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Math Help - Units of Local Rings

  1. #1
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    Units of Local Rings

    Hi, I'm trying to prove that the units of a local ring R with maximal ideal M are exactly the elements of R-M. I checked Google and there are proofs available, so I don't need a full proof, I just have 2 questions about the proof:

    1. It requires an assumption that any proper ideal I of R is contained in M. If I were asked to prove this assumption, would I have to appeal to Zorn's lemma and show the hypotheses for the lemma hold in this case?

    2. During the proof, this:
    ab = 1 + x, some x in R

    follows from this:
    (a + M)(b + M) = 1 + M.

    I see how we have ab + M = 1 + M, but then how do we get from that, that ab = 1 + x, some x\in R?
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  2. #2
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    Quote Originally Posted by electricjaguar View Post
    Hi, I'm trying to prove that the units of a local ring R with maximal ideal M are exactly the elements of R-M. I checked Google and there are proofs available, so I don't need a full proof, I just have 2 questions about the proof:

    1. It requires an assumption that any proper ideal I of R is contained in M. If I were asked to prove this assumption, would I have to appeal to Zorn's lemma and show the hypotheses for the lemma hold in this case?

    2. During the proof, this:
    ab = 1 + x, some x in R

    follows from this:
    (a + M)(b + M) = 1 + M.

    I see how we have ab + M = 1 + M, but then how do we get from that, that ab = 1 + x, some x\in R?
    Let R be a local ring and M be a maximal ideal.

    If a \in R-M, the ideal generated by a and M is (1). This implies that there exists b \in R and t \in M such that ab + t = 1. Thus, ab=1-t=1+x, where t \in M and  x \in R and x = -t.

    For 1, I think you need a Zorn's lemma for general cases. If your ring is Noetherian, you don't need a Zorn's lemma by the definition of a Noetherian ring.
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