberx and thereby treat R as a subset ofC. Let us henceforth write x for x,O. Now j2 0,1. This is the common notation for complex numbers. Then a and b are called the real part and the imaginary part of z, respectively. Fields Introductory algebra is concerned with the operations of addition and multiplication of real or complex numbers as well as the related opposite operations of subtraction and division.
On the other hand, abstract algebra is concerned with deriving properties of an abstract system such as 37 Fields the systems of integers, real numbers, or complex numbers from its axioms in a formal, rigorous fashion. The purpose of this section is to introduce the richest algebraic structure, for which the systems of rational, real, and complex numbers are concrete examples.
The multiplicative operation distributes over the additive operation.. Examples of fields are the systems of rational numbers, real numbers, and complex numbers under the usual operations of addition and multiplication. The system of integers Z is not a field under the usual operations because each nonzero integer is not invertible in Z. Z is an example of another algebraic structure—"commutative ring"—discussed later in the book.
Not every field is infinite. We close this section by giving an example of a finite field. Integers, real numbers, and complex nwnhers 38 Example integers modulo n.
Let ii be a fixed positive integer, and let 3. We proceed to show that this is indeed the case. This proves that addition in Similarly, is well defined. That addition satisfies field axioms i — iv for is trivial. The zero element is 0, and the negative of is i. It is also trivial that multiplication satisfies axioms v , vi , viii , and ix ; but axiom vii need not hold. Let 0 IE Z,. This implies that pfx. We shall return to the algebraic structure. They are a rich source of examples of algebraic structures, which we shall study later.
Let F be a field and m and n positive integers. ThenA i,j ,the image of the ordered pair i,j under the mapping A, is called the i,j entry or element of the matrix A and is denoted by A11 or the latter being more common. Note that it is really the ordered pair m,n that is the size. The list m-tuple A, denoted by A,, and exhibited as a,1 a,2 a1, ,av , An m X I matrix is called a column matrix or a column vector. Similarly, a I X n matrix is called a row matrix or a row vector.
An n X n matrix A is called a square matrix of order n. A matrix that is not necessarily square is said to be rectangular. If every entry mA that is not in its diagonal is zero, then A is called a diagonal matrix and denoted by diag d1, A diagonal matrix diag d1 , A strictly lower triangular matrix is defined similarly. The set The set of all m X n matrices over a field F is denoted by of all square matrices of order n over F is, for the sake of conveof all nience, written F,,.
Again, for the sake of convenience, the set I X n matrices row vectors and the set F"3. We shall later see further examples of these relations. The following theorem indicates the importance of permutation groups. Every group is isomorphic to a permutation group. Proof Let G be a group. More- Permutation groups 87 over, forall is an injective homomorphism.
Therefore, G is isomorphic to a 0 SubgroUp of Hence, The above isomorphism is called the lefi regular representation of G. Similarly, we have a right regular representation. Groups of Symmetries We now describe an important class of permutation groups known as groups of symmetries.
Let Xbe a set of points in space, so that the distance d x,y between points x,y is given for all x,y E X. In other words, a symmetry is a permutation that preserves distance between every two points. Let T1. This proves that is a subgroup of and, therefore, rcr' E is itself a group under composition of mappings.
It is called the group of symmetries of X. Consider, in particular, the case when Xis the set of points constituting a polygon of n sides in a plane. It is clear from geometrical considerations that any symmetry ofXis determined uniquely by its effect on the vertices of the polygon. Therefore we need consider only the symmetries of the set of vertices, which can be labeled as I ,2, Thus, the group of symmetries of a polygon of n sides is a subgroup of Hence, DefinitIon.
Let us now consider the general case of a regular polygon see the if and only is asymmetry figure. It is clear that a permutation a if a takes any two adjacent vertices of to adjacent vertices; that is, if and Groups 88 3 2 n only if a I ,a 2 , Thus, the symmetries of Pn can be classified into two kinds: those that preserve the cyclic order I ,2, Let a be a symmetry that preserves the cyclic order.
