The study of local rings is central to commutative algebra. As we will see, life is easier in a local ring, so much so that we often want to localize so we can be in a local ring. A lot of the things we will say in this chapter have graded analogues: in some ways, \(\N\)-graded \(k\)-algebras and their homogeneous ideals behave like a local ring, where the homogenous maximal ideal plays the role of the maximal ideal.
Definition10.2.Local Ring.
A ring \(R\) is a local ring if it has exactly one maximal ideal. We often use the notation \((R,\fm)\) to denote \(R\) and its maximal ideal, or \((R,\fm,k)\) to also specify the residue field \(k=R/\fm\text{.}\)
Some people reserve the term local ring for a noetherian local ring, and call what we have defined a quasilocal ring; we will not follow this convention here.
Lemma10.3.Local if Nonunits Form an Ideal.
A ring \(R\) is local if and only if the set of non-units of \(R\) forms an ideal.
Proof.
If \(R\) is a local with maximal ideal \(\fm\text{,}\) then every nonunit must be in \(\fm\text{,}\) and \(\fm\) contains no nonunits, so \(\fm\) must be the set of nonunits. Conversely, if the set of nonunits is an ideal, that must be the only maximal ideal, since any other element in \(R\) is a unit.
Example10.4.\(\Z/(p^n)\) is Local.
The ring \(\Z/(p^n)\) is local with maximal ideal \((p)\text{.}\)
Solution.
Let \(R = \mathbb{Z}/(p^n)\) for some prime \(p\) and \(n\geqslant 1\text{.}\) We want to show that \(R\) has a unique maximal ideal.
Recall that a proper ideal \(M\) in \(R\) is maximal if and only if \(R/M\) is a field. Let \(M\) be a proper ideal of \(R\text{.}\) Since \(\mathbb{Z}\) is a PID, it follows that \(M = (m)\) for some \(0 \leqslant m \leqslant p^n-1\text{.}\)
Now, note that \(R/M\) is an integral domain since \(M\) is a prime ideal of \(R\text{.}\) By the [provisional cross-reference: cite] [[Mathematics/Number Theory/Results/Theorem - Sunzi’s Remainder Theorem|Chinese Remainder Theorem]], we have the following isomorphism of rings:
where \(n_1,\ldots,n_k\) are positive integers such that \(n_1 + \cdots + n_k = n\text{,}\) and \(k\) is the number of distinct prime factors of \(m\text{.}\)
Since \(R/M\) is a product of fields, it follows that \(M\) is maximal in \(R\text{.}\) Hence, \(R\) has a unique maximal ideal \((p)\text{,}\) since \((p)\) is the only proper ideal of \(R\) that contains \(p\text{.}\)
Example10.5.Power Series Ring over Field is Local.
The ring of power series \(k[[ \vx ]]\) over a field \(k\) is local.
Solution.
Indeed, one can show that a power series has an inverse if and only if its constant term is nonzero; this can be done explicitly, by writing down the conditions for a power series to be the inverse of another. The unique maximal ideal is \((\vx)\text{.}\) More generally, \([[k x_1, \ldots, x_d ]]\) is local with maximal ideal \((x_1, \ldots, x_d)\text{.}\)
Example10.6.Z_(p) is Local.
The ring \(\Z_{(p)} = \{ \frac{a}{b} \in \Q \ | \ p\nmid b \text{ when in lowest terms}\}\) is a local ring with maximal ideal \((p)\text{.}\)
Example10.7.Polynomial Rings are not Local.
A polynomial ring over a field is certainly not local; we have seen it has so many maximal ideals!
Definition10.8.Ring Characteristic.
Let R be a commutative ring. The characteristic of \(R\text{,}\) written \(\char(R)\text{,}\) is the unique non-negative generator of the kernel of the unique ring homomorphism \(\varphi: \Z \to R\text{.}\) (Recall \(\varphi(n) = n \cdot 1_R\text{.}\))
Equivalently, \(\char(R)\) is the smallest positive integer \(n\) such that \(n\cdot 1_R=\underbrace{1_R+\cdots+1_R}_{n}=0_R\text{,}\) if such and integer exists, and \(\char(R) = 0\) otherwise.
Proposition10.9.Characteristic in Local Rings.
Let \((R,\fm,k)\) be a local ring. Then one of the following holds:
\(\char(R)=\char(k)=0\text{.}\) We say that \(R\) has equal characteristic zero.
\(\char(R)=0\text{,}\)\(\char(k)=p\) for a prime \(p\text{,}\) so \(R\) has mixed characteristic\((0,p)\)
\(\char(R)=\char(k)=p\) for a prime \(p\text{,}\) so \(R\) has equal characteristic\(p\text{.}\)
\(\char(R)=p^n\text{,}\)\(\char(k)=p\) for a prime \(p\) and an integer \(n>1\text{.}\) If \(R\) is reduced, then one of the first three cases holds.
Proof.
Since \(k\) is a quotient of \(R\text{,}\) the characteristic of \(R\) must be a multiple of the characteristic of \(k\text{,}\) since the map \(\Z\longrightarrow k\) factors through \(R\text{.}\) We must think of \(0\) as a multiple of any integer for this to make sense. Now \(k\) is a field, so its characteristic is \(0\) or \(p\) for a prime \(p\text{.}\) If \(\char(k)=0\text{,}\) then necessarily \(\char(R)=0\text{.}\) If \(\char(k)=p\text{,}\) we claim that \(\char(R)\) must be either \(0\) or a power of \(p\text{.}\) Indeed, if we write \(\char(R)=p^n \cdot a\) with \(a\) coprime to \(p\text{,}\) note that \(p\in \fm\text{,}\) so if \(a \in \fm\text{,}\) we have \(1\in (p,a)\subseteq \fm\text{,}\) which is a contradiction. Since \(R\) is local, this means that \(a\) is a unit. But then, \(p^n a=0\) implies \(p^n=0\text{,}\) so the characteristic must be \(p^n\text{.}\)
Remark10.10.
If \(R\) is an \(\N\)-graded \(k\)-algebra with \(R_0 = k\text{,}\) and \(\fm = \bigoplus_{n > 0} R_0\) is the homogeneous maximal ideal, \(R\) and \(\fm\) behave a lot like a local ring and its maximal ideal, and we sometimes use the suggestive notation \((R,\fm)\) to refer to it. Many properties of local rings also apply to the graded setting, so given a statement about local rings, you might take it as a suggestion that there might be a corresponding statement about graded rings — a statement that, nevertheless, still needs to be proved. There are usually some changes one needs to make to the statement; for example, if a theorem makes assertions about the ideals in a local ring, the corresponding graded statement will likely only apply to homogeneous ideals, and a theorem about finitely generated modules over a local ring will probably translate into a theorem about graded modules in the graded setting.
