Question 3 - iso,homo,theorems

Question 1

Definition : Ring Homomorphism

Answer 1

Let R and S be rings. A function f : R→S
is a (ring) homomorphism if
f(x+ y) = f(x) + f(y) for all x,y∈R,
f(x·y) = f(x)·f(y) for all x,y ∈R,
f(1_R) = 1_S

Question 2

Definition : Ring isomorphism

Answer 2

A ring isomorphism is a bijective ring homomorphism.

Question 3

Definition : Image and Kernel

Answer 3

Let R and S be rings and
f : R →S a homomorphism.
The image of f is
im(f) = {f(x) : x∈R}.
The kernel of f is
ker(f) = {x∈R: f(x) = 0}.

Question 4

Kernel and Image Lemma

Answer 4

Let R and S be rings and f : R→S a homomorphism.
(1) im(f) is a subring of S, and f is surjective if and only if im(f) = S.
(2) ker(f) is an ideal of R, and f is injective if and only if ker(f) = {0_R}.

Proof.

(1) We apply the subring test. Since f(1) = 1, we have 1 ∈im(f). Since f(x)−f(y) = f(x−y), im(f) is closed under subtraction.
Since f(x)f(y) = f(xy),
im(f) is closed under multiplication.
Thus im(f) is a subring by the subring test.
Obviously f is surjective if and only if im(f) = S.

(2) Since f(0_R) = 0S we have 0_R ∈ker(f). If x,y∈ker(f) then f(x−y) = f(x)−
f(y) = 0_S−0_S = 0_S, so x−y∈ker(f), so ker(f) is closed under subtraction. Thus
ker(f) is a subgroup. If r∈R and x∈ker(f) then f(rx) = f(r)f(x) = f(r)·0_S = 0_S,
so rx∈ker(f). Similarly xr∈ker(f). Thus ker(f) ◁R.
The fact that f is injective if and only if ker(f) = {0_R}we saw in both linear algebra and group theory. We repeat that argument here.
Assume f is injective. Let r ∈ker(f), we see that f(r) = 0_S = f(0_R) so by
injectivity of f we must have r= 0_R. For the converse, assume that the kernel of
f is {0_R}and that f(r_1) = f(r_2). We see that 0S = f(r_1)−f(r_2) = f(r_1−r_2), so
that r_1−r_2 ∈ker(f) = {0_R}. Thus r_1 = r_2 so f is injective.

Question 5

Isomorphism Theorems pf of 1 on exam !!!!!

Answer 5

Photo on phone pf of 1

Question 6

Correspondence theorem - pf on exam !!!!!!

Answer 6

Let R be a ring and let I ◁R be an ideal.
Then there is a bijective order-preserving correspondence S ↔S/I between subrings
of R containing I and subrings of R/I, and likewise J ↔J/I between ideals of R
containing I and ideals of R/I. pf on phone

Question 7

Definition: Integral Domain

Answer 7

An integral domain is a commutative ring
R<> {0}with the property that
xy= 0 = ⇒ x= 0 or y= 0 (x,y∈R).
An element x ∈R such that xy = 0 or yx = 0 for some y<> 0 is called a zero divisor. Therefore an integral domain is a commutative ring without non-zero zero divisors.

Question 8

Wedderburn’s little theorem pf omitted

Answer 8

Any finite domain is a field

Question 9

Definition : Maximal ideal

Answer 9

Let R be a ring. A proper ideal I◁R is maximal
if for any ideal J ◁R such that I ⊆J ⊆R we have J= I or J= R.

Question 10

Definition : Prime Ideal

Answer 10

Let R be a commutative ring. A proper ideal I ◁R
is prime if, for all x,y∈R,
xy∈I= ⇒ x∈I or y∈I.

Question 11

Maximal and prime proposition pf on exam !!!!!!!

Answer 11

Let R be a commutative ring and let I ◁R be a proper ideal.
(1) I is maximal if and only if R/I is a field.
(2) I is prime if and only if R/I is an integral domain.
pic of pf on phone

Question 12

Every maximal ideal is a prime ideal

Answer 12

Let I ◁ R be a maximal ideal. Then R/I is a field, and in particular an integral domain. Therefore I is a prime ideal.

Question 13

Definition : Simple ring

Answer 13

A ring is simple if it has no non-zero proper ideals.