Now a 1 can have any one of the values 1 , Therefore, there are exactly n symmetries that preserve the cyclic order of the vertices. Let us call them a1 ,a2 where a, is the symmetry taking I to i. By the same argument there are exactly n symmetries that reverse the cyclic order, where r, takes I to i. Clearly, is the identity we easily permutation e, and 02 is the cycle 1 Moreover, r2 takes Ito 1 and preserves the order.
Hence r2 — e. Now consider the product ra. The dihedral group D4 is called the octic group. As an example of the group of symmetries of a nonregular polygon, we shall consider the symmetries of a rectangle in Example 5.
Example 5. Thus, the group of symmetries of a rectangle is the Klein four-group. Problems 1. Find all subgroups of 53 and S4. Let k i,j. Then arXi taXi , so at to-. Let a and b be elements of order 2 in a group G.
Suppose Show that I o ab 4. Show that the subgroup generated by a and b is D4. Hence, we are naturally interested in a minimal subset Xof the group that generates it, and also in the set of equations satisfied by elements in the generating set that would suffice to give us all products of elements in the group. Let G be a group generated by a subset X of G.
The set X is called a set of generators. The system called a presentation of the group. Since multiplication in G is associative, the induced multiplication of subsets of G is also associative. Let G be a group. Trivially, the subgroups e and G itself are normal subgroups of G. If G is abelian, then every subgroup of G is a normal subgroup. But the con- verse is not true: A group in which every subgroup is normal is not necessarily abelian [Example 1.
G and the order of each element of GIG"' is finite. Show that the dihedral group can be expressed as a split Show that the quaternion group see Problem 4, Section 1, Chapter 4 cannot be expressed as a split extension of two Let L G be the set of normal subgroups of a group G. Show that L G is a modular lattice. Show also that the lattice of all subgroups need not be modular. We saw in Theorem 1. The first isomorphism theorem which is also known as the fundamental theorem of homomorphisms proves the converse; that is, every homomorphic image of G is isomorphic to a quotient group of G.
Let 4: homomorphism of groups. Since IL" is obviously suijective, we conclude that. Im 4 is the isomorphism obtained in the theore,n, andj: Im 4 —' G' is the inclusion map. The inclusion diagram shown below is helpful in visualizing the theorem.
Because of this, the theorem is also known as the "diamond isomorphism theorem. Then G,,IG, for and and x Solution. Let a,b G. Because there are only two conjugate classes, these must be e and C a. Therefore, if e b E G, then b E C a. Let mln.
Hence, n is prime. This implies G is abelian. But, because G has only two Solution. Then Normal i ii Conjugacy defined in is an equivalence relation. Follows from Theorem 4. Then the orbit of and the stabilizer is H fl N A. A E X is the conjugate class The direct argument is as follows.
First, i is obvious. For the proof of ii , consider the mapping a: where a a is also , for let h1 ,h2 E H. The example which follows gives a beautiful illustration of the Burnside Theorem. Let X be the set of all possible necklaces. Then X is a G-set under the action Conjugacy. Clearly the action of any fixed element a given necklace yields the same necklace. The beads are just being permuted cyclically. So the number of orbits is the same as the number of different necklaces. We now compute Xg for each gEG.
Then Let g e. Therefore, Xg consists of all those necklaces which are unchanged by any permutation and these are precisely those which are of one color.
The Burnside Theorem then gives the number of different necklaces as p -. I timcs I ]. Find the number of conjugates of the element 1 3 in D4. Determine the conjugate classes of the symmetric group of degree 3 and verify that the number of elements in each conjugate class is a divisor of the order of the group.
Let H be a proper subgroup of a finite group G. Show that G contains at least one element that is not in any conjugate off!. Show that every element of order dividing INI is contained in N. Prove that in any group the subset of all elements that have only a finite number of conjugates is a subgroup. If the commutator group G' ofa group G is of order m, prove that each element in G has at most m conjugates. Let G be the group of symmetries of a cube.
Prove that each of the following sets is a G-set: i Set of vertices. Show G acts transitively on X if and only if there is only one orbit. Also show if H 2. Use the Burnside Theorem to Show that there exists a homomorphism G—. Find the number of necklaces with six beads and two colors.
