# Commutative Algebra, Notes 3: Local Rings

Definition 1.  The Jacobson radical of a commutative ring $R$ is defined as the intersection of all maximal ideals of $R.$

Proposition 2.  Let $R$ be a ring and $J$ be the set of all nonunits of $R.$  Then the following are equivalent.

1. $J$ is an ideal.
2. $R$ has a unique maximal ideal.

Proof.  If $J$ is an ideal, then it is maximal since any bigger ideal would contain a unit, and hence equal the whole ring.  Uniqueness follows by adjoining nonunits.  Conversely, note that the product of a unit and a nonunit, as well as the product of two nonunits, must be a nonunit.  If $R$ has a unique maximal ideal, then note that if $x,y\in J$ and $x+y$ is a unit, then there is a unit $u$ such that $ux+uy=1.$  Hence the ideals generated by $x$ and $y$ are relatively prime.  Hence no maximal ideal can contain both $(x)$ and $(y),$ so uniqueness is violated.

Definition 3.  $R$ is a local ring if it satisfies the above conditions.

If $R$ is a local ring, then of course $J$ is the Jacobson radical of $R.$  Note that if $\mathfrak{p}$ is a prime ideal in $R,$ then its complement is a multiplicative subset.

Definition 4.  We define $R_\mathfrak{p}=(R-\mathfrak{p})^{-1}R.$  We may call $R_\mathfrak{p}$ the localization of $R$ at $\mathfrak{p}.$  We define the globalization of $R$ as the ring

$\displaystyle G(R)=\bigoplus_{\mathfrak{m}\subset R}R_\mathfrak{m}$

where $\mathfrak{m}$ is a maximal ideal in $R.$

So if $R$ is a local ring, then its globalization is just $R$ since the one factor $R_J$  would invert already invertible elements.

Proposition 5.  The set $\Omega(R)$ of maximal ideals of $R$ is finite iff $R/J(R)$ is a finite direct sum of fields (where $J(R)$ is the Jacobson radical of $R$).

Proof.  $R/\mathfrak{m}$ is a field for each maximal ideal $\mathfrak{m}$ in $R.$  Moreover if $|\Omega(R)|<\infty,$ then $R$ maps onto $\oplus_{i=1}^nR/\mathfrak{m}_i$ via the canonical mapping.  The kernel is clearly $J(R),$ so we have the quotient as a finite sum of fields.  Conversely, if $R/R(J)$ is a finite direct sum of fields, then it has a finite number of ideals, and hence a finite number of maximal ideals.  Since each maximal ideal in $R$ contains $R(J),$ these are just the preimages of the maximal ideals of $R/R(J)$ under the map $R\to R/R(J),$ of which there are finitely many.

Definition 6.  A  ring $R$ is semilocal if it is a finite direct sum of local rings.

Hence if $|\Omega(R)|<\infty,$ then $R/R(J)$ is semilocal, as fields are local.  Now suppose $R=\oplus_{i=1}^nM_i$ is a semilocal ring.  So each $M_i$ is local, with a maximal ideal $\mathfrak{m}_i.$  We have a canonical epimorphism

$\displaystyle f:R\to\bigoplus_{i=1}^nM_i/\mathfrak{m}_i$

whose kernel is $J(R).$  Hence if $R$ is semilocal, then $R/R(J)$ is semilocal (and in particular semisimple (as an $R$-module), which is the traditional definition of $R$ being semilocal).

Proposition 7.  Let $\Pi$ be a finite set of prime ideals of $R.$  Let us define

$\displaystyle S_\Pi=\bigcap_{\mathfrak{p}_i\in\Pi}(R-\mathfrak{p}_i).$

Then if $R_\Pi=R[S_\Pi^{-1}],$ $R_\Pi/J(R_\Pi)$ is semilocal (i.e. $R_\Pi$ is semilocal in the traditional sense), and $R_\Pi$ has maximal ideals $\mathfrak{q}_i[S_\Pi^{-1}]$ where $\mathfrak{q}_i$ are the maximal elements of $\Pi.$

Proof.  A maximal ideal in $R_\Pi$ must be contained in $\cup_{\mathfrak{p}_i\in\Pi}\mathfrak{p}_i,$ otherwise it contains a unit.  Hence it must be contained in one of the $\mathfrak{p}_i,$ and hence it must be a $\mathfrak{q_i}.$  So $\mathfrak{q}_i[S_\Pi^{-1}]$ are the maximal ideals of $R_\Pi.$  Hence by Proposition 5 $R_\Pi/J(R_\Pi)$ is semilocal.

[1]  Bourbaki, N.  Commutative Algebra, Ch 1-7.  Springer-Verlag.  1989.