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 $V$.

  • $\begingroup$ 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

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 $A$.

  • $\begingroup$ 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

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