Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:
    Let R be a commutative ring and I an ideal of R. Show that if I is maximal then R/I is a field. I'm a bit stuck on how to start this. Any would would be appreciated. Thanks.
    Offline

    0
    ReputationRep:
    (Original post by JBKProductions)
    Let R be a commutative ring and I an ideal of R. Show that if I is maximal then R/I is a field. I'm a bit stuck on how to start this. Any would would be appreciated. Thanks.
    Suppose R/I is not a field. Then there exists x+I in R/I with no multiplicative inverse, so...
    Offline

    14
    Ok. Pick x in R with x not in I (else x is just 0 in R/I). Now consider the ideal generated by x and I: what can we say about this since I is maximal?
    • Thread Starter
    Offline

    0
    ReputationRep:
    Unless I misunderstood something, I'm not sure why if x is in I then x = 0 in R/I? Thanks for the replies btw.
    Offline

    2
    ReputationRep:
    If you know the correspondence theorem this is immediate (i.e. if R/I has a non-trivial proper ideal J then consider the corresponding ideal J' in R. The ideal J' contains I so must be either R or I by maximality of I. The first contradicts the fact that J was proper, the second contradicts non-triviality)

    I am assuming therefore, that you don't know and/or aren't expected to know the correspondence theorem. In that case; it is a bit harder to think up.

    Hint: For each non-zero element x+I in R/I and consider the ideal J = I + Rx.

    Spoiler:
    Show

    Since x isn't in I (else x+I would be zero in R/I), I is strictly contained in J, whence by maximality of I we have that J = R. Thus in particular - the identity element 1 of R is in J and so we may write 1 = i + rx for some r in R, i in I. It then follows that (x+I)(r+I) = xr +I = xr + i + I = 1 + I so that (x+I) is invertible as required.
    Offline

    2
    ReputationRep:
    (Original post by JBKProductions)
    Unless I misunderstood something, I'm not sure why if x is in I then x = 0 in R/I? Thanks for the replies btw.
    By pure definition: If x = i for some i in I then the image of x under the projection from R to R/I is equal to the coset 0 + I

    Look up the definition and construction of the quotient ring to refresh yourself.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Jake22)
    By pure definition: If x = i for some i in I then the image of x under the projection from R to R/I is equal to the coset 0 + I

    Look up the definition and construction of the quotient ring to refresh yourself.
    Ah ok, I see. I'll have a go at the rest of it now. Thanks.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Has a teacher ever helped you cheat?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Write a reply...
    Reply
    Hide
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.