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Proving R/I is a field iff I is maximal.

I can prove it one way but I can't seem to prove the converse, here is what I have so far any suggestions for the converse?

Theorem:

Let R\displaystyle R be a commutative ring with one and let I\displaystyle I be an ideal of R\displaystyle R then R/I\displaystyle R/I is a field if and only if I\displaystyle I is maximal.

Proof:

Let a∈R\displaystyle a \in R and a∉I\displaystyle a \notin I so that a+I\displaystyle a+I is a non-zero element in R/I\displaystyle R/I we want to show there exists an element r+I∈R/I\displaystyle r+I \in R/I such that (a+I)(r+I)=1+I\displaystyle (a+I)(r+I)=1+I (the one in R/I\displaystyle R/I).

Now let J=(a)+I\displaystyle J=(a)+I and note J\displaystyle J is in fact an ideal and I⊂J\displaystyle I \subset J (a proper subset since a∉I\displaystyle a \notin I but a∈J\displaystyle a \in J) and so J=R\displaystyle J=R since I\displaystyle I is maximal.

Hence the one in R\displaystyle R is also in J\displaystyle J so 1∈J=(a)+I\displaystyle 1 \in J=(a)+I and 1=ar+i\displaystyle 1=ar+i for some r∈R,i∈I\displaystyle r\in R, i \in I and so (a+I)(r+I)=ar+I=1+I\displaystyle (a+I)(r+I)=ar+I=1+I which is the one in R/I\displaystyle R/I so indeed R/I\displaystyle R/I is a field.

Stuck on how to prove the converse: If R/I\displaystyle R/I is a field then I\displaystyle I is a maximal ideal of R\displaystyle R.
Reply 1
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Zacken
22
Original post by poorform

Stuck on how to prove the converse: If R/I\displaystyle R/I is a field then I\displaystyle I is a maximal ideal of R\displaystyle R.


Okay, so we have that R/I\displaystyle R/I is a field and suppose BB is an ideal of RR that properly contains II. Suppose also that b∈Bb \in B but b∉Ib \notin I. What can you say about b+Ib + I?

Once you've done that, you should be able to state that there exists some element c+Ic + I such that (c+I)(b+I)=1+I(c + I)(b+I) = 1 + I.

Does this help at all? More in the spoiler:

Spoiler

Reply 2
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poorform
OP
10
got it now thanks for help!

the only step I couldn't see was to say 1-bc is in I (so in B) and bc in B (property of ideals) then ((1-bc)+bc) is in B since it is an ideal and closed under addition hence the one of the ring is in B and then it almost trivial to show B is a subset of R and R is a subset of B so B=R and so I is maximal.
(edited 8 years ago)
Reply 3
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Zacken
22
Original post by poorform
got it now thanks for help!


Great. No problem!
Reply 4
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poorform
OP
10
Original post by Zacken
Great. No problem!


A very nice proof :smile:.

It won't let me rep you tho.
Reply 5
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Zacken
22
Original post by poorform

the only step I couldn't see was to say 1-bc is in I (so in B) and bc in B (property of ideals) then ((1-bc)+bc) is in B since it is an ideal and closed under addition hence the one of the ring is in B and then it almost trivial to show B is a subset of R and R is a subset of B so B=R and so I is maximal.


Ah, so you were basically pretty much all there except for one single step. Very nice proof indeed. :-)
Original post by Zacken
Okay, so we have that R/I\displaystyle R/I is a field and suppose BB is an ideal of RR that properly contains II. Suppose also that b∈Bb \in B but b∉Ib \notin I. What can you say about b+Ib + I?

Once you've done that, you should be able to state that there exists some element c+Ic + I such that (c+I)(b+I)=1+I(c + I)(b+I) = 1 + I.

Does this help at all? More in the spoiler:

Spoiler



When you said that you had an offer for maths at Cambridge, did you mean a job offer?
Reply 7
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TeeEm
19
Original post by atsruser
When you said that you had an offer for maths at Cambridge, did you mean a job offer?


He will eventually replace DFranklin ....
Reply 8
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Krollo
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Original post by TeeEm
He will eventually replace DFranklin ....


Speaking of him, I haven't seen him around for ages. I hope he's alright.

Posted from TSR Mobile
Reply 9
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TeeEm
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Original post by Krollo
Speaking of him, I haven't seen him around for ages. I hope he's alright.

Posted from TSR Mobile


DFranklin
gone

ghostwalker
gone

Davros
gone

Firegarden
gone

smaug123
gone

TenOfThem
gone

Noble
gone

James........some number which I do not remember
gone


Maybe there is a "Bermuda triangle" in here somewhere, and I begin to worry if I will be gone soon ...
Reply 10
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poorform
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10
Original post by TeeEm
DFranklin
gone

ghostwalker
gone

Davros
gone

Firegarden
gone

smaug123
gone

TenOfThem
gone

Noble
gone

James........some number which I do not remember
gone


Maybe there is a "Bermuda triangle" in here somewhere, and I begin to worry if I will be gone soon ...


Even after you are gone from this site you will live on.

Many many years from now people will be using your legendary materials to practise Stokes, flux integrals, divergence theorem, Fourier series, PDE's etc.
Reply 11
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poorform
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10
Original post by atsruser
When you said that you had an offer for maths at Cambridge, did you mean a job offer?


zacken just kills it tbf. how long until he leaves this site for mathstack then stackoverflow. :frown:
@TeeEm The new will replace the old. It is a fact of life.

Original post by poorform
zacken just kills it tbf. how long until he leaves this site for mathstack then stackoverflow. :frown:

And Quora :wink:
(edited 8 years ago)
Original post by Kvothe the arcane
The new will replace the old. It is a fact of life.


Oldie power! We will resist! :shakecane:
Reply 14
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Zacken
22
Original post by poorform
zacken just kills it tbf. how long until he leaves this site for mathstack then stackoverflow. :frown:


To be fair: http://math.stackexchange.com/users/161779/zain-patel :tongue:
A cleaner method is by contrapositive:

Let ℘:R→R/I\wp : R \to R/I be the natural projection. If R/IR/I is not a field, let α∈R/I\alpha \in R/I non-zero and non-invertible. Then αR/I\alpha R/I is a proper non-trivial ideal of R/IR/I so ℘−1(αR/I)\wp^{-1}(\alpha R/I) is a proper ideal of RR which properly contains II so II is not maximal.

Conversely, if I⊊J⊊RI\subsetneq J\subsetneq R then ℘(J)\wp (J) is a non-trivial proper ideal of R/IR/I so R/IR/I is not a field.

Original post by Zacken


Hey what is my integral doing there :biggrin:
Reply 16
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Zacken
22
Original post by Lord of the Flies
A cleaner method is by contrapositive: [...]


That is gorgeous. :love:

Hey what is my integral doing there :biggrin:


t'was a really cool one, was looking for more approaches. Hope you don't mind. :colondollar:
Original post by Zacken
Hope you don't mind.


Not at all!
Reply 18
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Magrina
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How to prove the same using banach algebra.?

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