Ideals , homomorphisms and quotient rings

Question 1

Ideal

Answer 1

Let R be a ring and let I be a subset of R. We say I is an ideal of R if

• I is subring of R
• for all a e I and r e R, we have ar e I and ra e I

Question 2

The ideal test

Answer 2

Let R be a ring, and let I be a subset of R. Then I is an ideal of R provided

• 0 e I
• for all a, b e I, we have a-b e I
• for all a e I and r e R, we have ra e I and ar e I

Question 3

The principal ideal

Answer 3

Let R be a commutative ring with one and a e R. The principal ideal of R generated by a is defined to be <a> = {ar : r e R}</a>

Question 4

All ideal in Z are principal

Answer 4

Let I be an ideal of Z. Then there exists m e N0 s.t. I =

Question 5

Using ideals to get new ideals

Answer 5

• The intersection of ideals is an ideal

- The addition of two ideals is an ideal

Question 6

The ideal generated by a1, a2, a3, …, am

Answer 6

Let R be a commutative ring with one and let a1, a2, …, am eR. We define = {r1a1, r2a2, …, rmam: r1, r2, …, rm eR}.

Question 7

Lemma about principal ideals

Answer 7

Let R be a commutative ring with one, let I be an ideal of R and let a, b e R

1) Suppose that a e I. Then <a> C_ I
2) </a><a> = <b> iff a=bx for some xeR
3) Suppose that R is an integral domain. Then <a> = <b> iff a=bu for some ueR</b></a></b></a>

Question 8

Homomorphism

Answer 8

Let R and S be rings and let *: R->S be a function. We say that * is a homomorphism provided it satisfies
H1 for all a,b,e R, we have (a+b)=(a) + *(b)
H2 for all a,b e R, we have *(ab) = (a)(b)

Question 9

Isomorphism

Answer 9

• is a isomorphism if it is a homomorphism and a bijection

Question 10

Elementary properties of homomorphisms

Answer 10

Let R, S and T be rings, let * : R->S and #: S->T be homomorphisms and a,b e R.

1) *(0) =0
2) (-a)=-(a)
3) *(a -b) = *(a) - *(b)
4) The composition of *.#:R->T is a homomorphism
5) Suppose *: R->S is a isomorphism. Then *^(-1):S->R is a homomorphism and therefore and isomorphism

Question 11

Kernel

Answer 11

Let R and S be rings and let *:R-> S be a homomorphism
The kernel of * is defined to be
ker * = { r e R: *(r) =0}

Question 12

Image

Answer 12

Let R and S be rings and let :R-> S be a homomorphism
The image of * is defined to be
im = {*(r) : r e R}

Question 13

Kernel is the ideal of R

Image is a subring of S

Answer 13

Let R and S be rings and let : R-> S be a homomorphism. The ker * is an ideal of R
The im is a subring of S

Question 14

Injective homomorphism

Answer 14

Let R and S be rings and let *:R-> S be a homomorphism and a,b e R. then
a) *(a) = *(b) iff (a-b) e ker * and therefore * is injective iff ker * ={0}

Question 15

Surjective homomorphism

Answer 15

Let R and S be rings and let :R-> S be a homomorphism and let a,b, e R. Then
* is surjective iff im=S

Question 16

Coset of I w.r.t a

Answer 16

[a]I = {a +x: x e I}

Question 17

Coset representative

Answer 17

a is the coset representative of [a]I

Question 18

Set of cosets of I in R

Answer 18

R/I = { [a]I : a e R}

Question 19

Quotient ring

Answer 19

Let R be a ring an let I be an ideal of R. We define addition and multiplication on R/I to be
[a] I + [b]I = [a+b] I
[a] I [b]I = [ab]I
for a, b e R.
The set R/I with addition and multiplication is called the quotient ring of R by I.

Question 20

Isomorphism theorem

Answer 20

Let R and S be rings and let : R-> S be a homomorphism. Then R/ker is isomorphic to the im*.