Ring/Ideal related question.

[daon]

Q: Let R be the set of continuous real-valued functions on [0,1] and I the set of functions which vanish at a neighborhood to the right of zero. Show that I is an ideal. Is it prime? Is it maximal?


- What does "vanish at a neighborhood to the right of zero" mean? I'm thinking it means "is zero when evaluated at all real numbers greater than zero."
- I being prime means given f(x) and g(x) in R then if the product f(x)g(x) happens to be in I, then either f(x) or g(x) must be in I. This may be an easier task if I knew the answer to my first question.
- I being maximal means there is no ideal different from I and R that contains I and is contained in R. That is, I not maximal in R if there is a proper ideal J of R which contains all of I, and J is not equal to I.

Thanks
-Daon

 

[pka]

I am not certain of the exact definition. It should have been stated in the text.
But I have done a similar problem.
\(\displaystyle \L f \in I \Leftrightarrow \left( {\exists \varepsilon > 0} \right)\left[ {x \in [0,\varepsilon ) \Rightarrow f(x) = 0} \right]\)

 

[daon]

Thanks pka, with that definition the ideal proof was a sinch, with one minor issue. Do we know that C[0,1] (The continuous functions on 0,1) is an Euclidean domain? Don't I need this to show that a(x)b(x)=0 on [0,epsilon) implies a(x)=0 or b(x)=0 on that interval. (to show that I is prime).

Also, to show it is not maximal, I think the set {f(x) | f(x)=0 for a<=x<b, a and b are reals in [0,1]}. If that is not clear, what I mean is all functions that vanish on any interval within [0,1]. I hope my intuition is correct here.

Thanks

 

[pka]

I have thought about this problem on and off since you first posted it. It seems to me that the most difficult part is proving that the ideal is prime (that is of course if it is). My own reaction is to think that it maximal. Do you know the book RINGS OF CONTINUOUS FUNCTIONS by Gillman and Jerison. That book would be in a mathematics library. I have not found anything like this problem in it, however there are a great many techniques in the book that may help you.

BTW: What is the source of this problem? Does it appear in any textbook?

 

[daon]

I have thought about this problem on and off since you first posted it. It seems to me that the most difficult part is proving that the ideal is prime (that is of course if it is). My own reaction is to think that it maximal. Do you know the book RINGS OF CONTINUOUS FUNCTIONS by Gillman and Jerison. That book would be in a mathematics library. I have not found anything like this problem in it, however there are a great many techniques in the book that may help you.

BTW: What is the source of this problem? Does it appear in any textbook?

The source is from my Professor's own notes.

Also, a classmate showed it was not maximal by use of the following Ideal: Consider \(\displaystyle A = \{ f(x) \in C[0,1] \,\,| \,\, f(0) = 0 \}\). Clearly for all functions in I, they are zero when evaluated at zero. However, the function f(x)=x is not in I, but is in A. Finally the function f(x)=2 on [0,1] is in R but not in A, so A is a proper Ideal. So A lies strictly in between I and R.

My professor, today, said it was Prime but wants us to work it out.

I will see if my school has that book in its library, thanks.

 

[pka]

Ideal: Consider \(\displaystyle A = \{ f(x) \in C[0,1] \,\,| \,\, f(0) = 0 \}\).

What an obvious counter-example. I am ashamed of myself, but it was thirty years ago I had a seminar from Gillman’s book. I would like to see the proof of primacy.