Show that there are six different necklaces with four beads and two colors and eight different ones with five beads and two colors. Suppose X is a G-set. A composition series of a group G is a normal series G0,. We often refer to a normal series G0, G1,. If G is a simple group, then e C G is the only composition series of G. Every finite group has a composition series. We remark that a composition series is not necessarily unique. For example, another composition series for the octic group 1 4 is e C e,t C e,a2,r,a2r C However, it is shown in the next section that any two composition series of the same group are "equivalent," in the following sense.
Evidently, — is an equivalence relation. Any two composition series ofafinite group are equivalent. By Corollary 2. Therefore, 53 — S4. Further, by the induction hypothesis, S1 S2 —S4. This proves that S1 and S2 are equivalent. Examples 1. Therefore, an abelian group G has a composition series ifand only if , is finite. Moreover the composition factors of G are determined by the prime factors of JGJ. By Theorem 4.
Hence, G has a unique composition series if and only if Let G be a cyclic group of order n. Suppose n has two factonzations into positive primes, say q3. By the Jordan — HOlder theorem the two series must be equivalent. Then G has a composition series such that all its composition factors are of order p.
But since a group of prime power order has a nontrivial center, and each composition factor is a simple group, it follows that each composition factor is a simple abelian group and, therefore, of order p.
Problems Write down all the composition series for the quatern ion group. Write down all the composition series for the octic group. Write down a composition series for the Klein four-group. Find the composition factors of the additive group of integers modulo 8. Verify that they are equivalent. Let G be a finite group and N 1. A group G is said to be solvable integer k.
Thus, trivially, every abelian group is solvable. Then for any subProof Let G be a solvable group, so that for every positive integer n. This proves that every homo. Because GIN is a homomorphic image of G under the natural homomorphism for every positive integer n. Therefore, e 2. Further, a finite group is solvable if and only if its composition factors are cyclic groups of prime orders.
Then by Theorem 1. I Now suppose G is a finite solvable group. Then G has a normal series is abelian. Now C Inserting the corresponding subgroups of C H,, we and H,in the normal series C H, between the terms get a composition series of G in which every factor is a cyclic group of prime order.
The converse follows from the first part of the theorem. We shall see in the next chapter that S4 is solvable, but n 5, is not solvable. Then M is cyclic of order p. We may first show that M is an abelian p-group.
Show that the group of all upper triangular matrices of the form lab o i c is solvable. Show that a simple group is solvable if and only if it is cyclic. Normal series Let A, B be normal solvable subgroups of a group G. Show that AB is also solvable. Additional problems on solvable groups are given in the problem set at the end of Chapter 8. Nilpotent groups 3 We define inductively the nih center of a group Gas follows.
E C The ascending series of subgroups of a group G is called the upper central series of G. Thus, trivially, every abelian group is nilpotent. For a nontrivial example we prove 3. A group of order p" p prime is nilpotent. Proof By Theorem 4. Hence, x does not appear in any f3,. Ifx is in some y1, then a x x.
Hence, x must be in some fi,. But then a' x occurs in both y, and for every r E Z. Hence, y, and are identical cycles. This proves that the two decompositions written earlier are identical, except for the ordering of factors.
Every permutation can be expressed as a product of 1. Proof We need only show that every cycle is a product of transpositions. It is easily verified that aia D Note that decomposition into transpositions is not unique, and the factors need not be pairwise disjoints.
Let us write the decomposition of a into disjoint cycles, including cycles of length 1, and write the factors in order of increasing the cycles , Thus,a respectively. Then n1 n2 determines a partition n ,. Conversely, given a partition n1 , Hence, any Iwo conjugate permutations in have the same cycle structure. C'onversely, any two permutations in with the same cycle structure are conjugate. Hence, the cycle decomposition of t is obtained by substituting a x forx everywhere in the decomposition of a.
Thus, a and have the same cycle structure. Cyclic decomposition Conversely, suppose that a and r have the same cycle structure p,q, Hence, a and r are conjugate. There is a one-to-one correspondence between the set of conjugate classes of S, and the set of partitions of n.
Proof Obvious. Then start with any other element not in 1 4 , say 2. Starting with 2, we get a cycle 2 6 7 5 3. These two cycles exhaust the list 1,2,3,4,5,6,7. Compute By Theorem 1. The interested reader may actually verify this by first computing a'.
