0
$\begingroup$

This is a proposition from Artin's Algebra (1st edition, Proposition 5.12):

The presentation matrix $A$ of an $R$-module is the same as deleting a column of zeros of $A$.

He proves it by stating that a column of zeros corresponds to the trivial relation which can be deleted.

I am quite confused.

1) To talk about relation, we must first have a set of elements in $V$, the relation here refers to what set of elements in $V$? My guess is the generating set of $V$, but in his theorem he is talking about a general module $V$ which might not have a generating set.

2) Even though we confine ourselves to finitely generated $R$-module $V$, does $A$ always correspond to a complete relation of some generating set in $V$? I do not see why this must be true if we are only given an isomorphism between $R^m/AR^n$ and $V$.

$\endgroup$
  • $\begingroup$ Please state the proposition in your question. Not everyone has this textbook. $\endgroup$ – André 3000 Mar 10 '16 at 21:13
  • $\begingroup$ The title already states the proposition. $\endgroup$ – Keith Mar 10 '16 at 21:21
  • 1
    $\begingroup$ The title should never be needed to understand a question. Please make the body of the question self-contained. $\endgroup$ – Tobias Kildetoft Mar 10 '16 at 21:35
1
$\begingroup$

In the proposition, $V$ is a fixed finitely-generated module and $A$ is a presentation matrix for $V$. As explained in the above paragraphs, this means that $V \cong R^m /AR^n$.

  1. The generators are the images under the above isomorphism of the (cosets of the) $m$ standard basis vectors in $R^m$. Also note that any module always has a generating set, namely, take the whole module as a generating set.

  2. Yes, $A$ gives you a complete set of relations. Any relation in $V$ corresponds, under the above isomorphism, to an element of $R^m$ (a relation vector) that belongs in $AR^n$. This means precisely that the relation vector is a linear combination of columns of $A$.

$\endgroup$
  • $\begingroup$ So if V can be presented by a matrix, it should always be finitely generated? $\endgroup$ – Keith Mar 10 '16 at 21:43
  • $\begingroup$ @Keith Yes, that is correct. This is because $V \cong R^m/AR^n$, which is a quotient of a finitely generated module and thus finitely generated (by the cosets of the generators). $\endgroup$ – Alex Provost Mar 10 '16 at 21:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.