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More number theory

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  • #1
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For i) I said for a in [0,1], then the group of units are = {f in R | f(a) =/= 0}

i.e a continous function f on [0,1] would have a continous function g on [0,1] such that f.g=1

but the function would have to be g = 1/f, but this wouldnt be continous if f(a) = 0

for ii) I have to show for a,b in R then if ab=0 then either a = 0 or b = 0

but I havn't gotton any further here since I'm having a hard time finding functions a and b where the product is 0

for iii) I believe I have to construct a surjective ring homomorphism ( f: R -> F ) and show that the kernal is equal to the Ideal.

But I'm not sure how what field F to choose for this

Or I have to show that the quotient group R/I is a field.

Thanks for any help!
 
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  • #2
Dick
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The statement of (i) doesn't have any 'a' in it. You aren't thinking very abstractly. Yes, the inverse of f would be 1/f. What conditions on f do you need to ensure that exists and is continuous?
 
  • #3
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The statement of (i) doesn't have any 'a' in it. You aren't thinking very abstractly. Yes, the inverse of f would be 1/f. What conditions on f do you need to ensure that exists and is continuous?
the condition that f is a polynomial?
 
  • #4
Dick
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the condition that f is a polynomial?
Is that just a random guess?
 
  • #5
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Is that just a random guess?
No it wasn't, I looked up a more precise definition of the conditions for continuity and it stated that all polynomials are continuous.

Otherwise I'm not sure what you mean sorry :/
 
  • #6
Dick
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No it wasn't, I looked up a more precise definition of the conditions for continuity and it stated that all polynomials are continuous.

Otherwise I'm not sure what you mean sorry :/
Ok. Maybe I'm just being overly fussy about your original answer because you didn't specify what 'a' is supposed to be. Is it some point in [0,1], or is it ANY point in [0,1]?
 
  • #7
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Ok. Maybe I'm just being overly fussy about your original answer because you didn't specify what 'a' is supposed to be. Is it some point in [0,1], or is it ANY point in [0,1]?
Any point.
 
  • #8
Dick
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Any point.
Ok. Be careful when you get to part (iii). There 'a' means 'some point'. So f is a nonzero continuous function on [0,1]. So 1/f exists. If you can say why 1/f is continuous then you've show f is a unit.
 
  • #9
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For part iii) I've said

that I is a proper ideal since the constant function f(x)=1 is not an element of the ideal.

So R/I is isomorphic to the image of the functions, and the Im(f) = the real numbers

Since the real numbers are a field then I is maximal. Is this ok?
 
  • #10
Dick
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It's not very clear. What exactly is the ring homomorphism from R (the ring) to F (the reals) that I is the kernel of?
 
  • #11
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It's not very clear. What exactly is the ring homomorphism from R (the ring) to F (the reals) that I is the kernel of?
phi : R -> R

f -> f(a)=0

Pretty sure that's a ring homomorphism
 
  • #12
Hurkyl
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"f(a)=0" is a truth-value (well, a truth-value-valued equation) not a real number....
 
  • #13
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"f(a)=0" is a truth-value (well, a truth-value-valued equation) not a real number....
hmm so what could I change it to, just 0? Or do I need to change my co-domain?
 
  • #14
Dick
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hmm so what could I change it to, just 0? Or do I need to change my co-domain?
You just aren't expressing yourself very well. What is the real number phi(f)?
 
  • #15
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You just aren't expressing yourself very well. What is the real number phi(f)?
I'm having a hard time getting my head around the ring of functions, usually it's something simpler.

Perhaps it's just

phi : R -> R

f -> f(a)
 
  • #16
Dick
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Yes, it is. Is I the kernel? You said before you were only 'pretty sure' that it was a homomorphism. Can you check that it satisfies the properties of a homomorphism, so you can say you are 'really sure'?
 
  • #17
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For part ii) I can only think of the fact that

'Theorem: Let R be an integral domain and let R[x] be the ring of polynomials in powers of x whose coefficients are elements of R. Then R[x] is an integral domain if and only if R is.'

These polynomials R[x] are functions that are most likely continous on [0,1], so if I can choose an R that isn't an integral domain then by the theorem the continuous functions on [0,1] are not an integral domain?

Just a thoguht as I'm having trouble with this..
 
  • #18
Dick
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The problem is not about polynomials, it's about continuous functions, which don't have to be polynomials. Surely you can think of two nonzero continuous functions on [0,1] whose product is zero.
 
  • #19
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You make it sound trivial but nothings jumping out at me
 
  • #20
Dick
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You make it sound trivial but nothings jumping out at me
How about defining the functions piecewise?
 
  • #21
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hmm if they're piecewise then how to make sure they both stay continuous on [0,1] so that the product is zero?

is it still continuous if the endpoints match up?

so f(x) = x for x<0.5

and g(x) = x for x>=0.5

is sufficient?
 
  • #22
Dick
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You haven't defined what f(x) is for x>=0.5. If you mean f(x)=0 for x>=0.5, then f(x) isn't continuous. Making sure the pieces match up so it's continuous is YOUR job.
 
  • #23
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You haven't defined what f(x) is for x>=0.5. If you mean f(x)=0 for x>=0.5, then f(x) isn't continuous. Making sure the pieces match up so it's continuous is YOUR job.
how about for some small epsilon e > 0

f(x) = e for x<0.5 and f(x) = 0 for x>=0.5, would that be continuous?
 
  • #24
Dick
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how about for some small epsilon e > 0

f(x) = e for x<0.5 and f(x) = 0 for x>=0.5, would that be continuous?
What do you think? How small would e have to be to make it continuous? This isn't supposed to be anything horribly exotic. Just pick a nonzero function on [0,.5] such that f(.5)=0.
 
  • #25
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What do you think? How small would e have to be to make it continuous? This isn't supposed to be anything horribly exotic. Just pick a nonzero function on [0,.5] such that f(.5)=0.
oh so something like f(x) = -x+0.5 for x<0.5, f(x)=0 for x>=0.5

g(x) = x - 0.5 for x>0.5, g(x) = 0 for x<=0.5
 

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