Units / Zero divisors in Comm Ring

[RJLiberator]

Homework Statement

Let F(ℝ) = {ƒ:ℝ->ℝ}
define (f+g)(x) = f(x)+g(x)
(f*g)(x) = f(x)*g(x)
F(ℝ) is a commutative ring.
ƒ_0(x) = 0 and ƒ_1(x) = 1

a) Describe all units and zero divisors
b) Find a function f such that ƒ≠ƒ_0, ƒ≠ƒ_1, and ƒ^2 = ƒ

Homework Equations

A unit is an element r ∈ R, which has a multiplicative inverse s∈R with r*s = 1.
A zero divisor is an element r ∈R such that there exists s∈R and rs = 0 (or sr=0).

The Attempt at a Solution

This is a review problem for my upcoming abstract algebra exam. The professor stated that this or the other problem he gave us, will be on the exam.
So I want to get this one down.

a) The definitions of units and zero divisors are obvious, but the way the problem is stated makes it difficult to understand.

So in this initial statement, the problem is saying that functions from ℝ sent to ℝ ?

Would the units be all functions such that ƒ_1(x) = 1 ? I mean, I guess I need to go into more detail then this, correct?

Are there any examples someone can give me of what the problem actually is saying?

The problem will likely be easy for me to solve if I know this element of it as I understand what the questions are asking / definitions.

 

[Samy_A]

Homework Statement

Let F(ℝ) = {ƒ:ℝ->ℝ}
define (f+g)(x) = f(x)+g(x)
(f*g)(x) = f(x)*g(x)
F(ℝ) is a commutative ring.
ƒ_0(x) = 0 and ƒ_1(x) = 1

a) Describe all units and zero divisors
b) Find a function f such that ƒ≠ƒ_0, ƒ≠ƒ_1, and ƒ^2 = ƒ

Homework Equations

A unit is an element r ∈ R, which has a multiplicative inverse s∈R with r*s = 1.
A zero divisor is an element r ∈R such that there exists s∈R and rs = 0 (or sr=0).

The Attempt at a Solution

This is a review problem for my upcoming abstract algebra exam. The professor stated that this or the other problem he gave us, will be on the exam.
So I want to get this one down.

a) The definitions of units and zero divisors are obvious, but the way the problem is stated makes it difficult to understand.

So in this initial statement, the problem is saying that functions from ℝ sent to ℝ ?

Would the units be all functions such that ƒ_1(x) = 1 ? I mean, I guess I need to go into more detail then this, correct?

Are there any examples someone can give me of what the problem actually is saying?

The problem will likely be easy for me to solve if I know this element of it as I understand what the questions are asking / definitions.

Yes, the ring $F(\mathbb R)$ consists of all functions from $\mathbb R$ to $\mathbb R$.
Addition and multiplication in $F(\mathbb R)$ are defined as pointwise addition and multiplication of the functions.

A unit is a function $f$ such that there exists a function $g$ satisfying $fg=f_1$. This means that for all $x \in \mathbb R:\ f(x)g(x)=f_1(x)=1$.

A zero divisor is a function $f$ such that there exists a function $g \neq f_0$ satisfying $fg=f_0$.
This means that for all $x \in \mathbb R:\ f(x)g(x)=f_0(x)=0$.

 

[RJLiberator]

Excellent, that helps out a lot.

So, to describe the unit and zero divisor:

A unit here means the inverse of a the function f(x) exists and is g(x).
A zero divisor here implies f(x) = 0. Since g(x) =/= 0, we need f(x) = 0.

Is that a safe answer to 'describe' those two elements?

For b) f(x)f(x) = f(x), but f(x) can't be equal to 0 or 1 here.
I'm not sure how this one works, still need some time to think this over.

 

[Samy_A]

Excellent, that helps out a lot.

So, to describe the unit and zero divisor:

A unit here means the inverse of a the function f(x) exists and is g(x).

What exactly do you mean by "inverse"? If by inverse you mean the inverse function (as in: if $f(x)=y$, then $g(y)=x$), then no, this is not correct.

A zero divisor here implies f(x) = 0. Since g(x) =/= 0, we need f(x) = 0.

No, there are more functions that are zero divisors.
Take as an example the following function $f$: $f(x)=0$ for $x\leq0$, $f(x)=1$ for $x>0$.
Define $g$ as follows: $g(x)=0$ for $x>0$, $g(x)=1$ for $x\leq0$.
What is $fg$?

For b) f(x)f(x) = f(x), but f(x) can't be equal to 0 or 1 here.
I'm not sure how this one works, still need some time to think this over.

