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edited Jul 15, 2023 at 18:30

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Disclaimer: I'm a relative beginner in this area. I'm trying to prove that if one has a commutative ring $R$ and a prime number $p$, then there is an exact sequence of the form $$\DeclareMathOperator\THH{THH} 0 \rightarrow \pi_n(\THH(R)_{h\mathbb{T}})/p\pi_n(\THH(R)_{h\mathbb{T}}) \rightarrow \pi_n(\THH(R)_{h\mathbb{T}})\otimes \mathbb{F}_p \rightarrow \pi_{n-1}(\THH(R)_{h\mathbb{T}})[p] \rightarrow 0. $$$$\DeclareMathOperator\THH{THH} 0 \rightarrow \pi_n(\THH(R)_{h\mathbb{T}})/p\pi_n(\THH(R)_{h\mathbb{T}}) \rightarrow \pi_n(\THH(R; \mathbb{F}_p)_{h\mathbb{T}}) \rightarrow \pi_{n-1}(\THH(R)_{h\mathbb{T}})[p] \rightarrow 0. $$ My basic intuition is that arguing formally should go quite far. It would suffice to show that $$ \THH(R)_{h\mathbb{T}} \xrightarrow{\times p} \THH(R)_{h\mathbb{T}} \rightarrow \THH(R)_{h\mathbb{T}} \otimes \mathbb{F}_p $$$$ \THH(R)_{h\mathbb{T}} \xrightarrow{\times p} \THH(R)_{h\mathbb{T}} \rightarrow \THH(R; \mathbb{F}_p)_{h\mathbb{T}} $$ is a fibration, for one could then take the long exact sequence in homotopy groups and then do some manipulations to quickly obtain the result. I'm not really sure how to show that one gets a fibration out of this however, or if one should take an entirely different approach. It does resemble the standard exact sequence one gets in the Bockstein spectral sequence, except $\THH(R)_{h\mathbb{T}}$ is far from needing to be torsion-free and that wouldn't help give a fibration anyway. Thanks in advance.

Disclaimer: I'm a relative beginner in this area. I'm trying to prove that if one has a commutative ring $R$ and a prime number $p$, then there is an exact sequence of the form $$\DeclareMathOperator\THH{THH} 0 \rightarrow \pi_n(\THH(R)_{h\mathbb{T}})/p\pi_n(\THH(R)_{h\mathbb{T}}) \rightarrow \pi_n(\THH(R)_{h\mathbb{T}})\otimes \mathbb{F}_p \rightarrow \pi_{n-1}(\THH(R)_{h\mathbb{T}})[p] \rightarrow 0. $$ My basic intuition is that arguing formally should go quite far. It would suffice to show that $$ \THH(R)_{h\mathbb{T}} \xrightarrow{\times p} \THH(R)_{h\mathbb{T}} \rightarrow \THH(R)_{h\mathbb{T}} \otimes \mathbb{F}_p $$ is a fibration, for one could then take the long exact sequence in homotopy groups and then do some manipulations to quickly obtain the result. I'm not really sure how to show that one gets a fibration out of this however, or if one should take an entirely different approach. It does resemble the standard exact sequence one gets in the Bockstein spectral sequence, except $\THH(R)_{h\mathbb{T}}$ is far from needing to be torsion-free and that wouldn't help give a fibration anyway. Thanks in advance.

Disclaimer: I'm a relative beginner in this area. I'm trying to prove that if one has a commutative ring $R$ and a prime number $p$, then there is an exact sequence of the form $$\DeclareMathOperator\THH{THH} 0 \rightarrow \pi_n(\THH(R)_{h\mathbb{T}})/p\pi_n(\THH(R)_{h\mathbb{T}}) \rightarrow \pi_n(\THH(R; \mathbb{F}_p)_{h\mathbb{T}}) \rightarrow \pi_{n-1}(\THH(R)_{h\mathbb{T}})[p] \rightarrow 0. $$ My basic intuition is that arguing formally should go quite far. It would suffice to show that $$ \THH(R)_{h\mathbb{T}} \xrightarrow{\times p} \THH(R)_{h\mathbb{T}} \rightarrow \THH(R; \mathbb{F}_p)_{h\mathbb{T}} $$ is a fibration, for one could then take the long exact sequence in homotopy groups and then do some manipulations to quickly obtain the result. I'm not really sure how to show that one gets a fibration out of this however, or if one should take an entirely different approach. It does resemble the standard exact sequence one gets in the Bockstein spectral sequence, except $\THH(R)_{h\mathbb{T}}$ is far from needing to be torsion-free and that wouldn't help give a fibration anyway. Thanks in advance.