User:J D Bowen/Math735 HW3

1) Let $F \ $ be a field of finite order $q \ $, and $n\in\mathbb{Z}_+ \ $.

Observe that the determinant is a homomorphism $\Delta:GL_n(F)\to F \backslash \left\{{0}\right\} \ $. Then by the corollary to the first isomorphism theorem,

$[GL_n(F):\text{ker}(\Delta)] = |\Delta(GL_n(F))| \ $.

But $\text{ker}(\Delta) = SL_n(F) \ $ and $\Delta(GL_n(F))=F\backslash\left\{{0}\right\} \ $.

So $[GL_n(F):SL_n(F)]=q-1 \ $.

'3.3.4) Clearly $C\times D < A \times B \ $. We have $(a,b)(C,D)=(aC,bD)=(Ca,Db)=(C,D)(a,b) \ $.

Define $\phi:(A\times B)/(C\times D) \to A/C \times B/D \ $ as $\overline{(a,b)}\mapsto (\overline{a},\overline{b}) \ $.

Then $\phi(\overline{(a,b)}\cdot\overline{(x,y)} = \phi(\overline{(ax,by)}=(\overline{ax},\overline{by})=(\overline{a},\overline{b})\cdot(\overline{x},\overline{y}) = \phi(\overline{(a,b)})\cdot\phi(\overline{(x,y)}) \ $, so $\phi \ $ is a homomorphism.

We also have $\text{ker}(\phi) = \left\{{ (a,b)(C\times D) : \phi((a,b)(C\times D)) = 1 \cdot C\times D }\right\} \ $

$=\left\{{ (a,b)(C\times D) : aC\times bD = 1 \cdot C\times D }\right\} = \left\{{(A,b)(C\times D): (a,b)\in C\times D }\right\} = C\times D \ $.

7.1.29) Let $R=\left\{{\phi:A_+\to A_+: \phi \ \text{is hom.} }\right\} \ $.

We easily see for $\alpha,\beta\in R, \alpha+\beta=\beta+\alpha\in R \ $. We also have the map $0_A \ $ defined $x\mapsto 0 \ $, for an identity element. Since composition is associative, and compositions of homomorphisms is another homomorphism, composition and pointwise addition give $R \ $ a ring structure.

Suppose $\phi\in U(R) \ $. Then $\phi:A_+\to A_+ \ $ is a homomorphism and so is $\phi^{-1}:A_+\to A_+ \ $ is also homomorphism. Therefore, $\phi \ $ is one-to-one, which implies $\phi \ $ is an automorphism.

Now suppose $\phi\in\text{Aut}(A) \ $. Then $\phi \ $ is 1-1 and a homomorphism. So there is a homomorphism $\phi^{-1} \ $ such that the composition is the identity map. Then $\phi\in U(R) \ $.

So $U(R)=\text{Aut}(A) \ $.

PROOF PHI INVERSE IS HOM:

Since $\phi \ $ is 1-1, exists unique inverse $\phi^{-1} \ $.

$a+b= \ $

$\phi(\phi^{-1}(a)) + \phi(\phi^{-1}(b)) = \ $

$\phi(\phi^{-1}(a)+\phi^{-1}(b)) \ $, since $\phi \ $ is a homomorphism.

Now we can $\phi^{-1} \ $ both sides of this, to get

$\phi^{-1}(a+b) = \phi^{-1}(a)+\phi^{-1}(b) \ $

so the inverse is a homomorphism, too.

3.1.11) Let F be a field and let G = {

A) Prove the map ϕ: → a is surjective and a homomorphism from G to Fx.


 * a, b, c ∈ F, ac ≠ 0} ⊆ GL2(F).

Let a ∈ Fx. Then ∃ a surjective.

in G such that ϕ:

= a. Thus ϕ is

Now, Let

∈ G. Then ϕ:

,

= a, and ϕ:

definition of ϕ. So ϕ: [

ad= ϕ:

The Kernel of ϕ is { element in F x.

]=ϕ(

= a*d. And


 * ϕ:

. Thus ϕ is a homomorphism.

∈ G | b, c ∈ F}, since 1 is the neutral

The Fibers are cosets where ∀ a ∈ Fx, aK =

B) Prove the map ψ: → (a,c) is surjective and a homomorphism from G to Fx X Fx.

Let a, c ∈ Fx. Then ∃ a is surjective.

for some b, c ∈ F.

in G such that ψ:

= (a,c). Thus ψ

Now, let

(d, f). So ψ: [

∈ G. Then ψ:

,

] = ψ:

= (a, c) and ψ:

= (ad, cf) = (a, c)(d, f) and

(a, c)(d, f) = ψ:

The Kernel of ψ is {

F x.

ψ:

. Thus ψ is a homomorphism.


 * ∀ b ∈ F}, since 1 is the neutral element in

The Fibers are cosets where ∀ a, c ∈ Fx, aK =

C) Let H={ We want to show that H is a homomorphism and is bijective.
 * b ∈ F}. Prove H is isomorphic to the additive group F.

Define ψ:

for some b ∈ F.

→ b. Let b ∈ F. Then ∃ a

in G such that

ψ:

= b. Thus ψ is surjective.

Let

,

∈ H. Then ψ:

= b and ψ:

= c. So,

ψ: [

b + c = ψ: + ψ: . Thus ψ is a homomorphism. Now we want to show that H is injective. We want to show that if

] = ψ:

= c + b = b + c, since F is a field. And

ψ:

= ψ:

that b = c. It follows that since b = c, then = . Thus ψ is injective. Therefore, H is isomorphic to the additive group F.

= ψ:

, and since ψ:

, then

=

. So we have, ψ:

= b and ψ:

= c, we can say

7) To show a subset of a ring is a subring, it suffices to show it is nonempty and closed under subtraction and multiplication.

Let $x,y\in\text{Cen}(R) \ $. Then $\forall r \in R, xr=rx, yr=ry \ $. Then $(x-y)r=xr-yr=rx-ry=r(x-y) \implies (x-y)\in\text{Cen}(R) \ $.

Next, note that $(xy)r=x(yr)=x(ry)=(xr)y=(rx)y=r(xy) \implies xy\in\text{Cen}(R) \ $.

So, $\text{Cen}(R) \ $ is a subring of $R \ $.

Suppose $R \ $ is a division ring with identity. We know $\text{Cen}(R)$ is a commutative ring containing 0 and 1. To show it is a field, we need only show it is a division ring. So suppose $a\in\text{Cen}(R) \ $.

Then $\forall r\in R, \ raa^{-1}=aa^{-1}r \implies ara^{-1}=aa^{-1}r\implies ra^{-1}=a^{-1}r\implies a^{-1}\in \text{Cen}(R) \ $.

So $\text{Cen}(R) \ $ is a commutative division ring with identity, aka a field.