Proving N is Maximal Ideal of G in Real Functions

[Ceci020]

Given:
Denote a ring G = { all functions from ℝ --> ℝ }
And set N = { all functions from ℝ --> ℝ such that for any x in ℝ, f(x) = 0 }

Want to prove:
To prove that N is the maximal ideal of G by showing that the quotient ring G / N is isomorphic to the set of real numbers ℝ

** What I'm still getting stuck at this point is the following:

1/ What should an element of G / N look like. Since G contains all real function, is it true that an element in G / N should be something like f(x) + N, where f(x) is in ℝ ? or is it something else ?

2/ My approach is to try to define a map, say f, from G/N to ℝ. But then I get stuck on what (in ℝ) should I send elements of G/N to ?

For first approach, I try to send everything in G/N to f(0). But then I recognize that by the mean of "all functions", G may have some functions which are not defined at 0, say f(x) = 1/x, so f(0) doesn't work.

For second try, I try to send everything in G/N to f(1). But again, the same issue as my first approach comes up again.


Would someone please help me on this problem?
Thanks in advance.

 

[SteveL27]

Given:
Denote a ring G = { all functions from ℝ --> ℝ }
And set N = { all functions from ℝ --> ℝ such that for any x in ℝ, f(x) = 0 }

Your description of N isn't right. You're saying that a function f is in N if for any x, f(x) = 0. So N consists of the zero function and nothing more. That's not what's intended here. You're misinterpreting what the problem says.

Your first step would be to re-read the problem and clarify what N is supposed to be.Want to prove:

To prove that N is the maximal ideal of G

You mean "a" maximal ideal. I'm being picky because these kinds of exercises are designed to teach you to think precisely about the concepts. A ring may have many maximal ideals.

by showing that the quotient ring G / N is isomorphic to the set of real numbers ℝ

In other words, by invoking the fact that a ring modulo an ideal is a field iff the ideal is maximal. Is that part already clear to you?

1/ What should an element of G / N look like. Since G contains all real function, is it true that an element in G / N should be something like f(x) + N, where f(x) is in ℝ ? or is it something else ?

Another important bit of pickiness. f is a function. f(x) is the value of the function at the point x. We'll sometimes casually ignore the difference, and there's no harm in it. But this is one of those times when the distinction is important. f(x) is a real number, not a function. A coset would look like f + N. What you need to do is first clarify the definition of N; then go through the exercise of proving that it is indeed an ideal. Once you do that, the rest of this will become more clear.

Also, by the way, is G supposed to be all functions? Or possibly all continuous functions? I've usually seen the result you're trying to prove stated with continuous functions; and I don't happen to know if it works out for arbitrary functions.

 

[mathwonk]

you are right that this problem is easy if stated correctly, and just as easy for all functions as for continuous ones. the less easy direction is the converse, that all maximal ideals have this form, and that uses continuity and compactness.

 

[Ceci020]

Thanks for your responses.

In other words, by invoking the fact that a ring modulo an ideal is a field iff the ideal is maximal. Is that part already clear to you?

Yes, I'm familiar with this idea. I think the goal of this problem is to show this, since ℝ is a field.

** By how the person, who put up this question, restates the question for me, I have some updates :

1/ Ring G indeed contains all real functions. There is no notion of continuous, invertible ...
2/ I have an interpretation that set N is not the set of zero functions. It contains functions f(x) such that when I take a number x in ℝ, then I get 0 when evaluate the function at x.

So I think an example is if I take x = 3, then the function x - 3 is in N, because when I evaluate it at 3, the result is 0. But x - 3 is not the zero function.

Using f(x) = x + 3 as an example, is it true that when I mod this element with N, to get an element in G/ N, it will look something like (x + 3) + N ? But then if I want to send this element to ℝ, what should the output be ? I'm thinking about sending to f(x), but ...

I'm a bit confused

 

[HallsofIvy]

Thanks for your responses.Yes, I'm familiar with this idea. I think the goal of this problem is to show this, since ℝ is a field.

** By how the person, who put up this question, restates the question for me, I have some updates :

1/ Ring G indeed contains all real functions. There is no notion of continuous, invertible ...
2/ I have an interpretation that set N is not the set of zero functions. It contains functions f(x) such that when I take a number x in ℝ, then I get 0 when evaluate the function at x.

So I think an example is if I take x = 3, then the function x - 3 is in N, because when I evaluate it at 3, the result is 0. But x - 3 is not the zero function.

This is still very awkwardly stated. But I think you mean either that N is "the set of all functions such that f(x)= 0 for some x", not for "all x", or that N is "the set of all functions such that f(x0)= 0 for some specific x0".

Using f(x) = x + 3 as an example, is it true that when I mod this element with N, to get an element in G/ N, it will look something like (x + 3) + N ? But then if I want to send this element to ℝ, what should the output be ? I'm thinking about sending to f(x), but ...

I'm a bit confused

 

[Vargo]

To me it looks like you need to clarify whether or not the definition of N depends on some pre-chosen point x_0. If so, then your notation should reflect the difference between the "constant" x_0 and the "variable" x.

 

[mathwonk]

try letting N be the set of functions such that f(0) = 0.

 

[DonAntonio]

Thanks for your responses.


Yes, I'm familiar with this idea. I think the goal of this problem is to show this, since ℝ is a field.

** By how the person, who put up this question, restates the question for me, I have some updates :

1/ Ring G indeed contains all real functions. There is no notion of continuous, invertible ...
2/ I have an interpretation that set N is not the set of zero functions. It contains functions f(x) such that when I take a number x in ℝ, then I get 0 when evaluate the function at x.

So I think an example is if I take x = 3, then the function x - 3 is in N, because when I evaluate it at 3, the result is 0. But x - 3 is not the zero function.

Using f(x) = x + 3 as an example, is it true that when I mod this element with N, to get an element in G/ N, it will look something like (x + 3) + N ? But then if I want to send this element to ℝ, what should the output be ? I'm thinking about sending to f(x), but ...

I'm a bit confused

Either you're having a hard time trying to understand your own question's terms (and the material and etc.), or else you're having

some difficulties with the language.

I guess that, just as in stachexchange, if you post your question in your own language it may be a chance somebody will translate

it and make it clear to everybody.

Finally, if you meant $$N_x:=\{f:\mathbb{R}\to\mathbb{R}\,\,|\,\,\text{for some fixed}\,x\in\mathbb{R}\,,\,f(x)=0\}$$ then you only need to check the quotient ring $G/N_x$ is a field (and a rather common one)...

DonAntonio