Problems I. If a is a cycle of length r, then a' is the identity permutation. Show that the order ofany c E is the least common riiultiple of the lengths of its disjoint cycles. Any two cycles in Sn are conjugate if and only if they are of the same length. Alternating group We showed in Section I that every permutation in be expressed as a product of transpositions, and the number of transpositions in any given product is not necessarily unique.
However, we have the following important result showing that the number of factors is always even or always odd. If a permutation c E Sn isa product of r transpositions and also a product of s transpositions, ihen rand s are either both even or both odd. Hence, 4 is surjective. But Ker 4 is precisely the set A,, of all even permutations.
The subgroup A,, of all even permutations in S, is called the alternating group of degree n. Hence, a b cXa b d c The only possible even permutations inS4 are 3-cycles and products of two transpositions, in addition to the identity.
Note that two elements of A,, may be conjugate in S,, but not in A,,. For example, a and a2, being of the same length, are conjugate in 54 but not in A4. Consequently, A4 has no subgroup of order 6. This example shows that the converse of Lagrange's theorem does not hold. Moreover, since a, b, and c constitute a conjugate class, H is normal. It is Simplicity of clear from the relations satisfied by a, h, and c that H is the Klein fourgroup. Let K be any normal subgroup of A4.
Finally, any subgroup of order 6 of 44, being of index 2, would be a normal subgroup. Hence, A4 has no subgroup of order 6. Now every 3-cycle is an even permutation and, hence, an element of A,,. Conversely, every element in A,, is a product of an even number of transpositions. Consider the product of any two transpositions a and r. Thus, every even permutation is a product of 3-cycles.
Hence, A,, is generated by the set of all 3-cycles in Sn. The derived group of S,, is A,,. Because Sn, On the other 1 2 3. Hence, every 3-cycle; therefore, An C hand, every commutator in an even permutation. Hence, C A,,. We first prove that H must contain a 3-cycle. Let a e be a permutation in H that PermutatIon groups the least number of integers in n.
Being an even permutation, a cannot be a cycle of even length. Consider first case I. Because a cannot be a 4-cycle, it must move at least two more elements, say d and e. Thus, r moves fewer elements than a. But r E H, a contradiction. Consider now case 2. Thus, flE Hand moves fewer integers than a, a contradiction. Hence, we conclude that a must be a 3-cycle. Let r be any 3-cycle inS,,. Because any two cycles of the same length are a transpoa conjugate in Sn,? Then are disjoint.
Therefore, r E H. Thus, H contains every 3-cycle in 5,,. Hence, is not trivial. Consequently, 0 proves that S,, is not solvable. SimplidtyofA, Solution. CHAPTER 8 Structure theorems of groups Direct products I If a group is isomorphic to the direct product of a family of its subgroups known as summands whose structures are known to us, then the structure of the group can generally be determined; these summands are like the "building blocks" of a structure.
Theorem 1. Let H, , Then any subgroup of G of order ptm is called a Sylow p-subgroup of G. Let p be prime. A group G is called a p-group if the order of every element in G is some power ofp.
Likewise, a subgroup H ofany group G is called a p-subgroup of 0 if the order of every element in H is some power of p.
Lemma Cauchy's theorem for abelian groups. Jfp divides IAI. Suppose that the result is true for all groups of orders m. K is normal by Corollary 4. Hence, G is generated by the element xi', and, thus, G is cyclic. Structure theorems of groups g Any group of order 15 is cyclic.
Follows from f. Consider a nonabelian group G of order 8. Then G contains no element of order 8; otherwise G is abelian. If each element of G is of order 2, then G is abelian. Thus G contains an element a of order 4. Let b E G such that b [a]. Because [a] is of index 2 in G, [a] I and R is not a trivial ring, then is not commutative. We denote by the matrices in whose i,J entry is I and whose other entries are zero. The i. I This representation of matrices in terms of matrix units is often a conve- nient tool in certain problems involving algebraic manipulations.