$f(x)f(x)=f(x)$ implies $f(x)(f(x)-1)=0$. That should give you a clue.

 

[RJLiberator]

Ah, for part B, that makes sense.
Is it safe to say that from there, either f(x) = 0 or f(x) = 1 and therefore there is no function that exists since f(x) =/= 0,1 based on the question?

What exactly do you mean by "inverse"?

Well, in commutative rings we know that it is not necessary for multiplicative inverses to exist.
So I can't just say there there the inverse.

So, what I mean is f(x) = x and y(x) = x^(-1), is what I am trying to show/conclude.

No, there are more functions that are zero divisors.

I see. So it would be safe to say, that all such functions such that one of f(x) or g(x) is equal to 0.
The piecewise function you described is = 0 as for any x value, one of the corresponding parts is equal to 0.

 

[Samy_A]

EDIT: by mistake I used $L(\mathbb R)$ instead of $F(\mathbb R)$ in this post.

Ah, for part B, that makes sense.
Is it safe to say that from there, either f(x) = 0 or f(x) = 1 and therefore there is no function that exists since f(x) =/= 0,1 based on the question?

Indeed, either f(x)=0 or f(x)=1. But your conclusion is not correct. Of course there are such functions (for example the functions $f,\ g$ in my previous post).

Well, in commutative rings we know that it is not necessary for multiplicative inverses to exist.
So I can't just say there there the inverse.

So, what I mean is f(x) = x and y(x) = x^(-1), is what I am trying to show/conclude.

This is not very clear (although you may be correct).
The question is: what condition(s) must a function $f$ satisfy in order to have a multiplicative inverse in $L(\mathbb R)$?

I see. So it would be safe to say, that all such functions such that one of f(x) or g(x) is equal to 0.
The piecewise function you described is = 0 as for any x value, one of the corresponding parts is equal to 0.

So what condition(s) must $f$ satisfy in order that there is a $g\neq 0$ for which $fg=f_0$?

Maybe a general comment: you have to be careful in this kind of exercise to clearly make the distinction between ring elements ($f \in L(\mathbb R$)), and the value of such an element in a specific $x \in \mathbb R\ , f(x)$, a real number. In the same vein, $f_0$ and $0$ are different things, even though $f_0(x)=0$ for every $x \in \mathbb R$. Similarly, $f_1$ and $1$ are different, even though $f_1(x)=1$ for every $x \in \mathbb R$

 

[RJLiberator]

Indeed, either f(x)=0 or f(x)=1. But your conclusion is not correct.

Oh? But the question states that f =/= f_0 and f =/= f_1?
Is it because we defined f as a piecewise function that it does not fall under those definitions and thus, can be used?

The question is: what condition(s) must a function f

the conditions being f(g(x)) = 1 and g(f(x)) = 1, is what you are implying?

So what condition(s) must $f$ satisfy in order that there is a $g\neq 0$ for which $fg=f_0$?

That f must be equal to f_0, is what I am getting.

 

[Samy_A]

EDIT: by mistake I used $L(\mathbb R)$ instead of $F(\mathbb R)$ in this post.

Oh? But the question states that f =/= f_0 and f =/= f_1?
Is it because we defined f as a piecewise function that it does not fall under those definitions and thus, can be used?

Well, a piecewise defined function is a perfectly valid element of $L(\mathbb R)$, so as long as it is different from $f_0, f_1$, it would be a valid answer for part b)

the conditions being f(g(x)) = 1 and g(f(x)) = 1, is what you are implying?

No, that is what I feared. Ring multiplication in $L(\mathbb R)$ is just pointwise multiplication, not composition.
So no, what you apparently meant with inverse is not the correct answer to the question about units.

That f must be equal to f_0, is what I am getting.

But this is a wrong answer. There are a lot of zero divisors in $L(\mathbb R)$. I showed you two in post #4.

Maybe you should take the questions one at the time. That will make it easier for you to solve the exercise.

 

[RJLiberator]

So, essentially, most of this question boils down to an understanding of piecewise functions.

We can answer Part b with a piecewise function as you presented in post #4.
This piecewise function is not f_0 or f_1 but has the properties of such, and when you have two of them like you do in post 4, f(x) and g(x), then we can easily determine a situation where part b is answered.

For part a), finding units and zero divisors
A unit is one such that f(x)g(x) = 1
A zero divisor is one such that f(x)g(x) = 0.

For zero divisors, we conclude that we have a zero divisor when either f(x) or g(x) is equal to 0, and piecewise functions play a role here, but ultimately, one of these two elements need to equal 0.

A unit can be found when f(x)g(x) = 1 and so they are the inverse of one another. Not in the sense of composition, but in the sense of pointwise multiplication.

