Homework Help: More number theory

1. Mar 5, 2010

Firepanda

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!

Last edited: Mar 5, 2010
2. Mar 5, 2010

Dick

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. Mar 6, 2010

Firepanda

the condition that f is a polynomial?

4. Mar 6, 2010

Dick

Is that just a random guess?

5. Mar 6, 2010

Firepanda

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. Mar 6, 2010

Dick

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. Mar 6, 2010

Any point.

8. Mar 6, 2010

Dick

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. Mar 6, 2010

Firepanda

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. Mar 6, 2010

Dick

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. Mar 6, 2010

Firepanda

phi : R -> R

f -> f(a)=0

Pretty sure that's a ring homomorphism

12. Mar 6, 2010

Hurkyl

Staff Emeritus
"f(a)=0" is a truth-value (well, a truth-value-valued equation) not a real number....

13. Mar 6, 2010

Firepanda

hmm so what could I change it to, just 0? Or do I need to change my co-domain?

14. Mar 6, 2010

Dick

You just aren't expressing yourself very well. What is the real number phi(f)?

15. Mar 6, 2010

Firepanda

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. Mar 6, 2010

Dick

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. Mar 6, 2010

Firepanda

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. Mar 6, 2010

Dick

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. Mar 6, 2010

Firepanda

You make it sound trivial but nothings jumping out at me

20. Mar 6, 2010

Dick

How about defining the functions piecewise?

21. Mar 6, 2010

Firepanda

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. Mar 6, 2010

Dick

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. Mar 6, 2010

Firepanda

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. Mar 6, 2010

Dick

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. Mar 6, 2010

Firepanda

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