0
$\begingroup$

I'm reading a paper of Aidan Schofield "General Representations of Quivers" and he defines quiver representation over any commutative ring. See the below image.enter image description here

Towards the end, he has this representation $R_p\bigotimes\Lambda$ of the quiver $Q$. My questions are:

  1. is the tensor product taken over the field $\mathbb{K}$?
  2. what are the $\Lambda$-modules assigned at each vertex and what are the $\Lambda$-module morphisms assigned for each arrow for this representation $R_p\bigotimes\Lambda$?

(Here, $R_p$ is the usual representation of the quiver $Q$ where for each vertex you have a $\mathbb{K}$-vector space and for each arrow you have a $\mathbb{K}$-linear map between the corresponding vector spaces)

Improve this question
$\endgroup$
1
  • 2
    $\begingroup$ Cross-posted on MSE. Please don't post the same question on both sites at the same time, particularly without linking to the version on the other site. $\endgroup$
    – Jeremy Rickard
    Commented Aug 11, 2022 at 18:37

1 Answer 1

Reset to default
2
$\begingroup$

I think that $\Lambda$ should be a $k$-algebra and the tensor product should be over $k$. Then $R_p\otimes_k \Lambda$ makes sense as a representation of $Q$ over $\Lambda$, or equivalently as a $\Lambda Q$-module.

Namely, the Grassmannian $\mathcal{Gr}(R_p,\alpha)$ doesn't make sense as a scheme, but only as a $k$-scheme, determined by a functor from $k$-algebras to sets. In this case the functor sends a $k$-algebra $\Lambda$ to the set of summands of $R_p \otimes_k \Lambda$ of rank $\alpha$ (i.e., localizing at any maximal ideal of $\Lambda$, the projective $\Lambda$-module corresponding to a vertex $v$ should give a free module over the localization of rank $\alpha(v)$).

Improve this answer
$\endgroup$
3
  • $\begingroup$ Sorry, but localization part is not clear to me. Why are we localizing at a maximal ideal? Also, do we know what the $\Lambda-$ module map for each arrow of the quiver look like, given that we know the representation $R_p$? $\endgroup$
    – It'sMe
    Commented Aug 15, 2022 at 17:00
  • 2
    $\begingroup$ I should have said prime ideals. If $\Lambda$ is a commutative ring, the rank of a finitely generated projective $\Lambda$-module $P$ is defined to be $n$ if for each prime ideal $\mathfrak{p}$ in $\Lambda$, the module $P_{\mathfrak{p}}$ is free of rank $n$ over $\Lambda_\mathfrak{p}$. See for example math.stackexchange.com/questions/133333/… $\endgroup$
    – wcb
    Commented Aug 16, 2022 at 6:49
  • 2
    $\begingroup$ For the second question: $R = R_p$ is a representation of the quiver over $k$, so given by a vector space $R(v)$ for each vertex $v$ and a linear map $R(a):R(v)\to R(w)$ for each arrow $a:v\to w$. For a commutative $k$-algebra $\Lambda$, the representation $R\otimes_k \Lambda$ is given by the free $\Lambda$-module $R(v)\otimes_k \Lambda$ for each vertex and the $\Lambda$-module map $R(a)\otimes_k \Lambda:R(v) \otimes_k \Lambda \to R(w) \otimes_k \Lambda $ for each arrow $a:v\to w$. $\endgroup$
    – wcb
    Commented Aug 16, 2022 at 6:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .