Question

Prove that the map
\[Z_2\rightarrow Z_2\\[r]\rightarrow [r]^2\]
for \(r\in Z\), is a ring homomorphism
Please, help

Answer 1

Now what?
hahah

Answer 2

Hmm... it's not showing up. I guess you mean
\[\Large \left.\begin{matrix} \mathbb{Z}_2 &\rightarrow & \mathbb{Z}_2\\ r & \mapsto & r^2\end{matrix}\right.\]

Answer 3

Is that it?

Answer 4

[r] not just r

Answer 5

hahah

Answer 6

congruence class

Answer 7

Oh...

Answer 8

Let me see if I can learn what in the blazes a congruence class in within a minute.
That is, if you don't educate me first, LOSER :P

(haha... peace ^^)

Answer 9

\[Z_2 =\{[0],[1]\}\]

Answer 10

So what is a bloody congruence class, then? :D

Answer 11

\[\Large \left.\begin{matrix} \mathbb{Z}_2 &\rightarrow & \mathbb{Z}_2\\ [r] & \mapsto & [r]^2\end{matrix}\right.\]

Answer 12

in Z_2, [r]^2= Z_2 itself

Answer 13

Yeah, but it would help if you tell me what [r] really means, you know, its mathematical definition?

Answer 14

[r]= {km +a | k, m in Z, 0<= a<m}

Answer 15

and a = r (mod m)

Answer 16

for example in mod 2, [1]= {1,3,5,......}

Answer 17

because 2|3 remain 1
like 2|5 and so on
so that [r] = {1,3,5,......}

Answer 18

What's this?

Answer 19

Okay, so the gist of it is, in mod 2, which is in Z_2, [1] is all the numbers that give a remainder of 1 when divided by 2?

Answer 20

yes

Answer 21

@EngrChong that is group theory

Answer 22

I dont understand anything hahaha

Answer 23

Okay.... and what does [r]^2 mean?

Answer 24

mean {1^2,3^2,5^2......} but in mod 2, it is Z_2

Answer 25

because 3^2=9 and 2|9 remain 1 also

Answer 26

So [r] and [r]^2 are sets, yes?

Answer 27

[r]^2 = [r][r]=[r^2]

Answer 28

yes,

Answer 29

And how do you define [r] + [s] for instance?
And [r][s] ?

Answer 30

See, the way I was taught, Z_2 is simply the set {0,1}

Answer 31

[r]+[s]=[r+s]

Answer 32

but it is not equal r +s, right?

Answer 33

for example: [1] + [0]= [1+0]=[1] ,
but [1]\(\neq \)1 , [1] is a set of numbers whose remainder is 1 in mod2, 1 is just one of its element.

Answer 34

Oh, I see, since you study group theory while my course is abstract algebra.

Answer 35

So the members of Z_2 are sets?

Answer 36

yes

Answer 37

That's weird

Answer 38

Let's see...
Let [r] and [s] be elements of Z_2

Answer 39

f([r] + [s]) = f([r+s]) = [r+s]^2

Answer 40

(I still don't know what I'm doing with this, mind you)

Answer 41

Now, \[\Large [r+s]^2 = \left\{x^2 \quad | \quad x\equiv(r+s)(\mod 2)\right\}\]
right?

Answer 42

yes

Answer 43

for you to see what the definition is.

Answer 44

they always define Z_n elements are [ ] congruence class of remainders, not just remainders

Answer 45

Wait... backtrack...\[\Large [r+s]^2 =[(r+s)^2]= \left\{x \quad | \quad x\equiv(r+s)^2(\mod 2)\right\}\]

Answer 46

I think this is better

Answer 47

another proof :) at my mark

Answer 48

ok, next?

Answer 49

Okay, I can sort of see where this is going.... now, we'll examine
\[\Large [r]^2 + [s]^2 = [r^2 + s^2]\]

Answer 50

\[\Large [r^2+s^2]=\left\{x \quad | \quad x\equiv(r^2+s^2)(\mod 2)\right\}\]

Answer 51

it's true in this case only (mod 2)

Answer 52

It's always true. I haven't put in the specific mod 2 case yet.

Answer 53

I will in the next step. We can add any even integer to this bit, with no problem, right?
\[\Large [r^2+s^2]=\left\{x \quad | \quad x\equiv(\color{blue}{r^2+s^2})(\mod 2)\right\}\]

Answer 54

Because...
a = b(mod 2)
is equivalent to
a = (b+2k)(mod 2), for any integer k.

Answer 55

ok

Answer 56

So why don't we add 2rs? :P
\[\Large [r^2+s^2]=\left\{x \quad | \quad x\equiv(r^2\color{red}{+2rs}+s^2)(\mod 2)\right\}\]

Answer 57

Thus...
\[\Large [r^2+s^2]=\left\{x \quad | \quad x\equiv(r+s)^2(\mod 2)\right\}=[r+s]^2\]

Answer 58

because in mod 2 , 2|2rs so 2rs\(\equiv\) 0

Answer 59

Exactly.
So we have established that
\[\Large f\left([r]+[s]\right)=[r+s]^2 = [r^2+s^2]=f([r]) + f([s])\]
Right?
That's one bit of the ring homomorphism down haha

Answer 60

The other part is, I think, easier.

Answer 61

The first one is proving it is a ring, then this stuff, right?
Ok, got you, thanks a lot. You are...extremely good. hehehe.. I didn't post the problem until I saw you online. Thank you.

Answer 62

Again, with [r] and [s] being elements of Z_2, let's have a look at [r][s]
\[\Large f([r][s]) = f([rs]) = [rs]^2\]\[\Large [rs]^2 = [(rs)^2] = [r^2s^2] = [r^2][s^2]=[r]^2[s]^2 = f([r])f([s])\]

Whew :P

Answer 63

First one was proving that it was a group homomorphism. This next one proves that it is also a ring homomorphism.

Answer 64

:) Still have time? I need you help on other one. But it is ok if you don't have time. :)

Answer 65

Show it to me. If I don't have time, I'll just give you the best hint that I can :)

Answer 66

Let P denote the multiplicative group of positive real numbers. Prove that {1} is the ONLY finite subgroup of P.

Answer 67

multiplicative group?

Answer 68

group under multiplication operator

Answer 69

example?

Answer 70

for example:
\[Z_8 = \{[1].[3],[5],[7]\}\]
all elements 1,3,5,7 are relative prime with 8

Answer 71

I still don't quite get what P is... let me try to see

Answer 72

Let me give you the lattice of subgroup, it may trig something

Answer 73

o.O

Answer 74

I don't guarantee I'll be able to help with this one... o.O

Answer 75

It's perfectly ok, friend. I do appreciate your patience. :) I will ask my T.A tomorrow.
Thanks for the help.