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Since a one-sided maximal ideal A is not necessarily two-sided, the quotient R/A is not necessarily a ring, but it is a simple module over R. If R has a unique maximal right ideal, then R is known as a local ring, and the maximal right ideal is also the unique maximal left and unique maximal two-sided ideal of the ring, and is in fact the ...
In commutative algebra, the filtration on a commutative ring R by the powers of a proper ideal I determines the Krull (after Wolfgang Krull) or I-adic topology on R. The case of a maximal ideal I = m {\displaystyle I={\mathfrak {m}}} is especially important, for example the distinguished maximal ideal of a valuation ring .
Let R be a ring, and let I be a proper ideal of R. Then there is a maximal ideal of R containing I. The statement of the original theorem can be obtained by taking I to be the zero ideal (0). Conversely, applying the original theorem to R/I leads to this result. To prove the "stronger" result directly, consider the set S of all proper ideals of ...
For an example more geometrical in nature, take the ring R = {f/g : f, g polynomials in R[X] and g(0) ≠ 0}, considered as a subring of the field of rational functions R(X) in the variable X. R can be identified with the ring of all real-valued rational functions defined (i.e. finite) in a neighborhood of 0 on the real axis (with the ...
Let J(R) be the Jacobson radical of R. If U is a right module over a ring, R, and I is a right ideal in R, then define U·I to be the set of all (finite) sums of elements of the form u·i, where · is simply the action of R on U. Necessarily, U·I is a submodule of U. If V is a maximal submodule of U, then U/V is simple.
Still more generally, if A is a regular local ring, then the formal power series ring A[[x]] is regular local. If Z is the ring of integers and X is an indeterminate, the ring Z[X] (2, X) (i.e. the ring Z[X] localized in the prime ideal (2, X) ) is an example of a 2-dimensional regular local ring which does not contain a field.
Given a linear operator T on a finite-dimensional vector space V, one can consider the vector space with operator as a module over the polynomial ring in one variable R = K[T], as in the structure theorem for finitely generated modules over a principal ideal domain. Then the spectrum of K[T] (as a ring) equals the spectrum of T (as an operator).
An integral domain where every finitely generated ideal is principal (that is, a Bézout domain) satisfies (ACCP) if and only if it is a principal ideal domain. [4] The ring Z+XQ[X] of all rational polynomials with integral constant term is an example of an integral domain (actually a GCD domain) that does not satisfy (ACCP), for the chain of ...