Let Sbe the set of n X n matrices, all of whose entries below the main Types of rings diagonal are zero; that is, let S consist of matrices a11 as,, a12 o a ring with the usual addition and multiplication of matrices and is called the ring of upper triangular matrices. Similarly, we have the ring of lower triangular matrices. Polynomials may be added and multiplied to form a ring in the following manner. Thus, RIxj contains a carbon copy of R.
By identifying the element a E R with the sequence a,0,0, E R[x], we say that R is a subring of R[xJ. In what follows we assume I E R, and we denote by x the sequence 0,1,0, Then x2 0,0,1,0, Consider with finitely many nonzero entries. We call x an indeterminate and refer to the polynomials in R[xJ as polynomials in one indeterminate over R sometimes we also call them polynomials in a variable x. Types of rings c Ring of endomorphisms of an abelian group Let M be an additive abelian group. We show that the set End M together with the above binary operations of addition and multiplication forms a ring.
Let x,y C M. Also it is easily shown that the two mappings 0: x 0 zero endomorphism of M and identity endomorphism of U x of M into M are also endomorphisms of U. Then is a commutative ring with identity which is the whole set A. The zero element of this ring is the empty set. It is interesting to point out that any Boolean ring is known to be a subring ofthe ring of all subsets of a set.
Then S is called a subring [ S. We note that addition and multiplication of elements of S are to coincide with addition and multiplication of these elements considered as elements of the larger ring R. Similarly, we can define a subdivision ring of a division ring and a subfield of a field.
Let R be a ring with or without unity. A subring of R may have unity different from the unity of R [see Problem 10 b. The following result is frequently useful: 4. Proof Jfpart. Because the two distributive laws hold in R, they also hold in S.
D Definition. Let R be a ring. The center of a ring is a subring. Proof Let R be a ring. Then 0. Hence, ab E S. Therefore, by Theorem 4. Let S be a subset of a ring R. Then the smallest subring of R containing S is called the subring generated by S.
Because the intersection of a family of subrings is again a subring, it follows that the subring generated by a subset SofR is the intersection of all subrings of R containing S. R is said to have characteristic zero. The characteristic of R is denoted char R. For example, the ring of integers has characteristic zero. Let F he afield. Then the characteristic ofF is either 0 or a prune number p.
Because 0 and Dare also the only right or left ideals of D, it follows that Theorem 1. Ideals and homomorphisms Further, the following example shows that Theorem 1. Then R isa ring without unity. This implies I 0 or R2,since 0 or Rare the only ideals in R. But clearly I 0, and Ic R2. Hence, Theorem 1. Let A, ,EA be a family of right left ideals in a ring R. Then 0 because R E Let! The smallest right ideal of R containing a subset S is called a right ideal generated by S.
Simi- larly, we define the left ideal and the ideal generated by a subset S. A right ideal I of a ring R is called finitely generated a1 am p for some a, R, I i m.
A right ideal I of a ring R is called principal if! In a similar manner we define a finitely generated left ideal, a finitely generated ideal, a principal left ideal, and a principal ideal.
In this case the symbols RaR, aR, and Ra are also used for a , a 1, and a 1, respectively. A ring in which each ideal is principal is called a principal ideal ring PIR.
Similarly, we define principal righi left ideal rings. A commutative integral domain with unity that is a principal ideal ring is called a principal ideal domain ND. We note that all ideals in the ring of integers Z and the polynomial ring F x over a field Fare principal. Let I be a nonzero ideal in Z, and let n be the smallest positive integer in I note that I does contain positive integers.
Ideals and homomorphisms We can easily show that these binary operations are well defined. Because a — c, b — d E I, which is an ideal, we have ab — cd E 1. This proves that multiplication is well defined.
The ring RI! R, then RI! If n 0, then the quotient ring the usual ring of integers modulo n. Let I x be the ideal in R xJ consisting of the multiples of x. Then the quotient ring R[xJI! Ideals Solution. Then x 0. Then —1. Then forsomea,fiER. Then Solution. Case 1.
The proof also shows that A is a principal right ideal of R. Case 2. Therefore, in this case any right ideal of R is of the form where K is an additive subgroup of Q. Homolnorplllsm3 ii It is straightforward to verify that the nontrivial right ideals an additive subgroup of Q, A 2 of R are also left ideals.
