# Answer to Question #24790 in Abstract Algebra for Alanazi

Question #24790

Q 1)

Let I and J be (left or right or two-sided) ideals of a ring R. We define their product IJ to be a set {x1y1 + . . . + xnyn | xi ∈ I, yj ∈ J}.

Show that the set IJ is again an (left, right or two-sided) ideal. Moreover, show that (IJ)K = I(JK) for any ideal I, J, K of R

Q 2)

The composition ′◦′ satisfies the following two axioms

1. (associativity) If f: A → B, g: B → C and h: C → D, then h ◦ (g ◦ f) = (h ◦ g) ◦ f.

2. (identity) For every object A ∈ ob(C) there exists a map 1A : A → A called the identity map for A, such that for every morphism f : A → B we have 1B ◦ f = f ◦ 1A = f

Show using the axioms that the identity map is unique for every object.

Q3)

Check that R[G] satisfies the ring axioms.(R[G] is

a group ring of G over R).

Let I and J be (left or right or two-sided) ideals of a ring R. We define their product IJ to be a set {x1y1 + . . . + xnyn | xi ∈ I, yj ∈ J}.

Show that the set IJ is again an (left, right or two-sided) ideal. Moreover, show that (IJ)K = I(JK) for any ideal I, J, K of R

Q 2)

The composition ′◦′ satisfies the following two axioms

1. (associativity) If f: A → B, g: B → C and h: C → D, then h ◦ (g ◦ f) = (h ◦ g) ◦ f.

2. (identity) For every object A ∈ ob(C) there exists a map 1A : A → A called the identity map for A, such that for every morphism f : A → B we have 1B ◦ f = f ◦ 1A = f

Show using the axioms that the identity map is unique for every object.

Q3)

Check that R[G] satisfies the ring axioms.(R[G] is

a group ring of G over R).

Expert's answer

1

IJ = {x1y1 + . . . + xnyn | xi ∈ I, yj ∈ J}.

r * (x1y1 + . . . + xnyn) = (r*x1)y1 + . . . + (r*xn)yn ∈ IJ, since r*x_i ∈ I

x1y1 + . . . + xnyn + x'1y'1 + . . . + x'ny'n ∈ IJ

So, if I is left ideal then IJ os left ideal too.

Right case is similar.

Associativity I(JK)=(IJ)K follows from associativity of multiplication and can be proved on monoms, and then be extended to the whole

sums.

2

Suppose object A have two identity maps: 1 and 1'. Then

1=1 * 1' = 1' * 1 =1', so it is unique.

3

Elements of RG are sums of type x1g1 + ...+ xngn, where g_i are from group G.

If we have element xg, then its opposite will be (-x)g, and 0=0g is zero.

Axioms of abelian group now follows immediately.

Associativity of composition follows from associativity of elements in ring R and group G.

Distributive laws can be proved on monoms.

For example:

ag1(bg2+cg3)=abg1g2+acg1g3=ag1 * bg2 + ag1 * cg3.

So, RG will be a ring.

IJ = {x1y1 + . . . + xnyn | xi ∈ I, yj ∈ J}.

r * (x1y1 + . . . + xnyn) = (r*x1)y1 + . . . + (r*xn)yn ∈ IJ, since r*x_i ∈ I

x1y1 + . . . + xnyn + x'1y'1 + . . . + x'ny'n ∈ IJ

So, if I is left ideal then IJ os left ideal too.

Right case is similar.

Associativity I(JK)=(IJ)K follows from associativity of multiplication and can be proved on monoms, and then be extended to the whole

sums.

2

Suppose object A have two identity maps: 1 and 1'. Then

1=1 * 1' = 1' * 1 =1', so it is unique.

3

Elements of RG are sums of type x1g1 + ...+ xngn, where g_i are from group G.

If we have element xg, then its opposite will be (-x)g, and 0=0g is zero.

Axioms of abelian group now follows immediately.

Associativity of composition follows from associativity of elements in ring R and group G.

Distributive laws can be proved on monoms.

For example:

ag1(bg2+cg3)=abg1g2+acg1g3=ag1 * bg2 + ag1 * cg3.

So, RG will be a ring.

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## Comments

Assignment Expert21.02.13, 17:28You're welcome. We are glad to be helpful.

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Alanazi21.02.13, 17:19Hi

Thank you so much.

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