I have just learned the definition for $Ext_{A}^i(M,N)$ where $M,N$ are $A$-modules. However, in the book Quadratic Algebras by Polishchuk and Positselski they sometimes write $Ext_A^{ij}(M,N)$ (for graded $A,M,N$) and they call $j$ the internal grading but I don't understand what this $j$ does. Could someone help me understand what $Ext_A^{ij}(M,N)$ is?
Asked
Active
Viewed 571 times
1
-
If $M = \bigoplus_{i\in\mathbb Z}M_i$ and $N = \bigoplus_{i\in\mathbb Z} N_i$ are graded $A$-algebras, define $\hom^n_A(M,N) = {f\in \hom_A(M,N),|, f(M_i)\subseteq N_{i-n}\text{ for all $i\in\mathbb Z$}}$. This defines a graduation of $\hom_A(M,N) = \bigoplus_{n\in\mathbb Z} \hom^n_A(M,N)$. (Beware that there are different conventions, though.) This also induces a graduation on $\operatorname{Ext}_A^i(M,N)$ by taking graded projective/injective resolutions. – Claudius Feb 10 '17 at 15:14
-
Thank you! Some follow up questions , is a graded projective resolution a resolution in which all objects are graded modules? If so, can we always find a graded projective resolution for a graded module M? – frogorian-chant Feb 10 '17 at 15:21
-
I am not sure about the first question. You should define a graded module $P$ to be projective if the functor $\hom^0_A(P,-)$ is exact. This notion is equivalent with being a (graded) direct summand of a graded free module, where a graded free module is a direct sum of the form $\bigoplus_{i\in I}A[n_i]$ and $A[n_i]$ is the $n_i$-th shift, i. e. with $(A[n_i])j = A{j-n_i}$. There are enough graded free modules. – Claudius Feb 10 '17 at 15:45
-
Ok, thanks again! I will have to spend some time thinking about what you wrote=) – frogorian-chant Feb 10 '17 at 15:56
-
Your first question (whether a graded and projective module is also graded projective) is answered positively here. – Claudius Feb 10 '17 at 16:21