Question

Which of the following functions re homomorphisms and why?
a. f: Z to Z defined by f(x) = -x
b. f: Z2 to Z2 defined by f(x)= -x
c. g: Q to Q defined y g(x) = 1/ (x^2 + 1)
d. h: R to M(R) defined by h(a) = [ (-a, 0) (a, 0)]
e. f: Z12 to Z4 defined by f([x]12) = [x]4 where [u]n denotes the congruence class of the integer u in Zn.

Answer 1

Group theory again :D
You know what a homomorphism is?

Answer 2

for rings of course you need to check two conditions,
\[f(a+b)=f(a)+f(b)\] and \(f(ab)=f(a)f(b)\) so it is a little more stringent

Answer 3

oh actually it did not say rings did it? i assumed that from the last problem
are you viewing them as groups or rings?

Answer 4

I'm sure they are rings

Answer 5

You made me reconsider :/
Until the poster makes a specification, I'll be ambiguous :)


Let's start with the first one, because I think it's the easiest. :D
Let a and b be integers.
Consider
f(a+b) = -a-b = (-a) + (-b) = f(a) + f(b)
Therefore, it's a GROUP homomorphism :)
Let's check multiplication...
consider
f(ab) = -ab
Hmmm....
Now check
f(a)f(b) = (-a)(-b) = ab
is ab = -ab ?
Only when one of them is zero.
If none of them is zero, then ab is not equal to -ab
therefore
f(ab) is not always equal to f(a)f(b)
Therefore, it's NOT a RING homomorphism :P

Answer 6

second is even easier
in \(\mathbb{Z_2}\) there are only two elements, so \(f(x)=-x\) is the identity map

Answer 7

Okay, @satellite73
second is easier :)
But first is funner :D

Answer 8

(more fun?)
So, @urbanderivative
There's your template for proving... take two arbitrary elements of the domain, and check if the mapping preserves both addition and multiplication :)

Answer 9

off the top of my head i would say the third is not because a ring homo has to send 0 to 0

Answer 10

Fourth sends a real number to a matrix?

Answer 11

yeah, thats a matrix

Answer 12

\[\huge h: \mathbb{R} \rightarrow M_\mathbb{R}\]\[\huge a \rightarrow \left[\begin{matrix}-a & 0 \\ a & 0\end{matrix}\right]\]
this mapping?

Answer 13

yes thats correct, I'm thinking it's a homomorphism, but I'm wondering if you can disprove it

Answer 14

One sec... matrix multiplication makes me dizzy :D

Answer 15

that's what I'm saying!

Answer 16

Okay, clearly, it preserves addition, right?

Answer 17

yes

Answer 18

Not convinced? :D
Let a and b be real numbers. Then
\[\huge f(a+b) = \left[\begin{matrix}-a-b & 0 \\ a+b & 0\end{matrix}\right]=\left[\begin{matrix}-a & 0 \\ a & 0\end{matrix}\right]+\left[\begin{matrix}-b & 0 \\ b & 0\end{matrix}\right]\]

Answer 19

I'm pretty sure it* preserve multiplication, but I'm not sure on my matrix math

Answer 20

:/
replace f by h...

Answer 21

Okay, let's see... consider h(ab) first...
\[\huge h(ab)=\left[\begin{matrix}-ab & 0 \\ ab & 0\end{matrix}\right]\]

Answer 22

Now, let's have a look at h(a)h(b)
\[\huge h(a)h(b)=\left[\begin{matrix}-a & 0 \\ a & 0\end{matrix}\right]\left[\begin{matrix}-b & 0 \\ b & 0\end{matrix}\right]\]

Answer 23

And it's equal to
\[\huge \left[\begin{matrix}ab & 0 \\ -ab & 0\end{matrix}\right]\]

Answer 24

And once again, h(ab) = h(a)h(b)
if and only if a = 0 or b = 0
any other case, they won't be equal

Then, it wasn't a ring homomorphism after all :D

Answer 25

Thank you! can you show me the equation for matrix math

Answer 26

Equation for multiplication of matrices?

Answer 27

yes

Answer 28

I'm really not good with matrices, let alone explaining how the multiplication works, but here's the formula for 2x2 matrices :)
\[\large \left[\begin{matrix}a & b \\ c & d\end{matrix}\right]\left[\begin{matrix}p & q \\ r & s\end{matrix}\right]=\left[\begin{matrix}ap+br & aq+bs \\ cp+dr & cq+ds\end{matrix}\right]\]