Prove that I is a prime ideal iff R is a domain. Indeed, ker˚/Gso for every element g2ker˙˚ G, gker˚g 1 ˆ ker˚. (b) Prove that f is injective or one to one if and only… Thus Ker φ is certainly non-empty. Therefore a2ker˙˚. Then there exists an r2Rsuch that ˚(r) = sor equivalently that ’(r+ ker˚) = s. Thus s2im’and so ’is surjective. 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. (The values of f… K). Note that φ(e) = f. by (8.2). 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. We show that for a given homomorphism of groups, the quotient by the kernel induces an injective homomorphism. , 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… The homomorphism f is injective if and only if ker(f) = {0 R}. 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. e K) is the identity of H (resp. Solution for (a) Prove that the kernel ker(f) of a linear transformation f : V → W is a subspace of V . Functions can be injections (one-to-one functions), surjections (onto functions) or bijections (both one-to-one and onto). Furthermore, ker˚/ker˙˚. Therefore the equations (2.2) tell us that f is a homomorphism from R to C . This implies that ker˚ ker˙˚. 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. If there exists a ring homomorphism f : R → S then the characteristic of S divides the characteristic of R. Decide also whether or not the map is an isomorphism. . (4) For each homomorphism in A, decide whether or not it is injective. Suppose that φ(f) = 0. Exercise Problems and Solutions in Group Theory. If r+ ker˚2ker’, then ’(r+ I) = ˚(r) = 0 and so r2ker˚or equivalently r+ ker˚= ker˚. Given r ∈ R, let f be the constant function with value r. Then φ(f) = r. Hence φ is surjective. 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. . Moreover, if ˚and ˙are onto and Gis finite, then from the first isomorphism the- (3) Prove that ˚is injective if and only if ker˚= fe Gg. Thus ker’is trivial and so by Exercise 9, ’ is injective. The function f: G!Hde ned by f(g) = 1 for all g2Gis a homo-morphism (the trivial homomorphism). 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 … Let s2im˚. 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 … These are the kind of straightforward proofs you MUST practice doing to do well on quizzes and exams. Then Ker φ is a subgroup of G. Proof. Let us prove that ’is bijective. 2. 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. 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. The kernel of φ, denoted Ker φ, is the inverse image of the identity. If a2ker˚, then ˙˚(a) = ˙(e H) = e K where e H (resp. 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