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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.
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 ...
For any ideal I of a Boolean algebra B, the following are equivalent: I is a prime ideal. I is a maximal ideal, i.e. for any proper ideal J, if I is contained in J then I = J. For every element a of B, I contains exactly one of {a, ¬a}. This theorem is a well-known fact for Boolean algebras.
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 ...
In algebra and algebraic geometry, given a commutative Noetherian ring and an ideal in it, the n-th symbolic power of is the ideal = (/) ()where is the localization of at , we set : is the canonical map from a ring to its localization, and the intersection runs through all of the associated primes of /.