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Let $R$ be a commutative ring, $A$ a $R$-module and $n$ a natural number. Does there exist a CW complex $M(A,n)$ with $\tilde{H}_i(M(A,n),R)=0$ if $i\neq n$ and $\tilde{H}_n(M(A,n),R)\cong A$ as $R$-modules?

If $R=\mathbb{Z}$ one can choose a short exact sequence $$0\longrightarrow \bigoplus_{j\in J}\mathbb{Z}\longrightarrow\bigoplus_{i\in I}{\mathbb{Z}}\longrightarrow A\longrightarrow 0$$ and construct $M(A,n)$ by attaching $n+1$-cells indexed over $J$ to $\bigvee_{i\in I}S^{n}$ corresponding to the first homomorphism in the sequence.

If $R$ is any other ring, there occur several problems:

  1. If $R$ is not a PDI, submodules of free modules do not have to be free, so the existence of such a SES is not guaranteed.
  2. I don't know whether I can realize every ring element as $R$-degree of a map between spheres. It's easy for all elements of the subgroup generated by $1$, but I don't see how one can manage that for other elements.
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  • $\begingroup$ What $R$-module structure are you considering on $\tilde H_n(M(A,n))$? $\endgroup$ Jul 22 '15 at 8:01
  • $\begingroup$ (You can consider a free resolution of $A$ as an $R$-module and attach cells jus as you did, no?) $\endgroup$ Jul 22 '15 at 8:03
  • $\begingroup$ The usual one. Homology groups with coefficients in a ring $R$ have a canonical $R$-modules structure. I forgot to include the coefficients and edited the question accordingly. $\endgroup$
    – user256089
    Jul 22 '15 at 8:03
  • $\begingroup$ I don't see how. I addressed this in my two concerns in the question. $\endgroup$
    – user256089
    Jul 22 '15 at 8:04
  • $\begingroup$ "Irritated"?! Well, I guess I'll just leave you to it... $\endgroup$ Jul 22 '15 at 8:04
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Actually you cannot construct such $CW$-complex. Consider the map $d_k:C_k(X,R)\to C_{k-1}(X,R)$, the basis element ($k$-cell) $\alpha$ maps to $\sum [\alpha:\beta_i]\cdot\beta_i$, this is sum of $k-1$-cells with integer coefficients.

So, for example, if $R=\mathbb Z[t]$, you cannot get $H_n(X,R)=\mathbb Z[t]/(t)$.

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