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Steven Landsburg
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Regarding fields, you might already know this, but if you take $p=0$, then $Ch^{-q}(Spec(k),-q)=K_{-q}^M(k)$, whereas $H^0(Spec(k),-q)=K_{-q}^Q(k)$$H^0(Spec(k),\mathscr{K_{-q} })=K_{-q}^Q(k)$, where $K^M$ and $K^Q$ are Milnor and Quillen $K$-theory. These won't agree in general; for example, for finite fields the higher Milnor $K$-groups are all zero.

Regarding fields, you might already know this, but if you take $p=0$, then $Ch^{-q}(Spec(k),-q)=K_{-q}^M(k)$, whereas $H^0(Spec(k),-q)=K_{-q}^Q(k)$, where $K^M$ and $K^Q$ are Milnor and Quillen $K$-theory. These won't agree in general; for example, for finite fields the higher Milnor $K$-groups are all zero.

Regarding fields, you might already know this, but if you take $p=0$, then $Ch^{-q}(Spec(k),-q)=K_{-q}^M(k)$, whereas $H^0(Spec(k),\mathscr{K_{-q} })=K_{-q}^Q(k)$, where $K^M$ and $K^Q$ are Milnor and Quillen $K$-theory. These won't agree in general; for example, for finite fields the higher Milnor $K$-groups are all zero.

Source Link
Steven Landsburg
  • 23k
  • 5
  • 95
  • 153

Regarding fields, you might already know this, but if you take $p=0$, then $Ch^{-q}(Spec(k),-q)=K_{-q}^M(k)$, whereas $H^0(Spec(k),-q)=K_{-q}^Q(k)$, where $K^M$ and $K^Q$ are Milnor and Quillen $K$-theory. These won't agree in general; for example, for finite fields the higher Milnor $K$-groups are all zero.