Note that φ(e) = f. by (8.2). 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). e K) is the identity of H (resp. Moreover, if ˚and ˙are onto and Gis ﬁnite, then from the ﬁrst isomorphism the- (4) For each homomorphism in A, decide whether or not it is injective. Decide also whether or not the map is an isomorphism. 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. . Therefore a2ker˙˚. Furthermore, 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. , 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… (3) Prove that ˚is injective if and only if ker˚= fe Gg. Then there exists an r2Rsuch that ˚(r) = sor equivalently that ’(r+ ker˚) = s. Thus s2im’and so ’is surjective. If r+ ker˚2ker’, then ’(r+ I) = ˚(r) = 0 and so r2ker˚or equivalently r+ ker˚= ker˚. 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. Thus Ker φ is certainly non-empty. functions in F vanishing at x. We have to show that the kernel is non-empty and closed under products and inverses. 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. Thus ker’is trivial and so by Exercise 9, ’ is injective. If a2ker˚, then ˙˚(a) = ˙(e H) = e K where e H (resp. If there exists a ring homomorphism f : R → S then the characteristic of S divides the characteristic of R. Suppose that φ(f) = 0. These are the kind of straightforward proofs you MUST practice doing to do well on quizzes and exams. . The homomorphism f is injective if and only if ker(f) = {0 R}. The function f: G!Hde ned by f(g) = 1 for all g2Gis a homo-morphism (the trivial homomorphism). Then Ker φ is a subgroup of G. Proof. (b) Prove that f is injective or one to one if and only… φ is injective and surjective if and only if {φ(v1), . 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. 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. The kernel of φ, denoted Ker φ, is the inverse image of the identity. Prove that I is a prime ideal iﬀ R is a domain. 2. We show that for a given homomorphism of groups, the quotient by the kernel induces an injective homomorphism. Let s2im˚. Solution for (a) Prove that the kernel ker(f) of a linear transformation f : V → W is a subspace of V . Indeed, ker˚/Gso for every element g2ker˙˚ G, gker˚g 1 ˆ ker˚. 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. Therefore the equations (2.2) tell us that f is a homomorphism from R to C . Given r ∈ R, let f be the constant function with value r. Then φ(f) = r. Hence φ is surjective. This implies that ker˚ ker˙˚. 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 … K). 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 … Exercise Problems and Solutions in Group Theory. (The values of f… Is an isomorphism trivial and so by Exercise 9, ’ is and! 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