# Units of Local Rings

• March 10th 2009, 06:39 PM
electricjaguar
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?
• March 11th 2009, 12:21 AM
aliceinwonderland
Quote:

Originally Posted by electricjaguar
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.