Exercise Problems and Solutions in Group Theory. (b) Prove that f is injective or one to one if and only… Thus ker’is trivial and so by Exercise 9, ’ is injective. Therefore a2ker˙˚. If r+ ker˚2ker’, then ’(r+ I) = ˚(r) = 0 and so r2ker˚or equivalently r+ ker˚= ker˚. Then Ker φ is a subgroup of G. Proof. For an R-algebra (S,φ) we will frequently simply write rxfor φ(r)xwhenever r∈ Rand x∈ S. Prove that the polynomial ring R[X] in one variable is … Definition/Lemma: If φ: G 1 → G 2 is a homomorphism, the collection of elements of G 1 which φ sends to the identity of G 2 is a subgroup of G 1; it is called the kernel of φ. Therefore the equations (2.2) tell us that f is a homomorphism from R to C . Then there exists an r2Rsuch that ˚(r) = sor equivalently that ’(r+ ker˚) = s. Thus s2im’and so ’is surjective. Prove that I is a prime ideal iff R is a domain. Proof: Suppose a and b are elements of G 1 in the kernel of φ, in other words, φ(a) = φ(b) = e 2, where e 2 is the identity element of G 2.Then … Let s2im˚. e K) is the identity of H (resp. K). (3) Prove that ˚is injective if and only if ker˚= fe Gg. The homomorphism f is injective if and only if ker(f) = {0 R}. you calculate the real and imaginary parts of f(x+ y) and of f(x)f(y), then equality of the real parts is the addition formula for cosine and equality of the imaginary parts is the addition formula for sine. Suppose that φ(f) = 0. The kernel of f, defined as ker(f) = {a in R : f(a) = 0 S}, is an ideal in R. Every ideal in a ring R arises from some ring homomorphism in this way. These are the kind of straightforward proofs you MUST practice doing to do well on quizzes and exams. If His a subgroup of a group Gand i: H!Gis the inclusion, then i is a homomorphism, which is essentially the statement that the group operations for H are induced by those for G. Note that iis always injective, but it is surjective ()H= G. 3. Solution for (a) Prove that the kernel ker(f) of a linear transformation f : V → W is a subspace of V . Given r ∈ R, let f be the constant function with value r. Then φ(f) = r. Hence φ is surjective. Thus Ker φ is certainly non-empty. Moreover, if ˚and ˙are onto and Gis finite, then from the first isomorphism the- φ is injective and surjective if and only if {φ(v1), . If there exists a ring homomorphism f : R → S then the characteristic of S divides the characteristic of R. Note that φ(e) = f. by (8.2). . , φ(vn)} is a basis of W. C) For any two finite-dimensional vector spaces V and W over field F, there exists a linear transformation φ : V → W such that dim(ker(φ… (4) For each homomorphism in A, decide whether or not it is injective. . Indeed, ker˚/Gso for every element g2ker˙˚ G, gker˚g 1 ˆ ker˚. Let us prove that ’is bijective. Functions can be injections (one-to-one functions), surjections (onto functions) or bijections (both one-to-one and onto). The function f: G!Hde ned by f(g) = 1 for all g2Gis a homo-morphism (the trivial homomorphism). We have to show that the kernel is non-empty and closed under products and inverses. (The values of f… Informally, an injection has each output mapped to by at most one input, a surjection includes the entire possible range in the output, and a bijection has both conditions be true. If a2ker˚, then ˙˚(a) = ˙(e H) = e K where e H (resp. If (S,φ) and (S0,φ0) are two R-algebras then a ring homomorphism f : S → S0 is called a homomorphism of R-algebras if f(1 S) = 1 S0 and f φ= φ0. We show that for a given homomorphism of groups, the quotient by the kernel induces an injective homomorphism. Solution: Define a map φ: F −→ R by sending f ∈ F to its value at x, f(x) ∈ R. It is easy to check that φ is a ring homomorphism. 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