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If R is a unital commutative ring with an ideal m, then k = R/m is a field if and only if m is a maximal ideal. In that case, R/m is known as the residue field. This fact can fail in non-unital rings. For example, is a maximal ideal in , but / is not a field. If L is a maximal left ideal, then R/L is a simple left R-module.
A m is integrally closed for every maximal ideal m. 1 → 2 results immediately from the preservation of integral closure under localization; 2 → 3 is trivial; 3 → 1 results from the preservation of integral closure under localization, the exactness of localization , and the property that an A -module M is zero if and only if its ...
Frequently, is a local ring and is then its unique maximal ideal. In abstract algebra, the splitting field of a polynomial is constructed using residue fields. Residue fields also applied in algebraic geometry , where to every point x {\displaystyle x} of a scheme X {\displaystyle X} one associates its residue field k ( x ) {\displaystyle k(x ...
Assume the ideal M is maximal with respect to disjointness from the filter F. Suppose for a contradiction that M is not prime, i.e. there exists a pair of elements a and b such that a ∧ b in M but neither a nor b are in M. Consider the case that for all m in M, m ∨ a is not in F.
m is a minimal prime over (x 1, ..., x d). The radical of (x 1, ..., x d) is m. Some power of m is contained in (x 1, ..., x d). (x 1, ..., x d) is m-primary. Every local Noetherian ring admits a system of parameters. [1] It is not possible for fewer than d elements to generate an ideal whose radical is m because then the dimension of R would ...
The maximal ideal of () is the principal ideal generated by 2, i.e. (), and the "unique" irreducible element (up to units) is 2 (this is also known as a uniformizing parameter). Note that Z ( 2 ) {\displaystyle \mathbb {Z} _{(2)}} is the localization of the Dedekind domain Z {\displaystyle \mathbb {Z} } at the prime ideal generated by 2.