What does L(R) mean?

 

[Samy_A]

What does L(R) mean?

Oops, my mistake. Should have been $F(\mathbb R)$. Sorry.

Answer to the rest of your post follows.

 

[RJLiberator]

Yes, let's try one at a time.

Part a, zero divisors and units.

A zero divisor can be described such that f(x)*g(x) = 0. Here, we see that in all circumstances, either f(x) or g(x) must be the f_0 function (either or both).
A unit can be described such that f(x)*g(x) = 1. This implies that f(x) and g(x) are inverses of each other, in the sense that f(x)*g(x) = f(x)*f(x)^(-1) = 1.

Part b, find a function such that f^2 = f where f=/= f1 and f=/=f0.
We see f(x)*f(x) = f(x) ==> f(x)(f(x)-1) = 0
Here, f(x) must equal 1 or 0, and we found a function via post #4.
When x>= 0 f(x) = 1, when x< 0 f(x) = 0.
Here, f(x) is not the f1 function or the f0 function, but satisfies the requirements of the problem and is thus, the function we were looking for to answer the question.

 

[Samy_A]

Yes, let's try one at a time.

Part a, zero divisors and units.

A zero divisor can be described such that f(x)*g(x) = 0. Here, we see that in all circumstances, either f(x) or g(x) must be the f_0 function (either or both).
A unit can be described such that f(x)*g(x) = 1. This implies that f(x) and g(x) are inverses of each other, in the sense that f(x)*g(x) = f(x)*f(x)^(-1) = 1.

You are still not giving the conditions a function has to satisfy in order to be zero divisor or a unit.

I'll try it with examples: define two functions $f, \ g$ as follows:
$\forall x\in \mathbb R: f(x)=x²+1,\ g(x)=x+2$.
I claim that $f$ is a unit, and $g$ a zero divisor. Can you see why that is the case, and can you deduce the more general rule for units and zero divisors in $F(\mathbb R)$?

Part b, find a function such that f^2 = f where f=/= f1 and f=/=f0.
We see f(x)*f(x) = f(x) ==> f(x)(f(x)-1) = 0
Here, f(x) must equal 1 or 0, and we found a function via post #4.
When x>= 0 f(x) = 1, when x< 0 f(x) = 0.
Here, f(x) is not the f1 function or the f0 function, but satisfies the requirements of the problem and is thus, the function we were looking for to answer the question.

Correct. In fact, any function that takes only the values 0 and 1, but is different from $f_0,\ f_1$ is a correct answer.

 

[RJLiberator]

Excellent. So part b, I understand, we clearly have a function that works based on the requirements.

To analyze part a.
For g(x) = x+2 to be a 0 divisor, that means we have
g(x)*r(x) = 0
(x+2)*r(x) = 0.

We have x*r(x)+2*r(x) = 0 and so we have (x/2)*r(x) = -r(x)
Therefore r(x) is thus the f_0 function?

f(x) = x^2+1 is a unit is the claim.
f(x)*u(x) = 1
(x^2+1)*u(x) = 1

Let's see if we can work on the zero divisor first.

 

[Samy_A]

Excellent. So part b, I understand, we clearly have a function that works based on the requirements.

To analyze part a.
For g(x) = x+2 to be a 0 divisor, that means we have
g(x)*r(x) = 0
(x+2)*r(x) = 0.

We have x*r(x)+2*r(x) = 0 and so we have (x/2)*r(x) = -r(x)
Therefore r(x) is thus the f_0 function?

No, if $r$ must be $f_0$, then $g$ wouldn't be a zero divisor.
You can define $r$ as you like, as long as for every $x \in \mathbb R: (x+2)r(x)=0$ and $r \neq f_0$. Do you see how you can define such a function $r$?

 

[RJLiberator]

Is the trick here in the definitions?
We can't say r(x) is = f_0, but we can say r(x) is such that (x+2)*r(x) = 0 ? I can't think of a function off the top of my head for all x such that (x+2) * r(x) = 0 .

 

[Samy_A]

Is the trick here in the definitions?
We can't say r(x) is = f_0, but we can say r(x) is such that (x+2)*r(x) = 0 ? I can't think of a function off the top of my head for all x such that (x+2) * r(x) = 0 .

Yes you can!
$(x+2)r(x)=0$ implies that either $x+2=0$, or $r(x)=0$ (or both).
Now, $x+2=0$ only happens for $x=-2$, so if $x\neq-2$ we must set $r(x)=0$.
That leaves $r(-2)$. How could you define $r(-2)$ so that we still have $\forall x: g(x)r(x)=(x+2)r(x)=0$, but also $r\neq f_0$?

