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Proposition. nilpotent maps form a prime ideal [math-6W2V]

Definition. Coprimary [local-0]

Let AA be a Noetherian ring. A nonzero finitely generated AA-module MM is coprimary if for all aAa \in A, the multiplication map (reuse element aa to denote it)

Broken pipe (os error 32)

is injective or nilpotent.

If MM is coprimary, then the set

P:={aAa is nilpotent}P := \{ a \in A \mid a \text{ is nilpotent} \}

forms a prime ideal in AA.

Proof. [local-1]

To check PP is a prime ideal, we want to check that if a∉Pa \not\in P and b∉Pb \not\in P then ab∉Pab \not\in P.

Because MM is coprimary, so such a,ba, b are injective, and

(ab)(x)=a(b(x))(ab)(x) = a(b(x))

the composition of injective maps is injective, hence ab∉Pab \not\in P.