Therefore, we have an example of a ring that is not commutative in which each right ideal is an ideal. It is interesting to note that each left ideal of R is not an ideal. Let R be a commutative ring with unity. Suppose R has no nontrivial ideals.
Prove that R is a field. Prove the converse of Problem 1. Generalize Problem I to noncommutative rings with unity having no nontrivial right ideals. Find all ideals in a polynomial ring Fixi over a field F. Let L R be the set of all right left ideals in a ring R. Give an example to show that the lattice L R need not be distributive. The mapping is called a natural or canonical homomorphism.
Then f is called a homomorphism of R into S. In this casef is also called an e,nbedding of the ring R into the ring S or R is embeddable ins ; we also say that Scontains a copy of R, and R may be identified with a subring of S. The symbol R C. S means that R is embeddable in S. If a homomorphismffrom a ring R into a ring Sis both and onto, then there exists a homomorphism g from S into R that is also I-I and onto.
In this case we say that the two rings R and S are isomorphic, and, abstractly speaking, these rings can be regarded as the same. As stated above R S implies S R. Also, the identity mapping gives R R for any ring R. It is easy to verify that if! Thus, we have shown that Isomorphism is an equivalence relation in the class of rings.
Let us now list a few elementary but fundamental properties of homomorphisms. Then we have the following: 2. Proof i Let a R. For convenience, f 0 will also be written as 0.
Hence,f R is a subring of S. Ef—' O. If a,b c R, then is an ideal in R. Similarly, f l f a f a. Thus, f R is commutative.
Conversely, if the kernel of a homomorphismf: R S is 0 , thenf is follows that f a —b 0; thus, necessarily 1- I.
We record it this fact separately in Theorem. Letf R —' S be a homomorphism ofa ring R into a ring 2. Then Kerf— 0 if and only iff is I-I. Let N be an ideal in a ring R. It is an important fact that, in a certain sense, every homomorphic image of a ring R is of this type — a quotient ring of R modulo some ideal Also, if of R.
This is stated more precisely in the following Theorem fundamental theorem of homomorphisms. Let fbe a homomorphism of a ring R into a ring S with kernel N. Then 2. Next, g is a homomorphism. Clearly, gis an onto mapping. We showgis But then cib. This showsgis In this sense this theorem states that the only homomorphic images of a ring Rare the quotient rings of R.
Accordingly, if we know all the ideals N in R, we know all the homomorphic images of R. The fundamental theorem of homomorphisms between rings is also stated as follows. Theorem fundamental theorem of homomorphisms. Given a homomorphism of rings f R S. We remark is a mapping of subsets of S into subsets of R. If, however,fis a mapping of R onto S, then the is used to denote the mapping of S onto R defined byf' s r, where r is the unique element of R such thatf r — s.
The context will make it clear in which sense the symbolf' is used. Theorem correspondence theorem. Let j? Ii preserves ordering in the sensethat A B jfff A 2. Similarly, ra E f ' X. Hence, A f ' X is an ideal in R. For any ideal A in R it is easy to check thatf A is an ideal in S. So let x X. Ideal3 aDd We now show that the mappingFis We claim that f' f A — A.
Here it is tn vial that A Cf-' f A. Hence, XE A. Similarly, — B. Therefore, f A —f B impliesA —B. Thenf A cf B. But sincef A x—a we cannot have A — B.
The correspondence between the set of right or left ideals can be established exactly in the same way. This proves the theorem. Such ideals called "maximal" exist, in general, in abundance and play a very important role in the theory of rings and fields. We return to this discussion in Section 4. A concept opposite to that of homomorphism is that ofantihomomorphism, defined next. Let R andS be rings. A mappingf R —. An antibomomorphism that is both and onto is called an anti-isomorphism.
It is easy to Definition. Let f be a nonzero homomorphism. But the only nonzero idempotent 1 — a nonzero idempotent in in Z is 1.
A right or left ideal A in a ringR is called nilpotent 0 for some positive integer n. Clearly, every zero ideal is a nilpotent ideal, and every element in a nilpotent ideal is a nilpotent element.
However, the set of nilpotent elements in a ring R does not necessarily form a nilpotent ideal. Indeed this set may not even be an ideal. A ring R may have nonzero nilpotent elements, but it may not possess a nonzero nilpotent ideal, as the following example shows.