 

[RJLiberator]

Ah... piecewise functions.

So you are saying:

r(x), in this situation is the following
r(x) = 0 if x =/= -2, but r(x) =/= 0 if x = -2 would work.

Therefore r(x) is not f_1 or f_0.

And this is what you are getting at?

To describe the zero divisors we can use piecewise functions again, such that f(x) * r(x) = 0 where r(x) = 0 if x does not equal the roots of f?

 

[Samy_A]

Ah... piecewise functions.

So you are saying:

r(x), in this situation is the following
r(x) = 0 if x =/= -2, but r(x) =/= 0 if x = -2 would work.

Therefore r(x) is not f_1 or f_0.

And this is what you are getting at?

To describe the zero divisors we can use piecewise functions again, such that f(x) * r(x) = 0 where r(x) = 0 if x does not equal the roots of f?

Yes. Now the question remains: for which functions can we do this?
Look at the example $g(x)=x+2$. The key was that we could define $r(-2) \neq 0$. There was something specific about $g$ and $-2$.
Do you see the general rule for zero divisors in $F(\mathbb R)$ emerging?

 

[RJLiberator]

g(x) was equal to 0 when x = -2.

So, when g(x) can = 0 for certain x values, we have zero divisors.
Or, when g(x) has roots, we have zero divisors.

 

[Samy_A]

g(x) was equal to 0 when x = -2.

So, when g(x) can = 0 for certain x values, we have zero divisors.
Or, when g(x) has roots, we have zero divisors.

Correct. Any $g \in F(\mathbb R)$ that has a root (is 0 in at least one $x \in \mathbb R$), is a zero divisor in $F(\mathbb R)$. That's because we can construct $r$ by setting it to 0 in points where $g$ isn't 0, and to 1 (or another non zero value) in a root of $g$. As this $r\neq f_0$, this proves that $g$ is a zero divisor.

Now on to units. The example I suggested was $f(x)=x²+1$. Why is this a unit? How can you generalize this?

 

[RJLiberator]

Beautiful.

If f(x) = x^2+1, then
f(x) * g(x) = 1.

(x^2+1)*g(x) = 1
g(x)*x^2+g(x) = 1

when x is 0, g(x) = 1.
when x is 1, g(x) = 0.5
when x is 2, g(x) = 0.2
when x is 3, g(x) = 0.1

Hm.
Not sure if this is the proper way to attack it.

OH, perhaps when f(x) has no roots, it is a unit. That is what this question is testing us on.
Now, how to describe this situation, hm.
When f(x) has no roots (in real numbers here), then it cannot be a zero divisor, so it must be a unit. I guess that's one weak way of saying it since that was a lemma we proved.
But, when f(x) has no roots, it has an inverse is what this problem is saying.

 

[Samy_A]

Beautiful.

If f(x) = x^2+1, then
f(x) * g(x) = 1.

(x^2+1)*g(x) = 1
g(x)*x^2+g(x) = 1

when x is 0, g(x) = 1.
when x is 1, g(x) = 0.5
when x is 2, g(x) = 0.2
when x is 3, g(x) = 0.1

Hm.
Not sure if this is the proper way to attack it.

OH, perhaps when f(x) has no roots, it is a unit. That is what this question is testing us on.
Now, how to describe this situation, hm.
When f(x) has no roots (in real numbers here), then it cannot be a zero divisor, so it must be a unit. I guess that's one weak way of saying it since that was a lemma we proved.
But, when f(x) has no roots, it has an inverse is what this problem is saying.

In an infinite ring, it is possible for an element of the ring to be neither a unit nor a zero divisor. I assume the lemma you refer to concerned finite rings.
(As an example, take the ring $\mathbb Z$. $5$ is neither a unit nor a zero divisor.)

Back to $f(x)=x²+1$.
You want a function $g$ such that $f(x)g(x)=(1+x²)g(x)=1$.
What do you think about $g(x)=\frac{1}{1+x²}$?
Can you see how to generalize this in order to describe all the units of $F(\mathbb R)$?

 

[RJLiberator]

1 OVER f(x)!

g(x) = 1/f(x) where f(x) is not equal to 0.

Those are the units!
The f(x) that has a g(x) = 1/f(x)

boom!

 

[Samy_A]

1 OVER f(x)!

g(x) = 1/f(x) where f(x) is not equal to 0.

Those are the units!
The f(x) that has a g(x) = 1/f(x)

boom!

Voila!

The units are all the functions that have no roots.

 

[RJLiberator]

Thank you for your help. I understand this problem now.