# Units / Zero divisors in Comm Ring

1. Mar 1, 2016

### RJLiberator

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

2. Relevant 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).

3. 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.

2. Mar 2, 2016

### Samy_A

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 $x \in \mathbb R:\ f(x)g(x)=f_0(x)=0$.

3. Mar 2, 2016

### 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.

4. Mar 2, 2016

### Samy_A

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.

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

5. Mar 2, 2016

### 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?

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.

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.

6. Mar 2, 2016

### Samy_A

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

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).
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 $L(\mathbb R)$?
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 $x \in \mathbb R$

Last edited: Mar 2, 2016
7. Mar 2, 2016

### RJLiberator

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

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

8. Mar 2, 2016

### Samy_A

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

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)
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.
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.

Last edited: Mar 2, 2016
9. Mar 2, 2016

### 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?

10. Mar 2, 2016

### Samy_A

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

11. Mar 2, 2016

### 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.

12. Mar 2, 2016

### Samy_A

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 $F(\mathbb R)$?
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.

13. Mar 2, 2016

### 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.

14. Mar 2, 2016

### Samy_A

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 $r$?

15. Mar 2, 2016

### 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 .

16. Mar 2, 2016

### Samy_A

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 $r\neq f_0$?

17. Mar 2, 2016

### 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?

18. Mar 2, 2016

### Samy_A

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?

19. Mar 2, 2016

### 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.

20. Mar 2, 2016

### Samy_A

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?