Solving advanced algebra related problems

All multiplicative invertible elements in Z12 are as follows:

1 × 1 = 1
5 × 5 = 1
7 × 7 = 1
11 × 11 = 1

Therefore, all multiplicative invertible elements in Z12 are 1, 5, 7, and 11.

3, set {(0, 0)(0 a) | a ∈ R} about addition and multiplication of matrix is not the whole ring. First, it is easy to prove that the addition of this set with respect to matrices is an Abelian group. Secondly, the multiplication of matrices by this set also satisfies the associative and distributive laws. However, there is an element(0 0) in the set such that it has no inverse element, so the set is not a group. Therefore, the set is not a whole ring.

The multiplicative invertible elements in the remaining class ring Z12 of module 2 are 1 and 5. We can list the multiplication table for the remaining class ring Z12 of module 2 as follows:

2. The set 2Z of all even numbers. The group formed by the addition operation of 2Z with respect to numbers is cyclic group.

First, we can show that 2Z is closed for addition, i.e. for any a, b∈2Z, a+b also belongs to 2Z. Since a and b are both even, their sum is also even, so a+b∈2Z.

Next, we need to prove that 2Z is a group with respect to addition. The four axioms of groups are closure, associativity, existence unit and existence inverse. We have already shown that 2Z is closed to addition and therefore satisfies the closure and associative law. Next, we prove that there are identity elements and inverse elements.

Existence unit: For any a∈2Z, 0+a=a+0=a. So 0 is the identity of 2Z.

There is an inverse: for any a∈2Z, its additive inverse is -a, because a+(-a)=(-a)+a=0.

Since 2Z satisfies the four axioms of groups for addition operations, 2Z is a group for addition operations.

Next, we prove that 2Z is a cyclic group with respect to addition. Because 2 Z is generated by 2, 2 Z = 2 k | k ∈ {Z}. We can find that for any a∈2Z, it can be expressed in the form of 2k, i.e. a=2k. Then, for any b∈2Z, it can be expressed in the form of 2m, namely b=2m. So a+b=2k+2m=2(k+m) is also a member of 2Z. So we get that 2Z is closed. Since 2Z is an additive group, for any a∈2Z, its n times is na, i.e. n×2k=2(nk) also belongs to 2Z. So 2Z is the cyclic group of addition.

3. For any a, b∈R, [0 a]×[0 b]=[0(ab)], so matrix multiplication is also closed. Therefore, set {[0, 0], [a] 0 | a ∈ R} about matrix addition and multiplication is closed.

then, we need to prove that the set {[0, 0], [a] 0 | a ∈ R} about addition and multiplication of matrix satisfies the definition of the whole ring. The whole ring is defined as a commutative ring R if the following four conditions are met:

• R satisfies the associative law, the commutative law, and the existence of the unit and the inverse element.

  The multiplication in

• R satisfies the associative law, the commutative law and the existence of the identity element.

• Multiplication has distributive property over addition.

• R does not have a zero factor.

• For this set, we can verify that it satisfies the above four conditions:

• set satisfies the associative law, the commutative law, and the existence of the identity element and the inverse element, because it is an Abelian group.

• The multiplication in the set satisfies the associative law, the commutative law, and the existence of the identity element because it is matrix multiplication.

  < / li > < li >

  matrix multiplication meet distributive law, namely for any a, b, c ∈ R, [a] 0 x([0] b + c [0]) = [a] 0 x(b + c) [0] = [0(a)(b + c)]. And [a] 0 x [0 b] + [a] 0 x c [0] = [0](ab) +(ac) [0] = [0(a)(b + c)].

• There is no zero factor in the set because it contains only one non-zero element [0 1], and [0 a]×[0 b]=[0(ab)] is equal to zero only if a=0 or b=0, so there is no product of non-zero elements equal to zero. < br / > so, set {[0, 0], [a] 0 | a ∈ R} about addition and multiplication of matrix is the whole ring.

This answer quotes ChatGPT

The multiplicative invertible elements in the remaining class ring Z12 of module 2 are: 1, 5, 7, 11.

In the remaining class ring Z12 of module 2, any integer a if it is coprime with 12, then a exists multiplicative inverse in Z12, that is, there exists b such that ab≡1(mod 12). According to Euler's theorem, when a and 12 are prime, there is a^φ(12) ≡ 1(mod 12), where φ(12)=4 is the Euler function of 12.

Therefore, in Z12, integers that are prime to 1, 5, 7, and 11 are multiplicative invertible elements; they are 1, 5, 7, and 11, respectively.

Set 2Z The group formed by the operation of addition is not cyclic.

cyclic group means that there is an element a such that every element in the group can be represented as a power of a, i.e. the group is generated from A. For 2Z, its elements are of the form 2n, where n∈Z. Suppose 2Z is a cyclic group, then there is an element a such that 2n=a^n is true for all n. However, we can find that when n is odd, the left side is even and the right side is odd and not equal, so 2Z is not a cyclic group.

3, set {(0, 0)(0 a) | a ∈ R} about addition and multiplication of matrix is not the whole ring.

A whole ring is a set that satisfies both addition and multiplicative closure, associative law, commutative law, distributive law, existence of addition and multiplicative identity element, existence of multiplicative inverse element(for non-zero elements). Here we consider respectively set {(0, 0)(0 a) | a ∈ R} that addition and multiplication in the matrix is satisfy these properties.

For addition, the sum of any two matrices is still in the set and satisfies closure. Associative law of addition, commutative law of addition, existence of the identity element of addition are all true. However, there is no additive inverse in this set, because for any matrix(0 a), there is no other matrix(0 b) such that their sum is(0 0). Therefore, the addition of this set with respect to matrices is not a group and thus does not satisfy the definition of a whole ring.

For multiplication, the product of any two matrices in the set is also still in the set and satisfies closure. But multiplication is not associative. For example, for matrix(0),(0 b),(0 c), there are(a) 0((0 b),(0 c)) =(0, 0)(0 a) =(0, 0), and((a) 0(0(0 c) b) =(0, 0)(0 c) =(0, 0), they are not equal, the product of multiplication not meet the associative law. Therefore, the multiplication of this set with respect to matrices is not an integral ring.