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

[RJLiberator]

Homework Statement

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

Homework Equations

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.

The way I am going about this is the following:

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.

 

[fresh_42]

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.

 

[WWGD]

Homework Statement

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

Homework Equations

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? .

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

 

[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.

 

[fresh_42]

This seems like I am assuming that which I want to prove.

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.

 

[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.

 

[fresh_42]

$T(r)=$ (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 $0_S = x = T(0_R)$.

 

[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.

 

[fresh_42]

The term additive neutral element is another term for "zero" ?

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.

 

[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.

 

[fresh_42]

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.

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.

 

[RJLiberator]

Ohhh, so for B) perhaps I could start with:

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

 

[fresh_42]

Ohhh, so for B) perhaps I could start with:

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

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.

 

[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

 

[fresh_42]

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

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

 

[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?

 

[fresh_42]

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?

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 $T(-n)=T(-1+(-1)+...+(-1))=...$