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Rings

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.
Reply 1
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...
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?
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.:smile:
Reply 4
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

(edited 11 years ago)
Reply 5
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.:smile:


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.
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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