I was told that commutative ring category is not additive category all the time and it is thus not abelian category.

A category is additive if $Hom(A,B)$ is abelian group for all objects $A,B$ and morphisms satisfy distributive laws, it has zero objects and it has finite coproduct and products.

It seems commutative ring satisfies most of the requirement except $Hom(A,B)$ being abelian group.

$Hom(A,B)$ is not abelian group by taking any $f\in Hom(A,B)$. If $f$ has inverse $g$, I can check $g(a_1a_2)=f(a_1a_2)$ which never cancels out with $f(a_1a_2)$.

It definitely satisfies distributive laws of morphisms.(Wrong. Distributive law fails due to not fixing $1$.)

Zero object is 0 ring as identity element is $0$ in 0 ring.

Coproduct is tensor and product is direct sum.

Does commutative ring category fail only $Hom(A,B)$ being abelian group requirement?

  • $\begingroup$ What do you mean by "the morphisms satisfy distributive laws" I would interpret that as saying that for appropriate morphisms $f$, $g$, $g'$, $f(g+g')=fg+fg'$ and the symmetrical identity for a different set of appropriate morphisms. However, this assumes already that the hom sets are abelian groups. Hence my confusion, I don't see how morphisms can be distributive in a category which doesn't have hom sets that are abelian groups. $\endgroup$ – jgon Jun 24 '17 at 22:31
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    $\begingroup$ You can't add two morphisms because you want $f(1) = 1$ which is not an additive condition. $\endgroup$ – user171326 Jun 24 '17 at 22:33
  • $\begingroup$ Also the sentence that begins "$\operatorname{Hom}(A,B)$ is abelian group$\ldots$" confuses me. As in I do not understand at all what is meant. $\endgroup$ – jgon Jun 24 '17 at 22:33
  • $\begingroup$ @jgon I am not aware that distributive law implies $Hom$ sets are abelian group. I only thought $Hom$ is not abelian group as there may be no inverse elements. Sorry my bad. $Hom(A,B)$ is not abelian group for $A,B$ commutative rings. $\endgroup$ – user45765 Jun 24 '17 at 22:34
  • $\begingroup$ @jgon Yes. I think distributive law of morphisms does not work here for sure as $1$ needs to be fixed. Why distributive law alone would give rise to abelian group structure? $\endgroup$ – user45765 Jun 24 '17 at 22:37

The ring $0$ is not the zero object. Indeed a zero object is an object which is both terminal and initial but in rings with unit the initial object is $\mathbb{Z}$ and the terminal one is $0$. Moreover, products and coproducts do not coincide. This category will never be an additive category.


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