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Let $C$ be the ring of continuous functions $f:\Bbb R \to \Bbb R$ with addition and multiplication defined pointwise. Let $J=\{f \in C:f(s)=0\}$, where $s$ is some fixed integer. Then $J$ is an ideal. I want to show that $C/J$ is isomorphic to some well known ring. I know the First Isomorphism Theorem should be used.

I am having trouble in even defining a homomorphism from $C$ to some other ring, never mind finding a homomorphism for which $J$ is the kernel. Any help would be much appreciated.

Thanks.

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1 Answer 1

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Edited:

You want $\phi: C \to R$ so that $\phi(f)=0$ if and only if $f(s)=0$.

Isn't it obvious what $\phi(f)$ should be?

Added Once you realize that $\phi(f)$ should be $f(s)$, then your $R$ must contain all real numbers. Moreover, you want $\phi$ to be onto, thus $R$ must contain all real numbers and nothing more...$R= \mathbb R$...

So, to sum it up

$$\phi: C \to \mathbb R \,;\, \phi(f)=f(s)$$

is the function you need. Now check that this works....

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I think I wrote out my problem incorrectly. I meant to say that $f \in J$ if $f(n) = 0$ for some integer $n$. Not all integers. –  user61164 Feb 6 '13 at 3:40
    
@user61164 Then I am really curious how you proved that it is an ideal ;) Because if it is what you mean it is not closed under addition... –  N. S. Feb 6 '13 at 3:42
    
Would we not have the following to prove its closure under addition: suppose $f$ and $g$ are in $J$. Then we have $(f-g)(n) = f(n) - g(n) = 0 - 0 = 0$. So $f-g$ is in $J$. –  user61164 Feb 6 '13 at 3:47
    
Ok. I keep messing up how $J$ is defined: $ f \in J$ if $f(s) = 0$, where $s$ is a particular integer. –  user61164 Feb 6 '13 at 3:49
1  
@N.S.This is very embarrassing. Would we define $\phi$ by $\phi (f) = f(s)$? –  user61164 Feb 6 '13 at 4:02

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