Then R has nonzero nilpotent elements, such as eu, i ,"j, i,j n. Let I be a nilpotent right ideal in R with 0, where k is some positive integer. Then consider the ideal 5.
But RI cannot be equal to R, since R has unity 0. Hence,I— 0. A right or left ideal A in a ring R is called a nil ideal if each element of A is nilpotent. Clearly, every nilpotent right or left ideal is nil. However, the converse is not true, as the following example shows.
Let I be the set of all nilpotent elements. All three volumes have been planned as texts for courses. Download Algebra I books , Algebra is a compulsory paper offered to the undergraduate students of Mathematics. Download Basic Abstract Algebra books ,. Download Abstract Algebra Manual books , This is the most current textbook in teaching the basic concepts of abstract algebra.
The author finds that there are many students who just memorise a theorem without having the ability to apply it to a given problem. Therefore, this is a hands-on manual, where many typical algebraic problems are provided for students to be able to apply the theorems and to actually practice the methods they have learned.
Each chapter begins with a statement of a major result in Group and Ring Theory, followed by problems and solutions. Download Abstract Algebra books , Excellent textbook provides undergraduates with an accessible introduction to the basic concepts of abstract algebra and to the analysis of abstract algebraic systems.
Features many examples and problems. Download A First Graduate Course In Abstract Algebra books , Realizing the specific needs of first-year graduate students, this reference allows readers to grasp and master fundamental concepts in abstract algebra-establishing a clear understanding of basic linear algebra and number, group, and commutative ring theory and progressing to sophisticated discussions on Galois and Sylow theory, the structure of abelian groups, the Jordan canonical form, and linear transformations and their matrix representations.
The book's unique presentation helps readers advance to abstract theory by presenting concrete examples of induction, number theory, integers modulo n, and permutations before the abstract structures are defined.
Readers can immediately begin to perform computations using abstract concepts that are developed in greater detail later in the text. The Fourth Edition features important concepts as well as specialized topics, including: The treatment of nilpotent groups, including the Frattini and Fitting subgroups Symmetric polynomials The proof of the fundamental theorem of algebra using symmetric polynomials The proof of Wedderburn's theorem on finite division rings The proof of the Wedderburn-Artin theorem Throughout the book, worked examples and real-world problems illustrate concepts and their applications, facilitating a complete understanding for readers regardless of their background in mathematics.
A wealth of computational and theoretical exercises, ranging from basic to complex, allows readers to test their comprehension of the material. In addition, detailed historical notes and biographies of mathematicians provide context for and illuminate the discussion of key topics.
A solutions manual is also available for readers who would like access to partial solutions to the book's exercises. Introduction to Abstract Algebra, Fourth Edition is an excellent book for courses on the topic at the upper-undergraduate and beginning-graduate levels.
The book also serves as a valuable reference and self-study tool for practitioners in the fields of engineering, computer science, and applied mathematics. Download A History Of Abstract Algebra books , This book does nothing less than provide an account of the intellectual lineage of abstract algebra.
The development of abstract algebra was propelled by the need for new tools to address certain classical problems that appeared insoluble by classical means. A major theme of the book is to show how abstract algebra has arisen in attempting to solve some of these classical problems, providing a context from which the reader may gain a deeper appreciation of the mathematics involved.
Mathematics instructors, algebraists, and historians of science will find the work a valuable reference. Download A Book Of Abstract Algebra books , Accessible but rigorous, this outstanding text encompasses all of the topics covered by a typical course in elementary abstract algebra. Its easy-to-read treatment offers an intuitive approach, featuring informal discussions followed by thematically arranged exercises.
This second edition features additional exercises to improve student familiarity with applications. Download Basic Algebraic Systems books ,. Download Basic Abstract Algebra books , Thought-provoking and accessible in approach, this updated and expanded second edition of the Basic Abstract Algebra provides a user-friendly introduction to the subject, Taking a clear structural framework, it guides the reader through the subject's core elements.
A flowing writing style combines with the use of illustrations and diagrams throughout the text to ensure the reader understands even the most complex of concepts.
0コメント