# Ring Homomorphisms from Z to Z ... Lovett, Ex. 1, Section 5.

Gold Member

## Homework Statement

I am reading Stephen Lovett's book, "Abstract Algebra: Structures and Applications" and am currently focused on Section 5.4 Ring Homomorphisms ...

I need some help with Exercise 1 of Section 5.4 ... ... ...

## Homework Equations

The relevant definition in this case is as follows:

## The Attempt at a Solution

Thoughts so far ... ...

One ring homomorphism ##f_1 \ : \ \mathbb{Z} \rightarrow \mathbb{Z}## would be the Zero Homomorphism defined by ##f_1(r) = 0 \ \forall r \in \mathbb{Z}## ...

(##f_1## is clearly a homomorphism ... )

Another ring homomorphism ##f_2 \ : \ \mathbb{Z} \rightarrow \mathbb{Z}## would be the Identity Homomorphism defined by ##f_2(r) = r \ \forall r \in \mathbb{Z}## ...

(##f_2## is clearly a homomorphism ... )

Now presumably ... ... ??? ... ... ##f_1## and ##f_2## are the only ring homomorphisms from ##\mathbb{Z} \rightarrow \mathbb{Z}## ... ... but how do we formally and rigorously show that there are no further homomorphisms ... ...

Hope that someone can help ...

Peter

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fresh_42
Mentor
I think the crucial part is the image of ##\pm 1## under a ring homomorphism ##f##. Try to investigate whether it can be other than ##0## or ##\pm 1##. If not, then you have to show, that either ##f=f_1=0## or ##f=f_2=id_\mathbb{Z}##.
Assume you have ##f## and play with the properties in 5.4.1 to get some requirements for ##f##.

Math Amateur
Gold Member
Hi fresh_42 ...

Tried the following but no luck so far ...

Suppose ##\exists \ f_3 \ : \ \mathbb{Z} \rightarrow \mathbb{Z}## ... ... try to show no such ##f_3## different from ##f_1 , f_2## exists ... ...

Let ##f_3(2) = x## where ##x \in \mathbb{Z}## ...

Then ##f_3(4) = f_3( 2 \cdot 2 ) = f_3( 2) f_3( 2) = x^2##

and

##f_3(4) = f_3( 2 + 2) = f_3( 2) + f_3( 2) = x + x = 2x##

Then we must have ##x^2 = 2x## ... ... ... (1)

I was hoping that there would be no integer solution to (1) ... but ##x = 0## satisfies ... so ... problems ..

Was that the kind of approach you were suggesting ... ?

Peter

fresh_42
Mentor
Yes, only that I would have used ##1## instead of ##2##.
You have now ##f_3(2) \in \{0,2\}##. But what is ##f_3(y)## in those two cases? As ##\pm 1## are the only units, it might help to look at ##f(1\cdot z)##.

Gold Member
Hi fresh_42 ...

Tried the following but no luck so far ...

Suppose ##\exists \ f_3 \ : \ \mathbb{Z} \rightarrow \mathbb{Z}## ... ... try to show no such ##f_3## different from ##f_1 , f_2## exists ... ...

Let ##f_3(2) = x## where ##x \in \mathbb{Z}## ...

Then ##f_3(4) = f_3( 2 \cdot 2 ) = f_3( 2) f_3( 2) = x^2##

and

##f_3(4) = f_3( 2 + 2) = f_3( 2) + f_3( 2) = x + x = 2x##

Then we must have ##x^2 = 2x## ... ... ... (1)

I was hoping that there would be no integer solution to (1) ... but ##x = 0## satisfies ... so ... problems ..

Was that the kind of approach you were suggesting ... ?

Peter

============================================================================

OK thanks fresh_42 ...

Let ##f_3(1) = x## where ##x \in \mathbb{Z}## ...

##f_3(2) = f_3( 1 + 1) = f_3( 1) + f_3( 1) = x + x = 2x##

and we have ##f_3(2) = f_3( 2 \cdot 1 ) = f_3( 2) f_3( 1) = 2x \cdot x = 2x^2##

Then we must have ##2x^2 = 2x## ... ... ... (1)

That is, ##x^2 = x## where ##x \in \mathbb{Z}## ... ... so ##x = 0## or ##x = 1## ...

BUT ... where to from here ...

Note ... can you explain how you got ##f_3(2) in \{ 0, 2\}## ... ???

Hope you can help ...

Peter

fresh_42
Mentor
Note ... can you explain how you got ##f_3(2) \in \{ 0, 2\}## ... ???
Then we must have ##x^2 = 2x## ... ... ... (1)
This means ##x^2-2x=x(x-2)=0## and since ##\mathbb{Z}## is an integral domain, if follows that either ##x=0## or ##x=2##.
Let ##f_3(1) = x## where ##x \in \mathbb{Z}## ...

