# Tagged Questions

**4**

votes

**1**answer

128 views

### $ mult(R/I) = d_1 \cdots d_r \quad \Rightarrow \quad f_1,\dots,f_r \quad \text{is a $R$-regular sequence?}$

We define multiplicity of a module M of dimension $d>0$ as $$mult(M) := lc (P_M) (d-1)!$$ where $P_M$ denotes the Hilbert polynomial of M. Equivalently, we have $mult(M) = Q_M(1)$, where $HP_M (z) ...

**4**

votes

**0**answers

223 views

### Ext groups of affine scheme

Let $A$ be a commutative ring. $\textbf{Spec}(A)$ is the the
spectrum of $A$. $M$ and $N$ are $A$-modules. $\tilde{M}$ and
$\tilde{N}$ are sheaves associated on $\textbf{Spec}(A)$
respectively.
I ...

**1**

vote

**2**answers

202 views

### Decomposition of a quotient module

Let $R=k[v,x,y,z]/I$, with $I=\langle v^2,z^2,xy,vx+xz,vy+yz,vx+y^2,vy-x^2\rangle$,and let
$f:R^2 \rightarrow R^2$ denote the map given by the matrix
$$M=\begin{pmatrix}
v & y \\
x & z
...

**4**

votes

**2**answers

354 views

### Homological characterization of smooth maps

Let $A \to B$ be a finitely generated homomorphism between two commutative noetherian rings.
As far as I understand, in various generalizations of this situation, such a map is called smooth if $B$ ...

**0**

votes

**2**answers

152 views

### Is there a relationship between the right global dimensions of R and R[1/v]?

A few days ago I asked a similar question about Krull dimension and got fantastic answers. Unfortunately, for the application I have in mind (a question on ring spectra), Krull dimension doesn't ...

**30**

votes

**3**answers

3k views

### What the heck is the Continuum Hypothesis doing in Weibel's Homological Algebra?

On page 98 of Weibel's An Introduction to Homological Algebra he mentions that the ring $R = \prod_{i=1}^\infty \mathbb{C}$ has global dimension $\geq 2$ with equality iff the continuum hypothesis ...

**2**

votes

**2**answers

336 views

### Homological dimensions of module

$(A,\mathfrak{m})$ a Noetherian local ring, $M\neq 0$ a finitely generated $A$-module. As I understand, $\mbox{Ext }^{j}(A/\mathfrak{m}, M) = 0$ for $j<\mbox{depth }(M)$ and for $j>\mbox{inj. ...