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Recently I have seen two definition of a generator module:

1) A generator for a category $C$ is an object $G$ such that for any two parallel morphisms $f,g:X \rightarrow Y$ with $f \neq g$, then there is a morphism $h: G \rightarrow X$ such that $fh \neq gh$. If we choose $C$ to be a module category, we get a definition of a generator module;

2)A module $X$ over an algebra $A$ is called a generator if $add(A) \subseteq add(X)$.

So are the two definitions equivalent? Thank you.

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    $\begingroup$ What is $add(A)$? $\endgroup$
    – Qiaochu Yuan
    Commented Sep 12, 2016 at 4:51
  • $\begingroup$ @Qiaochu Yuan Given a module X, $add(X)$ consists of all direct summands of direct sums of finitely many copies of X. $\endgroup$
    – Xiaosong Peng
    Commented Sep 12, 2016 at 7:10
  • $\begingroup$ I think the second condition is stronger, but I don't have a counterexample. The problem is that it may be necessary to take infinite direct sums; I have in mind something like $A = \text{End}(V)$ for $V$ an infinite-dimensional vector space and $X = V$, but I haven't checked this example. $\endgroup$
    – Qiaochu Yuan
    Commented Sep 12, 2016 at 7:47
  • $\begingroup$ @ Qiaochu Yuan When A is a finite-dimensional k-algebra, and all modules of A is finitely generated, I find the second can induce the first. But I don't know the general case. But still thank you for your help. $\endgroup$
    – Xiaosong Peng
    Commented Sep 12, 2016 at 8:18

1 Answer 1

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They are equivalent.

If any object of $\textrm{add}(X)$ satisfies (1) then so does $X$, and $A$ satisfies (1), so (2) implies (1).

If $G$ satisfies (1) then let $I$ be the set of homomorphisms $\alpha:G\to A$, let $G^{(I)}$ be the direct sum of copies of $G$ indexed by $I$, and let $\beta:G^{(I)}\to A$ be the map which restricts to $\alpha$ on the copy of $G$ corresponding to $\alpha$. Then $\beta$ is surjective, or (1) is contradicted by the natural map $A\to A/\textrm{im}(\beta)$ and the zero map.

Taking a splitting map $\gamma:A\to G^{(I)}$ of $\beta$, the image of $\gamma$ is contained in $G^{(J)}$ for some finite $J\subseteq I$, and so $A$ is a direct summand of $G^{(J)}$ and is therefore in $\textrm{add}(G)$. So (1) implies (2).

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  • $\begingroup$ @ Jeremy Rickard Since A can be seen as a projective A-module, so we can take a splitting map $\gamma : A \rightarrow G^{(I)}$, but if the index set I is infinite, why the image of $\gamma$ is contained in $G^{(J)}$ for some finite J? $\endgroup$
    – Xiaosong Peng
    Commented Sep 12, 2016 at 9:49
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    $\begingroup$ @Penson $\gamma(A)$ is the submodule of $G^{(I)}$ generated by $\gamma(1)$, and $\gamma(1)\in G^{(J)}$ for some finite $J$. (The important point is that $A$ is finitely generated as an $A$-module: if a finitely generated module is contained in a direct sum then it is contained in a finite subsum.) $\endgroup$
    – Jeremy Rickard
    Commented Sep 12, 2016 at 9:54
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    $\begingroup$ Get it! Thank you very much for your help. $\endgroup$
    – Xiaosong Peng
    Commented Sep 12, 2016 at 9:59

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