Answer to Question #210321 in Abstract Algebra for KOMAL

Question

If R is a commutative noetherian ring and P is a prime ideal of R, then prove that an R-module is P-primary if and only if each nonzero submodule of M is subisomorphic to R/P.

Answer 1

Consider the family f of annihilators of ideals r(x) for non zero x ∈M. Being a family of ideals of noetherian ring R, f has a maximal element r(x) say. We will show that P=r(x) is prime ideal of R. For it let ab∈r(x), a∉r(x). As ab∈r(x) ⇒ (ab)x = 0. Since xa ≠ 0, therefore, b(xa) = 0 ⇒ b∈r(xa). More over for t∈r(xa) ⇒ t(xa)=0 ⇒ (ta)x=0 ⇒ r(xa) ∈ f. Clearly r(x) ⊆ r(xa). Thus the maximality of r(x) in f implies that r(xa)=r(x) i.e. b∈r(x). Hence r(x) is prime ideal of R. Define a mapping from R to xR by θ(r)=xr. Then it is an homomorphism from R to xR. Ker θ ={ r∈R | xr=0}. Then Kernal θ = r(x). Hence by fundamental theorem on homomorphism, R/ r(x) ≅ xR = R/P. Therefore R/P is embeddable in M. Hence [R/P]=[R/Q]. this implies that there exist cyclic submodules xR and yR of R/P and R/Q respectively such that xR≅yR. But then R/P≅R/Q, which yields P=Q. It proves the theorem.