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Tom Goodwillie
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$HZ^nHZ$ is trivial for $n<0$. $HZ^0HZ$ is infinite cyclic generated by the identity operation. For $n>0$ the group is finite. So you know everything if you know what's going on locally at each prime. For $n>0$ the $p$-primary part is not just finite but killed by $p$, which means that you can extract it from the Steenrod algebra $H(Z/p)^{*}H(Z/p)$ and Bocksteins.

EDIT Here's the easier part: The integral homology groups of the space $K(Z,n)$ vanish below dimension $n$, and by induction on $n$ they are all finitely generated. Also $H_{n+k}K(Z,n)$ is independent of $n$ for roughly $n>k$, so that in this stable range $H_{n+k}K(Z,n)$ is $HZ_kHZ$, which is therefore finitely generated. This plus the computation of rational (co)homology gives that $HZ_kHZ$ is finite for $k>0$. Here's the funny part: Of course one expects there to be some elements of order $p^m$ for $m>1$ in the (co)homology of $K(Z,n)$, and in fact there are; the surprise is that stably this is not the case.

$HZ^nHZ$ is trivial for $n<0$. $HZ^0HZ$ is infinite cyclic generated by the identity operation. For $n>0$ the group is finite. So you know everything if you know what's going on locally at each prime. For $n>0$ the $p$-primary part is not just finite but killed by $p$, which means that you can extract it from the Steenrod algebra $H(Z/p)^{*}H(Z/p)$ and Bocksteins.

$HZ^nHZ$ is trivial for $n<0$. $HZ^0HZ$ is infinite cyclic generated by the identity operation. For $n>0$ the group is finite. So you know everything if you know what's going on locally at each prime. For $n>0$ the $p$-primary part is not just finite but killed by $p$, which means that you can extract it from the Steenrod algebra $H(Z/p)^{*}H(Z/p)$ and Bocksteins.

EDIT Here's the easier part: The integral homology groups of the space $K(Z,n)$ vanish below dimension $n$, and by induction on $n$ they are all finitely generated. Also $H_{n+k}K(Z,n)$ is independent of $n$ for roughly $n>k$, so that in this stable range $H_{n+k}K(Z,n)$ is $HZ_kHZ$, which is therefore finitely generated. This plus the computation of rational (co)homology gives that $HZ_kHZ$ is finite for $k>0$. Here's the funny part: Of course one expects there to be some elements of order $p^m$ for $m>1$ in the (co)homology of $K(Z,n)$, and in fact there are; the surprise is that stably this is not the case.

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Tom Goodwillie
  • 55.9k
  • 7
  • 151
  • 240

$HZ^nHZ$ is trivial for $n<0$. $HZ^0HZ$ is infinite cyclic generated by the identity operation. For $n>0$ the group is finite. So you know everything if you know what's going on locally at each prime. For $n>0$ the $p$-primary part is not just finite but killed by $p$, which means that you can extract it from the Steenrod algebra $H(Z/p)^{*}H(Z/p)$ and Bocksteins.