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    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.
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    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...
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    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?
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    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.
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    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.
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    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.
  7. Offline

    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.

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