# Homework Help: Simple Abstract Algebra Proof: T(0_r) = 0_s

1. Feb 13, 2016

### RJLiberator

1. The problem statement, all variables and given/known data

Let T:R-> S be a homomorphism of rings. Show that T(0_r) = 0_s.
2. Relevant equations

3. The attempt at a solution

First off, the terminology used is kinda confusing. I take 0_r to be the zero in R. Is this correct? For some reason I recall my teacher quickly saying that it was the additive inverse or something. Perhaps I heard wrong, as that makes no sense.

Since T:R->S we take the 0 in r and map it to S.
I mean, is there much more to say? By the definition of homomorphisms, we are mapping a 0 in R to S. If we mapped (1-1) (assuming 1 is in R) to S it would be still be 0.

2. Feb 13, 2016

### Staff: Mentor

You likely are supposed to derive it from the definition of $0$ as neutral additive element and the definition of ring homomorphisms.
Normally $T(0_R) = 0_S$ isn't part of the definition but can be derived.

3. Feb 13, 2016

### WWGD

Be careful, remember there are two operations in a ring. And remember that $0_R+ 0_R= 0_R$ and that
$T(a+b)=....$

4. Feb 13, 2016

### RJLiberator

Let's try some things out then.

T(0_R+0_R) = T(0_R)+T(0_R) = 0_s+0_S = 0_S

Is, that how I show it? This type of proof is quite confusing to me as it is seemingly blatantly obvious.

We have those two operations, I use the operation of addition to prove that 0+0 = 0, but in this proof I assumed that T(0_r) = 0_s, so this actually can't be right...

Let me retry something:

Since 0_r is the 0 in R, we see that
R(0_R) = 0.

But this doesn't quite make sense or get me anywhere.

T(0_R) = 0 ? Is it safe to say this? This seems like I am assuming that which I want to prove.

5. Feb 13, 2016

### Staff: Mentor

It does not only seem so - it is so.
You have $T(r)=T(r+0_R)=T(r)+T(0_R)$. You don't know what $T(0_R)$ actually is but the equation should give you a strong hint, considering how $0_S$ is defined.

6. Feb 13, 2016

### RJLiberator

Since the equation states that this is a homomorphism of rings, can we state that the properties in R hold in S which means that :

Well, here is where I am confused.

T(r) = S(r_R)

T(r+0) = S(r_R) + S(0_r)

But, we can't say much can we?

It's almost like I just want to say the zero property holds based on the definition of homomorphism.

7. Feb 13, 2016

### Staff: Mentor

$T(r)=$ (by definition of $0_R$)
$T(r+0_R)=$ (by definition of a ring homomorphism)
$T(r)+T(0_R)$.

Setting $x=T(0_R) \in S$ for the term we do not know and want to calculate we then have:
$T(r) = T(r) + x$.

This means $x$ is an additive neutral element in $S$.
But there is only one neutral element for the addition in $S$ because it is an additive group.
Therefore $x$ must equal $0_S$. Replacing $x$ again we have $0_S = x = T(0_R)$.

8. Feb 13, 2016

### RJLiberator

Okay, every step of your proof made crystal clear sense to me.

This means, my problem was in the terminology of the question.

The term additive neutral element is another term for "zero" ?
This is where I must have heard my professor mention it, then.

9. Feb 13, 2016

### Staff: Mentor

It's the definition of zero, the element that does nothing if added.
I admit it's all about holding different things apart and being careful on definitions. If you like to exercise you can prove from the group laws why there can only be one zero. It's done similar.

10. Feb 13, 2016

### RJLiberator

well, perhaps I can quickly check my understanding with this similar question:

Let T: ℤ -> ℤ be a homomorphism of rings.
a) Show that either T(1) = 0 or T(1) = 1.
b) Conclude that either T(n) = 0, for all n ∈ℤ, or T is the identity map.

PF a) T(1)*T(1) = T(1*1) = T(1)
Therefore T(1)*T(1) - T(1) = 0
Distributive: (T(1)-1)*T(1) = 0
Either T(1) = 0 or T(1) = 1 to satisfy this solution.

PF b) Here is more difficulty for me.
Would I be able to stay that
If T(1) = 0 then this will force T(n) = 0.
or if T(1) = 1 then we know that T is the identity map.

Would the proof be along those lines? Breaking it up into 2 cases.

11. Feb 14, 2016

### Staff: Mentor

Yep. Part a) is ok. Perhaps you should mention that you use the fact that ℤ is an integral domain, i.e. $a \cdot b = 0 ⇒ a = 0 ∨ b = 0$.
This won't be true, e.g. in $ℤ_{12}$ the hours on a clock.
Part b) has the correct outline. Remember that $n=1+...+1=n \cdot 1$ to calculate $T(n)$ in both cases.

12. Feb 14, 2016

### RJLiberator

T(n) = 0
If we add T(n) "n" times to the left and add 0 "n" times to the right we get:
n*T(n) = 0

Now, either n = 0 or T(n) = 0
if n = 0 then we have the identity map, if T(n) = 0 then we can conclude it as such.

Correct? :D

13. Feb 14, 2016

### Staff: Mentor

You don't know $T(n)$ yet.
You have $T(1)=0$, i.e. $T(n)=T(1+...+1)=T(1)+....+T(1)=...$
or
you have $T(1)=1$, i.e. $T(n)=....$

Edit: If you want to be complete you will have to concern about negative $n$ as well.

14. Feb 14, 2016

### RJLiberator

Ah, indeed.

IF
T(1) = 0 then T(n) = T(1+...+1) = T(1)+...+T(1) = 0+...+0 = 0. So we can conclude any number n can be made up of 1's which means it would equal 0. (in essense, obviously a proper proof would look nicer)

OR
T(1) = 1 so T(n) = T(1+...+1) = T(1)+...+T(1) = 1+...+1 = n
And so we see the identity map :D

15. Feb 14, 2016

### Staff: Mentor

Yes. And now the negative $n$ (Start with $T(-1)$.)

16. Feb 14, 2016

### RJLiberator

We show that
T(-1) = T(-1)*T(1)

If T(1) = 0 then T(-1) = 0
If T(n) = n then T(1) =1 and T(-1) = -1 and thus it is -1.
Which works for all negative numbers.

Proper?

17. Feb 14, 2016

### Staff: Mentor

Halfway. If $T(1)=0$ then your equation shows indeed $T(-1)=0$. For $T(-n)$ do the same as before:
$T(-n)=T(-1+(-1)+...+(-1))=T(-1)+...+T(-1)=...$

In the case $T(1)=1$ you have $0=T(1+(-1))=T(1)+T(-1)=1+T(-1)$, i.e. $T(-1)=-1$ and $T(-n)=T(-1+(-1)+...+(-1))=....$