Question

Need help understanding problem:
Let S be a subset of a set X. Let R be the ring of real-valued functions on X, and let I be the set of real-valued functions on X whose restriction to S is zero. Show that I is an ideal in R.
Please, help

Answer 1

I don't understand : " WHOSE RESTRICTION TO S IS ZERO", what does it mean?
This is my interpret: \(S\subset X\) , \(Fun(X, R)\) , \(I\subset Fun(X,R)\)

Answer 2

If \(f\in I\), then \(\forall x \in S, f(x) =0\) .
Am I right?

Answer 3

HI!!

Answer 4

the second statement looks right to me, in other words L is the set of functions that vanish on S

Answer 5

should be pretty clearly an ideal, since if \(f\in I , g\in R\) then \(fg\in I \)

Answer 6

You mean S is kernel of Fun?

Answer 7

nope, f(S) is kernel, right?

Answer 8

i think you lost me there
R is the ring right?

Answer 9

yes

Answer 10

it is the ring of real valued function on X
I is a subset of R, it is the set of functions that vanish on S

Answer 11

I confused, can you visualize it? Like

Answer 12

that is not the ring, that is just the set on which the functions act

Answer 13

say X is the interval [-1,1] and S = {0}
then I would be the set of functions \(f\) on [-1,1] for which \(f(0)=0\)

Answer 14

I think I got it :) Let me interpret it and you check, ok?

Answer 15

the ring is the set of all real valued functions on X, I is the set of all real valued function on X that are zero on S
what you have to show is the following
a) (I,+) is a subgroup of R, i.e. if you add to functions in I you get another function in I etc
b) \(\forall f\in I, g\in R, fg\in I\)

Answer 16

R = {set of real-valued functions on X} =\(\{f_a, f_b, f_c, ........\}\) and \(I\subset R\) such that \(f_I (S) =0\)

Answer 17

i wouldn't bother to list them with indices
no one says they are countable, and you don't reallyl want to introduce some uncountable index set.
just say R is what it is, the set of all real valued functions on X

Answer 18

in fact they are uncountable if X is,
leave that part alone

Answer 19

So, to prove I is an ideal, I have to prove it is a subring first, right? then the properties of left/right ideal to get it is an ideal, right?

Answer 20

\[I=\{f\in R:\forall s\in S, f(s)=0\}\]

Answer 21

The reason I list R out is to easily define identities

Answer 22

oh no don't list anything
just pick two arbitrary ones
\[f,g\in I\implies f+g\in I \] etc

Answer 23

Thanks a lot. I will show my stuff, please check then. :)

Answer 24

ok
the point of this exercise is to see if you know exactly what you have to show to show a subset of a ring is an ideal in the ring.
list all the steps
showing them in this case should not be hard at all, the hard part is knowing what to show

Answer 25

1) I is closed under addition and additive inverse:
Let f, g in I, then f(s) =0 and g(s) =0
f(s) + g(s) = (f+g)(s) = 0+0=0 --> (f+g) in I
Moreover, f(s) + g(s) =0 --> f(s)= -g(s)
That is g(s) is additive inverse of f(s)
Hence, 1) follows
2) I is closed under multiplication
f(s)*g(s) = (f*g)(s) =0--> f*g is in I
Hence 2) follows
1), 2) --> I is a subring of R

Answer 26

Now, prove I is a left ideal
let g in R , \(g(s) \neq 0\), f is I
then g(s)f(s) = (gf)(s) =0 , hence g(s)f(s) \(\in I\)
similarly for right ideal
we get I is ideal
Am I right?

Answer 27

looks reasonable to me
i always forget whether you have to show that 0 is in it, but of course it is

if you want to be a bit more precise at the beginning you could say
" let \(f,g\in I\) then \(\forall s\in S, f(s)-g(s)=0\implies f-g\in I\)"

Answer 28

as for the "left, right" business, multiplication is commutative, so the is a commutative right, not necessary

Answer 29

In my book, they say a ring is abelian group, hence it automatically has additive identity 0

Answer 30

yes what i meant was you may need to show that 0 is in the set, but maybe not
if \(f-g\) is in it then \(f-f=0\) is in it, so that step is probably unnecessary

Answer 31

:)
Again, thanks a ton.