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.

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