##f_3(2) = f_3( 1 + 1) = f_3( 1) + f_3( 1) = x + x = 2x##

and we have ##f_3(2) = f_3( 2 \cdot 1 ) = f_3( 2) f_3( 1) = 2x \cdot x = 2x^2##

Then we must have ##2x^2 = 2x## ... ... ... (1)

That is, ##x^2 = x## where ##x \in \mathbb{Z}## ... ... so ##x = 0## or ##x = 1## ...
I think ##0=f(1 \cdot a)-f(a)=f(a)(f(1)-1)## would have been shorter, but o.k. So we have ##f(1)=x \in \{0,1\}##.
Make two cases now: a) ##f(1)=0## and b) ##f(1)=1##. What do you get for an arbitrary ##f(z)## in those cases?

Math Amateur
Gold Member

This means ##x^2-2x=x(x-2)=0## and since ##\mathbb{Z}## is an integral domain, if follows that either ##x=0## or ##x=2##.

I think ##0=f(1 \cdot a)-f(a)=f(a)(f(1)-1)## would have been shorter, but o.k. So we have ##f(1)=x \in \{0,1\}##.
Make two cases now: a) ##f(1)=0## and b) ##f(1)=1##. What do you get for an arbitrary ##f(z)## in those cases?

Thanks fresh_42 ... all clear now ...

We have for a ring homomorphism ##f \ : \ \mathbb{Z} \rightarrow \mathbb{Z}##

... either ...

(a) ##f(1) = 0##

or

(b) ##f(1) = 1##

Now it is clear that if ##f## is a ring homomorphism such that ##f(1) = 0## ...

then we must have ##f(r) = 0 \ \forall \ r \in \mathbb{Z}## ...

and ...

... if ##f## is a ring homomorphism such that ##f(1) = 1## ...

...
then we must have ##f(r) = r \ \forall \ r \in \mathbb{Z}## ...

Therefore there are only two ring homomorphisms ##f \ : \ \mathbb{Z} \rightarrow \mathbb{Z}##

Thanks for all the help ...

Peter

fresh_42
Mentor
(a) ##f(1) = 0##
A
then we must have ##f(r) = 0 \ \forall \ r \in \mathbb{Z}## ...
(b) ##f(1) = 1##
B
then we must have ##f(r) = r \ \forall \ r \in \mathbb{Z}## ...
Yes, but which is the equation placed at A and B which supports the conclusion? What is the step from ##f(1)## to ##f(r)##?

Math Amateur
Gold Member
A

B

Yes, but which is the equation placed at A and B which supports the conclusion? What is the step from ##f(1)## to ##f(r)##?
=================================================================================================

Hi fresh_42 ...

So, I will do the "proof" informally ... to do it formally we would simply use mathematical induction ... but the informal proof demonstrates what is going on ...

Case A

For a ring homomorphism ##f \ : \ \mathbb{Z} \rightarrow \mathbb{Z}## we have that f(1) = 0

We claim that this implies that ##f(r) = 0 \ \forall \ r \in \mathbb{Z}## ...

'Proof'

We have ##f(1) = 0##

##\Longrightarrow f(2) = f(1) + f(1) = 0 + 0 = 0##

##\Longrightarrow f(3) = f(2) + f(1) = 0 + 0 = 0##

... and so on ...

So for all positive integers ##r##, we have ##f(r) = 0## ... (could have done this formally via mathematical induction)

For the negative integers we proceed as follows:

Given that ##f## is a ring homomorphism we have that ##f(-1) = -f(1)## ...

... so ##f(-1) = -f(1) = -0 = 0##

##\Longrightarrow f(-2) = f( (-1) + (-1)) = f(-1) + f(-1) = 0 + 0 = 0##

##\Longrightarrow f(-3) = f( (-2) + (-1)) = f(-2) + f(-1) = 0 + 0 = 0##

and so on ...

... so for all the negative integers ##s = -r## we have ##f(s) = 0##

.. and ...

we also have that, given that ##f## is a ring homomorphism, ##f(0) = 0##

So we have for all integers ##r, f(r) = 0##

Case B

For a ring homomorphism ##f \ : \ \mathbb{Z} \rightarrow \mathbb{Z}## we have that f(1) = 1

We claim that this implies that ##f(r) = r \ \forall \ r \in \mathbb{Z}## ...

'Proof'
We have ##f(1) = 1##

##\Longrightarrow f(2) = f(1) + f(1) = 1 + 1 = 2##

##\Longrightarrow f(3) = f(2) + f(1) = 2 + 1 = 3##

... and so on ...

So for all positive integers r, we have f(r) = r ... (could have done this formally via mathematical induction)

For the negative integers we proceed as follows:

Given that ##f## is a ring homomorphism we have that ##f(-1) = -f(1)## ...

... so ##f(-1) = -f(1) = -1##

##\Longrightarrow f(-2) = f( (-1) + (-1)) = f(-1) + f(-1) = (-1) + (- 1) = -2##

##\Longrightarrow f(-3) = f( (-2) + (-1)) = f(-2) + f(-1) = (-2) + (- 1) = -3##

and so on ...

... so for all the negative integers ##s = -r## we have ##f(s) = s##

... and ...

we also have that, given that f is a ring homomorphism, ##f(0) = 0##

So we have for all integers ##r, f(r) = r##

Hope that is what was required ...

Peter

fresh_42
Mentor
Yes, that's fine. You could have also used a general "##n##" and some "##\ldots ##" with the exact same arguments.

Math Amateur