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

[Math Amateur]

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 ... ... ...

Exercise 1 reads as follows:

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

 

[fresh_42]

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]

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]

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 $f(1\cdot z)$.

 

[Math Amateur]

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 ...

Following your suggestion ... ... 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]

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

From your equation.

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]

From your equation.

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]

(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]

AB

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]

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