1
$\begingroup$

As described here and here the singular homology of a topological space $X$ with coefficients in a ring $R$ is given by a bunch of $R$-modules $H_n(X,R)$.

However, sometimes I see people talking about the homology groups $H_n(X,A)$ of a topological space $X$ with coefficients in an abelian group $A$. For instance, here it says:

In what follows, the coefficient group $A$ is sometimes not written. It is common to take $A$ to be a commutative ring $R$; then the cohomology groups are $R$-modules. A standard choice is the ring $\mathbb Z$ of integers.

My first confusion is: if $A$ is supposed to be an abelian group, how can one instantiate $A$ with a commutative ring $R$? A commutative ring is a completely different type of object than an abelian group.

Question: How does singular homology with coefficients in a ring relate to singular homology with coefficients in an abelian group? Is one of these concepts more general?

$\endgroup$
2
  • 2
    $\begingroup$ A ring is an abelian group under addition? $\endgroup$ Commented Dec 25, 2021 at 16:56
  • 1
    $\begingroup$ An Abelian group is a Z-module, so you take R to be Z. In general, for the definition of $C_n(X,\mathbb Z)$you define a simplex to be a map $\Delta^n \to X$ and then take free $\mathbb Z$-module on that set of simplices. More generally, you take the free $R$-module on those simplices and proceed exactly as before. The free R-module is formed by taking formal R linear combinations of those simplices $\endgroup$ Commented Dec 25, 2021 at 17:11

1 Answer 1

3
$\begingroup$

If $R$ is a ring, it has an underlying abelian group : just forget the multiplication and the $1$. It is common to denote this underlying abelian group by the same symbol (here $R$) and use phrases such as "we view $R$ as an abelian group".

In particular, singular homology with coefficients in a ring is a special case of singular homology with coefficients in an abelian group.

However, the extra structure on $R$ affords extra structure on $H_n(X;R)$, namely that of an $R$-module . More generally, if $M$ is an $R$-module, then it also has an underlying abelian group and you can take $H_n(X;M)$ : this has the extra structure of an $R$-module too.

Overall, people tend to remove "forgetful" functors from the notation (so use the same letter for an $R$-module and its underlying abelian group, a ring and its underlying abelian group, etc.), at least the common ones.

$\endgroup$
3
  • $\begingroup$ Thanks! That answers my question. One additional question: how do these two approaches (coefficients in abelian groups vs. in rings) relate to the machinery of homological algebra? It seems in homological algebra the object $T$ in an expression "$H_n(X,T)$" should always be an object of an abelian category. (See I.§6.1 and II.§5.1 in Gelfand and Manin's Methods of Homological Algebra.) However, the category of rings isn't an abelian category. So taking coefficients in rings is unnatural from the point of view of homological algebra, right? $\endgroup$ Commented Dec 27, 2021 at 11:32
  • 1
    $\begingroup$ $H_n(X;R)$ is an $R$-module, and does not have a natural ring structure . In fact, you can really think of $R$ as an $R$-module in $H_n(X;R)$, it's just the "most important one" (for an $R$-module $M$, $H_n(X;M)$ can somehow be "deduced" from $H_n(X;R)$ via things like the universal coefficient theorem when $R$ is a PID) $\endgroup$ Commented Dec 27, 2021 at 11:36
  • $\begingroup$ Thanks, that makes much sense! $\endgroup$ Commented Dec 27, 2021 at 11:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .