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Let $K$ be a finite extension of $\mathbb{Q}_p$. Let $F$ be a Lubin TateLubin–Tate formal group law defined over $K$ with endomorphism $f(T)$ corresponding to $\pi$ (a uniformizer of $K$). Then one can define the logarithm of $F$ to be $\lambda_F(T) = \lim_{n\rightarrow\infty}\pi^{-n}f^n(T)$. Then $\lambda_F(T) = T +$ higher degree terms$\lambda_F(T) = T + \text{higher-degree terms}$, so the logarithm is invertible under function composition. We define $\text{exp}_F(T)$$\operatorname{exp}_F(T)$ to be the inverse of $\lambda_F$ under composition. Then $\text{exp}_F(\lambda_F(T)) = T$$\operatorname{exp}_F(\lambda_F(T)) = T$. However I don't understand how this is possible if $\lambda_F$ is not one to one. In particular $\lambda_F$ sends all torsion points of $F$ to 0. Please let me know what is wrong here.

Let $K$ be a finite extension of $\mathbb{Q}_p$. Let $F$ be a Lubin Tate formal group law defined over $K$ with endomorphism $f(T)$ corresponding to $\pi$ (a uniformizer of $K$). Then one can define the logarithm of $F$ to be $\lambda_F(T) = \lim_{n\rightarrow\infty}\pi^{-n}f^n(T)$. Then $\lambda_F(T) = T +$ higher degree terms, so the logarithm is invertible under function composition. We define $\text{exp}_F(T)$ to be the inverse of $\lambda_F$ under composition. Then $\text{exp}_F(\lambda_F(T)) = T$. However I don't understand how this is possible if $\lambda_F$ is not one to one. In particular $\lambda_F$ sends all torsion points of $F$ to 0. Please let me know what is wrong here.

Let $K$ be a finite extension of $\mathbb{Q}_p$. Let $F$ be a Lubin–Tate formal group law defined over $K$ with endomorphism $f(T)$ corresponding to $\pi$ (a uniformizer of $K$). Then one can define the logarithm of $F$ to be $\lambda_F(T) = \lim_{n\rightarrow\infty}\pi^{-n}f^n(T)$. Then $\lambda_F(T) = T + \text{higher-degree terms}$, so the logarithm is invertible under function composition. We define $\operatorname{exp}_F(T)$ to be the inverse of $\lambda_F$ under composition. Then $\operatorname{exp}_F(\lambda_F(T)) = T$. However I don't understand how this is possible if $\lambda_F$ is not one to one. In particular $\lambda_F$ sends all torsion points of $F$ to 0. Please let me know what is wrong here.

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Neil Strickland
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Let $K$ be a finite extension of $\mathbb{Q}_p$. Let $F$ be a Lubin Tate formal group law defined over $K$ with endomorphism $f(T)$ corresponding to $\pi$ (a uniformizer of $K$). Then one can define the logarithm of $F$ to be $\lambda_F(T) = \lim_{n\rightarrow\infty}\pi^{-n}f^n(T)$. Then $\lambda_F(T) = T +$ higher degree terms, so the logarithm is invertible under function composition. We define $\text{exp}_F(T)$ to be the inverse of $\lambda_F$ under composition. Then $\text{exp}_F(\lambda_F(T)) = T$. However I don't understand how this is possible if $\lambda_F$ is not one to one. In particular $\lambda_F$ sends all torsion points of $F$ to 0. Please let me know what is wrong here.

Let $K$ be a finite extension of $\mathbb{Q}_p$. Let $F$ be a Lubin Tate formal group law defined over $K$ with endomorphism $f(T)$ corresponding to $\pi$ (a uniformizer of $K$). Then one can define the logarithm of $F$ to be $\lambda_F(T) = \lim_{n\rightarrow\infty}\pi^{-n}f^n(T)$. $\lambda_F(T) = T +$ higher degree terms, so the logarithm is invertible under function composition. We define $\text{exp}_F(T)$ to be the inverse of $\lambda_F$ under composition. Then $\text{exp}_F(\lambda_F(T)) = T$. However I don't understand how this is possible if $\lambda_F$ is not one to one. In particular $\lambda_F$ sends all torsion points of $F$ to 0. Please let me know what is wrong here.

Let $K$ be a finite extension of $\mathbb{Q}_p$. Let $F$ be a Lubin Tate formal group law defined over $K$ with endomorphism $f(T)$ corresponding to $\pi$ (a uniformizer of $K$). Then one can define the logarithm of $F$ to be $\lambda_F(T) = \lim_{n\rightarrow\infty}\pi^{-n}f^n(T)$. Then $\lambda_F(T) = T +$ higher degree terms, so the logarithm is invertible under function composition. We define $\text{exp}_F(T)$ to be the inverse of $\lambda_F$ under composition. Then $\text{exp}_F(\lambda_F(T)) = T$. However I don't understand how this is possible if $\lambda_F$ is not one to one. In particular $\lambda_F$ sends all torsion points of $F$ to 0. Please let me know what is wrong here.

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Alexey Ustinov
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