Question

Let R be a ring and X is a set.
a) Show that the set Fun(X,R) of functions on X with values in R is a ring.
b) Show that R is isomorphic to the subring of constant functions on X.
c) Show that Fun(X, R) is commutative iff R is commutative.
d) Suppose that R has an identity, show that Fun (X,R) has an identity and describe the units of Fun(X,R)
Please, help

Answer 1

@zzr0ck3r

Answer 2

start with a?

Answer 3

If you can, please give me all. I missed only 1 class and ..... blank

Answer 4

I am going to look at it on paper, and what ever I figure out I will post the paper.

Answer 5

Thanks a ton.

Answer 6

But give me a few, I am in topology mode.... so it could take a second to make the switch lol.

Answer 7

I am not in hurry, whenever you have time. :) (but not after Monday, I have another thing on Monday, can't miss this part before getting a new one, please)

Answer 8

so do you know the operations on the functions?

composition and addition?

Answer 9

No :(

Answer 10

I am guessing it is...

Answer 11

hmm

Answer 12

Can you give me any link to read for this part? I don't want to be blink like this. I want to know .....all. :)

Answer 13

what book are you using?

Answer 14

The free book online Algebra abstract and concrete 2.6edition
http://homepage.math.uiowa.edu/~goodman/22m121.dir/2005/sections6.1-6.3.pdf
part 6.1, 6.2 but I don't see the notation Fun (X,R) on it.

Answer 15

and the problem is 6.1.12

Answer 16

a)

For a ring \((R,+,*)\) we will use the inherited operations for the functions. i.e. \((f+g)(x)=f(x) +g(x)\text{ and } (f*g)(x) = f(x)*g(x)\). Let \(X\subset R\). Consider the set \(\text F=\text{Fun}(X,R)\).

Let \(f\in F \), and consider \(f_0:X\rightarrow R, f(x)=0\) where \(0\) is the additive identity in \(R\). Then \((f+f_0)(x) = f(x)+f_0(x) =f(x) + 0 =f(x) \ \ \ \forall \ x\in X \). Similarly \((f_0+f)(x) = f(x)\). So we have additive identity in \(F\).

if \(f:X\rightarrow R,f(x)=a\) is a in \(F\), then so is \(g:X\rightarrow R,g(x)=-a\) is also in \(F\). So \((f+g)(x) = a-a=0 \ \ \ \ \forall x\in X\). So \(g=-f\) So we have inverses.

Let \(f, g, h \in F\) then since there image is in \(R\), \((f+g)+h = f+(g+h)\), Similarly \(f+g=g+f\).

So \(F\) forms a group and we are halfway done with a).

you with me?

Answer 17

Can you show that
1) \(F\) is associative under multiplication (given that R is).
2) there is an identity function (given \(1\in R\)). note, it's not f(x) = x.
3) \(f*(g+h) = f*g+f*h\) for all \(x\).

Then we are done with part a). Since we are not in a field we don't need multiplicative inverses...

Answer 18

I know you are having inet probs, so I will wait before I go on. Since you are in no rush and I will be around for the next few days ;)

Answer 19

For your questions: 1) Let f, g, h ∈F , we have f(g(h(x) ) = fg(h(x) = fgh(x) , the same for approaching fro the right.
2) To be a ring, set doesn't need to have multiplicative identity, right? but if needed, the multiplication identity of F should be \(f_i =1\)
3) honestly, to me, it's the hardest part when I have to prove the trivial problem. Is it not that (g+h)(x) = g(x) +h(x) and f(g+h) (x) = f*g+f*h

Answer 20

Some definitions include multiplicative identity and some don't...So its fine to skip that part if you are working in a book that does not.

I actually think they wanted you to use the multiplication inherited from R, check out how I defined multiplication on \(F\).

In this case, since multiplication distributes in \(R\), function multiplication is trivial(the codomain is R, so your image is in R).

Answer 21

For the isomorphism Let \(C\) be the set of constant functions, then \(g:R\rightarrow C, g(a) = f_a\) where \(f_a\) is the constant function with output \(a\), is an isomorphism.

Answer 22

For commutative property, note that \(\forall \ f \in F \ \forall \ x\in X \ (f(x) \in R )\ \), so given any \(f,g\in F\) we have \( \ (f*g)(x) = f(x)*g(x)=g(x)*f(x)=(g*f)(x) \ \forall \ x\in X\).

Answer 23

So \(R\) being commutative implies \(F\) is commutative. Suppose \(F\) is commutative, than any subring is commutative. We proved \(C\) was a subring by showing it was isomorphic to ring \(R\)(surely its a subset of \(F\)). So \(R\) is commutative because \(C\) is a commutative subring.

Answer 24

Its easy to show that \(f_1\in C\) is the identity for multiplication in \(F\).

Answer 25

Up there where it says

In this case, since multiplication distributes in \(R\) , function multiplication is trivial(the codomain is \(R\), so your image is in \(R\)).

it should say

In this case, since multiplication distributes in R , function multiplication distributes(the codomain is R, so your image is in R).

Answer 26

how we doing?

Answer 27

for the last part, suppose F has identity, by definition of the unit in ring, we have set of the units in F(X,R) = { g(x) | g(x) = \(f^{-1}(x)\) x \(\in X\)}, right?

Answer 28

yeah that's fine

Answer 29

Make sure that you note that you mean multiplicative inverse not function inverse

i.e. \(f^{-1}=\frac{1}{f}\)

Answer 30

Question on part b)
C is a subring of R, hence g is one of endomorphism from R to itself, right?
all we need to prove is g is bijective to get g is isomorphism. Am I right?

Answer 31

for d) show that F has an identity
Question: from a) we proved that F is a ring, so that F is an addition abelian GROUP, which always has identity.
All we need is to define multiplication identity, right?