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Any module over $R=k[x_1,\dots, x_r]$ has a finite resolutions of projective modules, and projective modules are free. So, in the Grothendieck group, the class of the module is a multiple of the class …

There is a localization sequence, as given in the reference mentioned in Angelo's comments. However the map is not always surjective. Perhaps an easy example is $X=Spec(R)$ where $R=k[x,y]_{(x,y)}/(xy …

One can define Chow groups over any Noetherian scheme $X$. Let $Z_iX$ be the free abelian group on the $i$-dimensional subvarieties (closed integral subschemes) of $X$. For any $i+1$-dimensional subsc …

Dear Charles, here is a more general statement: Let $R$ be a reduced affine algebra over $k$ of dimension 1. Then the set of singular points of $Spec(R)$ is finite. Proof: Let $V$ be the set of si …

Perhaps it is worth recording a simple example: let $X_k= \text{Spec}(k[x,y]/(x^2+y^2-1))$. Then $\text{Pic}(X_{\mathbb R}) = \mathbb Z/(2)$ while $\text{Pic}(X_{\mathbb C}) = \mathbb 0$, see Fossum's …

Let $I\subset R = k[x,y,z]$ be the defining ideal of your curve. Then $R/I$ has dimension one and no embedded components, so has projective dimension $2$ by the Auslander-Buchsbaum formula. Therefore …

Here is a more algebraic perspective on your question. If $X=\text{Spec} (R)$ is affine and $R$ is a Cohen-Macaulay algebra over some field (the following is true in more general setting), then $K_X$ …

I believe it was first proved by Kleiman (you only need to assume fixed dimension and degree, no need to assume a subvariety of some fixed $\mathbb P^n$), see Corollary 6.11 S. Kleiman, Exp XIII in …

In Section 7 of this paper the authors gave an example of a smooth 3-fold $X$ with a divisor $D$ such that: $\lim_{n \rightarrow \infty} \frac{h^0(\mathcal O_X(nD))}{n^3} = 6+ \frac{2\sqrt 3}{9}$ …

There is a huge amount of work on these kinds of questions (I am aware of them since one of my colleagues, Satya Mandal, works on related topics). For example, this recent paper: http://128.84.158. …

Here is a quick proof that any complete local Cohan-Macaulay ring of dimension $1$ and multiplicity $2$ is a hypersurface, so the answer to your last question is always yes. Call such ring $R$ with m …

Work locally, assume that $f: R\to S$ is a local homomorphism. Let $\operatorname{cmd}(R) =\operatorname{dim} R-\operatorname{depth} R$ (this is the so-called Cohen-Macaulay defect of $R$). Claim: $\o …

Dear anonymous, Here is an expansion of what Georges said in the comment. I will assume, as you wrote, that you are a beginner in AG but not in math. And please do not feel too bad about diverietti's …

Pete's answer gave a very good reason, and indeed contained what I planned to say at the beginning of this post. However, let me provide some mildly positive news, namely some cases when we are forced …

Well, if you read on to Chapter 2, exercise 6.3, then it is stated that: $$Cl(A) \cong Cl(X)/\mathbb Z[H]$$ here $[H]$ represents the hyperplane section. So the answer is yes. There is a less well-kno …