Bourbaki

Exercise 11, § 2

Commutative Algebra, I, 43

Let EE be a right AA-module. EE is called pseudo-coherent if every finitely generated submodule of EE is finitely presented; every submodule of a pseudo-coherent module is pseudo-coherent. EE is called coherent if it is pseudo-coherent and finitely generated (and therefore finitely presented).

(a) Let 0→E′→E→E′′→00 \to E' \to E \to E'' \to 0 be an exact sequence of right AA-modules. Show that, if EE is pseudo-coherent (resp. coherent) and E′E' is finitely generated E′′E'' is pseudo-coherent (resp. coherent). Show that, if E′E' and E′′E'' are pseudo-coherent (resp. coherent), so is EE. Show that, if EE and E′′E'' are coherent, so is E′E' (use Exercise 6 and Lemma 9 of no. 8).

(b) Let EE be a coherent AA-module and E′E' a pseudo-coherent (resp. coherent) AA-module. Show that, for every homomorphism u:E→E′u : E \to E', Im⁡(u)\operatorname{Im}(u) and Ker⁡(u)\operatorname{Ker}(u) are coherent and that Coker⁡(u)\operatorname{Coker}(u) is pseudo-coherent (resp. coherent) (use (a)).

(c) Show that every direct sum (resp. every finite direct sum) of pseudo-coherent (resp. coherent) modules is a pseudo-coherent (resp. coherent) module.

(d) If EE is a pseudo-coherent module and M,NM, N are coherent submodules of EE, show that M+NM + N and M∩NM \cap N are coherent (use (a) and (c)).

(e) Suppose that AA is commutative. Show that, if EE is a coherent AA-module and FF a coherent (resp. pseudo-coherent) AA-module, Hom⁡A(E,F)\operatorname{Hom}_A(E, F) is a coherent (resp. pseudo-coherent) AA-module. (Reduce it to the case where FF is coherent and consider a finite presentation of EE, then use (b).)

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.