# Thread: Commutative ring and integral domain

1. ## Commutative ring and integral domain

(a) Let F(R) = all f: R -->R
For f,g belongs to F(R) define f+g: R -->R by x-->f(x)+g(x)
and f.g:R-->R by x-->f(x)g(x)
Show that F(R) is a commutative ring which is not an integral domain

(b) Define C(R), D(R) to be all f in F(R) which are respectively continuous and differentiable. Show that D(R)=<C(R)=<F(R)
Calculate F(R)* and U(F(R))

Note: I used R to denote for Real number not Ring R because I dont use Latex

In part a,I think we have to show somehow F(R) satisfy all the conditions of a ring under + and . ? how about integral domain, what do we have to prove?

In part b, I have no idea, please show me how to do it or give me some hints?

2. Originally Posted by knguyen2005
(a) Let F(R) = all f: R -->R
For f,g belongs to F(R) define f+g: R -->R by x-->f(x)+g(x)
and f.g:R-->R by x-->f(x)g(x)
Show that F(R) is a commutative ring which is not an integral domain

(b) Define C(R), D(R) to be all f in F(R) which are respectively continuous and differentiable. Show that D(R)=<C(R)=<F(R)
Calculate F(R)* and U(F(R))

Note: I used R to denote for Real number not Ring R because I dont use Latex

In part a,I think we have to show somehow F(R) satisfy all the conditions of a ring under + and . ? how about integral domain, what do we have to prove?

In part b, I have no idea, please show me how to do it or give me some hints?

For (a) choose two function f,g different from the zero function but its
product fg is zero. this must be pretty easy and in fact you can choose
f,g to be differentiable

For (b): I thought F(R)* denotes the units in F(R) but then I saw UF(R) and I am not sure anymore. Anyway, the units in F(R) are those functions which have a multiplicative inverse, so f is a unit <==> we can form 1/f.
Now, when can we form 1/f? Can you see it?

Tonio

3. Thanks Tonio for a quick reply

In part (a), we just need to use definition to show that F(R) isa commutative ring
i.e.
(i) (F(R),+) is an Abelian group
(ii) (F(R),.) is monoid
(iii) . is ditributive over +
And plus, for all f, g in F(R), f.g = g.f (Commutative)

Then we conclude that F(R) is a commutative ring. Is that enough ? Or we have to show that F(R) is not an integral domain? Or it automatically follow from the proof that F(R) is a commutative ring then F(R) is not an integral domain?

In part (b), To show that D(R)=<C(R)=<F(R). I can use the results from analysis (sandwich rule) to prove it but I forgot how to do it?
Also, we have to calculate F(R)* and U(F(R)). It is different, F(R)* is a ring with non-zero elements in it. And U(F(R)) is unit of the ring

I hope it does make sense to you

4. Originally Posted by knguyen2005
Thanks Tonio for a quick reply

In part (a), we just need to use definition to show that F(R) isa commutative ring
i.e.
(i) (F(R),+) is an Abelian group
(ii) (F(R),.) is monoid
(iii) . is ditributive over +
And plus, for all f, g in F(R), f.g = g.f (Commutative)

Then we conclude that F(R) is a commutative ring. Is that enough ? Or we have to show that F(R) is not an integral domain? Or it automatically follow from the proof that F(R) is a commutative ring then F(R) is not an integral domain?

In part (b), To show that D(R)=<C(R)=<F(R). I can use the results from analysis (sandwich rule) to prove it but I forgot how to do it?
Also, we have to calculate F(R)* and U(F(R)). It is different, F(R)* is a ring with non-zero elements in it. And U(F(R)) is unit of the ring

I hope it does make sense to you

the sandwich rule is for limits: it won't help you here. But remember that a derivalbe function (in D(R)) is always a continuous function (in C(R)) which is a function (in F(R))

F(R)* is easy and so is U(F(R))

Tonio

5. Can you just show me how to calculate F(R)* please?
I have no idea how to do it

6. Originally Posted by knguyen2005
Can you just show me how to calculate F(R)* please?
I have no idea how to do it

After re-reading your definitiom of F(R)* I realize it makes no sense: every ring HAS to have a zero element = the neutral element wrt addition, so F(R)* cannot be "A ring with non-zero lements".
Please check this since I don't know what you meant.

Tonio

7. I am sorry I made a mistake

Can I just point out that F(IR)* where IR denotes Real number, not a ring and for the whole question R is the real number, I should have type like this (IR) instead of R because it makes u confuse with rings R

So, the question is to calculate F(IR)* and U(F(IR))

Once again, sorry Tonio for bothering you

8. Originally Posted by knguyen2005
I am sorry I made a mistake

Can I just point out that F(IR)* where IR denotes Real number, not a ring and for the whole question R is the real number, I should have type like this (IR) instead of R because it makes u confuse with rings R

So, the question is to calculate F(IR)* and U(F(IR))

Once again, sorry Tonio for bothering you

Ooooh! Now I think I understand what you meant: IR is a field and thus the set of all non-zero elements in IR is a GROUP wrt multiplication!
STILL F(IR) is NOT a field and thus the set F(IR)* of all non-zero elements in F(IR) is NOT a group, leave alone a ring.
So if you jsut one THE SET F(IR)* then...well, then it is what it's been said before: the set of all non-zero functions in F(IR).

Tonio