Length of finitely generated module over $0$-dimensional Gorenstein local ring
Please see attached images: Can someone slowly walk me through the details of the proof of Lemma 4.1 using basic commutative algebraic arguments and definitions?
1) Why $R[\mathfrak{m}] \cong k$? (where $k$ is the residue field of $R$)
2) I know length is additive on direct sums, but how is the RHS of the inequality additive as well, since it is a product?
3) Why does the inequality coincide when $R$ is the module over itself?
4) How does $R$ being Gorenstein of dimension $0$ imply that $R$ is the only indecomposable injective module, and why is its injective hull free as a result?
5) Why $M \subseteq \mathfrak{m}F$?
6) Why does $M=0$ imply the desired result?
Jbuck
152
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- closed
- 606 views
- $44.00
Related Questions
- Motorcycle Valve Clearance Calculation and Spacer Size Word Problem
- Confused on this graph question, not sure how to reduce it to linear and It looks too wonky to draw a best fit line, probably won't take long
- Five times the larger of two consecutive odd integers is equal to one more than eight times the smaller. Find the integers.
- Three questions on Vectors
- Linear Algebra - Matrices (Multiple Choice Question) (1st Year College)
- Graph the pair of equations in the same rectangular coordinate system: Y=-2x ; y=-2
- Mechanical principle help (maths)
- Linearly independent vector subsets.
