A local ring is a commutative ring with 1 which has a unique maximal ideal.
Show that the group of units of a local ring is precisely the set of elements lying outside of \(M\).
Let \(F\) be a field, and \(| \ | : F \longrightarrow \mathbb{R}\) such that
\(|x| \geq 0\) for all \(x \in F\) and \(|x| = 0\) if and only if \(x = 0\);
\(|xy| = |x||y|\), for all \(x, y \in F\);
\(|x + y| \leq \max\{|x|, |y|\}\), for all \(x, y \in F\).
Show that the set \(R = \{x \in F : |x| \leq 1\}\) is a local ring.
- Let \(M\) be a maximal ideal of a commutative ring \(R\) with 1 and \(n\) be a positive integer. Show that \(R/M^n\) is a local ring.
Solution
1. If \(A\) is a local ring with unique maximal ideal \(M\), then\(A^\times=A\setminus M\).
A unit cannot lie in a proper ideal: if \(u\in I\) and \(u\) is a unit, then \(1=u^{-1}u\in I\). Thus no element of \(M\) is a unit.
Conversely, if \(a\in A\) is not a unit, then \((a)\) is proper. Every proper ideal of a commutative unital ring is contained in a maximal ideal. Order the proper ideals containing \((a)\) by inclusion. The union of any nonempty chain is an ideal containing \((a)\) and remains proper, since membership of \(1\) in that union would put \(1\) in a member of the chain. Zorn’s lemma gives a maximal element, which is a maximal ideal of \(A\) [@DF04]. By uniqueness it is \(M\), so \(a\in M\).
2. In part (b), \(R\) is local with unique maximal ideal\[ \mathfrak m=\{x\in F:|x|<1\}. \]
The axioms imply \(|0|=0\) and \(|1|>0\). Multiplicativity gives \(|1|=|1|^2\), hence \(|1|=1\). Also \(|-1|^2=|1|=1\) and \(|-1|\geq0\), so \(|-1|=1\) and \(|-x|=|x|\).
The set \(R\) contains \(0,1\). For \(x,y\in R\), \(|x-y|\leq\max\{|x|,|y|\}\leq1\) and \(|xy|=|x||y|\leq1\). Thus \(R\) is a commutative unital subring of \(F\). The set \(\mathfrak m\) contains zero, is closed under subtraction by the same inequality, and satisfies \(|rx|=|r||x|<1\) for \(r\in R\), \(x\in\mathfrak m\). It is an ideal of \(R\), proper because \(|1|=1\).
For \(x\ne0\), multiplicativity gives \(|x^{-1}|=1/|x|\). Consequently an element \(x\in R\) is a unit in \(R\) exactly when \(|x|=1\): in that case its inverse lies in \(R\), and conversely \(xy=1\) with \(x,y\in R\) forces \(|x|=|y|=1\). Thus every element of \(R\setminus\mathfrak m\) is a unit. An ideal strictly containing \(\mathfrak m\) contains such a unit and hence is all of \(R\), proving maximality. Every proper ideal contains no unit, so is contained in \(\mathfrak m\). This proves uniqueness. The argument also covers the trivial absolute value, when \(R=F\) and \(\mathfrak m=(0)\).
3. In part (c), the unique maximal ideal of \(R/M^n\) is \(M/M^n\).
Since \(n\geq1\), one has \(M^n\subseteq M\ne R\). Moreover, \[ (R/M^n)/(M/M^n)\cong R/M \] is a field, so \(M/M^n\) is maximal [@DF04].
Let \(\mathfrak n\) be any maximal ideal of \(R/M^n\), and let \(N\subset R\) be its inverse image under the quotient map. Then \(M^n\subseteq N\) and \(R/N\cong(R/M^n)/\mathfrak n\) is a field. In particular \(N\) is a proper prime ideal. For each \(m\in M\), one has \(m^n\in M^n\subseteq N\). Primality, applied repeatedly to this product of \(n\) copies of \(m\), gives \(m\in N\). Hence \(M\subseteq N\). Maximality of \(M\) and properness of \(N\) give \(N=M\), so \(\mathfrak n=M/M^n\). Thus no other maximal ideal exists.