Question
If $R=S[\Gamma]$, use Problem 4 to prove that $\mathrm{gl} \operatorname{dim} R \geq \mathrm{gldim} S$
Step 1
The global dimension of a ring R, denoted by gldim R, is the supremum of the projective dimensions of all R-modules. In other words, it is the largest projective dimension that can be achieved by any R-module. Show more…
Show all steps
Your feedback will help us improve your experience
Chai Santi and 67 other educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Show that Eq. (15.14), $$ g(r)=\frac{(N+r-1) !}{r !(N-1) !} $$ is approximately equal to $r^{N}$ in the limit $r \gg N \gg 1 .$ [Hint: Use Stirling's approximation: $\ln N ! \approx N \ln N-N$. You will need to make several approximations along the way, including, for instance, $\ln (N+r-1) \approx \ln (r) .]$
Statistical Mechanics
The Origin of the Boltzmann Relation
Prove that the sum of a convergent geometric series: $1+r+r^{2}+\cdots$ must be greater than $\frac{1}{2}$
Infinite Series
Repeat the preceding problem if $\mathrm{AB}$ is an adiabat and $\gamma=1.4$
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD