1
$\begingroup$

Let $\mathcal{A}$ be a small abelian category (additive category with AB1) and AB2)). We say $\mathcal{A}$ is artinian if for every $A\in\mathcal{A}$, every descending chain of subobjects of $A$ stabilize.

Is an artinian category necessarily noetherian? If this is not true, which conditions shall we impose to make this work?

Improve this question
$\endgroup$
7
  • 3
    $\begingroup$ It seems that this is not true. The abelian group $\mathbf Q_p/\mathbf Z_p$ has the property that every descending chain of subobjects stabilizes. The abelian subcategory of the category of abelian groups generated by this group seems to be provide a counterexample. (It is the full subcategory of abelian groups of finite rank which are $p$-torsion and divisible.) $\endgroup$
    – ACL
    Jan 21, 2017 at 15:36
  • $\begingroup$ @ACL or maybe just the subcategory generated by $\Bbb Z$ and ${\Bbb Z}_{(p)}/{\Bbb Z}$ - see an answer to a related question on math.SE $\endgroup$
    – მამუკა ჯიბლაძე
    Jan 21, 2017 at 16:33
  • 3
    $\begingroup$ Or the abelian category of finitely generated abelian groups is noetherian but not artinian, so its opposite category is artinian but not noetherian. $\endgroup$
    – Jeremy Rickard
    Jan 21, 2017 at 23:40
  • 2
    $\begingroup$ @მამუკაჯიბლაძე If $X_0<X_1<X_2<\cdots$ is a sequence of subobjects of $X$, then it induces a sequence $Y_0\twoheadrightarrow Y_1\twoheadrightarrow Y_2\twoheadrightarrow\cdots$ of quotients of $X$, where $Y_i=X/X_i$, and so in the opposite category a sequence $Y_0>Y_1>Y_2>\cdots$ of subobjects of $X$. $\endgroup$
    – Jeremy Rickard
    Jan 22, 2017 at 10:57
  • 1
    $\begingroup$ Thanks, I see now, so simple! Let me add that if one wishes your example can be viewed as the abelian subcategory of compact abelian groups (and continuous homomorphisms) generated by the circle group $\mathbb R/\mathbb Z$, made of products of finite groups and finite dimensional tori. $\endgroup$
    – მამუკა ჯიბლაძე
    Jan 22, 2017 at 12:52

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.