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Let $p$ be a prime number, $\mathbb{Q}_p$ the field of $p$-adic numbers, and $\mathbb{C}_p$ the completion of the algebraic closure of $\mathbb{Q}_p$. Let $U_p$ be the units $(1+\mathfrak{m})$ of $\mathbb{C}_p$, the $p$-adic version of Baker theorem is:

Theorem (Brumer): Let $\alpha_1, \cdots, \alpha_n$ be elements of principal units $U_p$ which are algebraic over the rationals $\mathbb{Q}$ and their $p$-adic logarithms $\log_p(\alpha_1), \cdots, \log_p(\alpha_n)$ are linearly independent over $\mathbb{Q}$. Then $\log_p(\alpha_1), \cdots, \log_p(\alpha_n)$ are then linearly independent over the algebraic closure $A$ of $\mathbb{Q}$ in $\mathbb{Q}_p$.

I am confused if the above result is the true the $p$-adic version of Baker theorem or not because wikipedia says the Baker's theorem in $p$-adic setting is an open problem.

Anyway, I want to construct an example where the above theorem(Brumer) holds:

Fix $p=2$, and assume, $K=\mathbb Q_2(\sqrt{-1})$ with uniformizer $\pi=-1+\sqrt{-1}$ and "restrict" ourselves on the principal units $U=1+\mathfrak{m}_K$, where $\mathfrak{m}_K$ is the maximal ideal of $K$. Further, assume we know a basis $\{2\pi, 4\}$ of $\log_2(U)$, where $\log_2$ is the $2$-adic logarithm. Take $\alpha_1, \alpha_2, \cdots, \alpha_n$ in the principal units $U_1=1+\mathfrak{m}_K$ of the quadratic extension $K$ satisfying the Brumer theorem above i.e., are algebraic over the rationals $\mathbb{Q}$ and their $2$-adic logarithm $\log_2(\alpha_1), \cdots, \log_2(\alpha_n)$ are linearly independent over $\mathbb{Q}$ for $n=2$. My question:

Does the above theorem holds in the given situation? i.e., Are $\log_1(\alpha_1), \log_2(\alpha_2)$ linearly independent over the algebraic closure $A$ of $\mathbb{Q}$ in $\mathbb{Q}_2$ ?

We have given the basis $\{2 \pi, 4\}$ of $\log_2(U)$, and therefore $\log_2(\alpha_1), \log_2(\alpha_2)$ all can be obtained by linear combination of the basis $\{2\pi, 4 \}$, where the uniformizer is $\pi=-1+\sqrt{-1}$. So we can write \begin{align} &\log_2(\alpha_1)=2 \pi a_1+4b_1, \log_2(\alpha_2)=2 \pi a_2+4b_2 \\ \Rightarrow &\log_2(\alpha_1)=(-2a_1+4b_1)+\sqrt{-1} \cdot 2a_1, \cdots, \log_2(\alpha_n2)=(-2a_2+4b_2)+ \sqrt{-1} \cdot 2a_2 \end{align} for scalars $a_i,b_i \in \mathbb{Z}_2$.

Since $\sqrt{-1} \notin \mathbb{Q}_2$, I think $\log_2(\alpha_1), \log_2(\alpha_2)$ are linearly independent over the algebraic closure $A$ of $\mathbb{Q}$ in $\mathbb{Q}_2$ because for scalars (not all zero) $\beta_1, \beta_2, \cdots, \beta_n \in \mathbb{Q}$ the linear combination \begin{align*} &\beta_1 \log_2(\alpha_1)+\cdots+\beta_2 \log_2(\alpha_2) \neq 0 \\ \Rightarrow & \beta_1((-2a_1+4b_1)+\sqrt{-1} \cdot 2a_1)+\cdots+2 \beta_2((-2a_2+4b_2)+\sqrt{-1} \cdot 2a_2) \neq 0, \end{align*} where $\beta_i \in \mathbb{Q}$ and $a_i,b_i \in \mathbb{Z}_2$.

The logarithms $\log_2(\alpha_1), \log_2(\alpha_2)$ will be "linearly independent" over $A=\bar{Q} \cap \mathbb Q_2$. The question remains

Can we talk about "algebraic independence" of the logarithms $\log_2(\alpha_1), \log_2(\alpha_2)$ ?

So it seems this example works. I appreciate your comments/answers.

Let $p$ be a prime number, $\mathbb{Q}_p$ the field of $p$-adic numbers, and $\mathbb{C}_p$ the completion of the algebraic closure of $\mathbb{Q}_p$. Let $U_p$ be the units $(1+\mathfrak{m})$ of $\mathbb{C}_p$, the $p$-adic version of Baker theorem is:

Theorem (Brumer): Let $\alpha_1, \cdots, \alpha_n$ be elements of principal units $U_p$ which are algebraic over the rationals $\mathbb{Q}$ and their $p$-adic logarithms $\log_p(\alpha_1), \cdots, \log_p(\alpha_n)$ are linearly independent over $\mathbb{Q}$. Then $\log_p(\alpha_1), \cdots, \log_p(\alpha_n)$ are then linearly independent over the algebraic closure $A$ of $\mathbb{Q}$ in $\mathbb{C}_p$.

I am confused if the above result is the true the $p$-adic version of Baker theorem or not because wikipedia says the Baker's theorem in $p$-adic setting is an open problem.

Anyway, I want to construct an example where the above theorem(Brumer) holds:

Fix $p=2$, and assume, $K=\mathbb Q_2(\sqrt{-1})$ with uniformizer $\pi=-1+\sqrt{-1}$ and "restrict" ourselves on the principal units $U=1+\mathfrak{m}_K$, where $\mathfrak{m}_K$ is the maximal ideal of $K$. Further, assume we know a basis $\{2\pi, 4\}$ of $\log_2(U)$, where $\log_2$ is the $2$-adic logarithm. Take $\alpha_1, \alpha_2, \cdots, \alpha_n$ in the principal units $U_1=1+\mathfrak{m}_K$ of the quadratic extension $K$ satisfying the Brumer theorem above i.e., are algebraic over the rationals $\mathbb{Q}$ and their $2$-adic logarithm $\log_2(\alpha_1), \cdots, \log_2(\alpha_n)$ are linearly independent over $\mathbb{Q}$ for $n=2$. My question:

Does the above theorem holds in the given situation? i.e., Are $\log_1(\alpha_1), \log_2(\alpha_2)$ linearly independent over the algebraic closure $A$ of $\mathbb{Q}$ in $\mathbb{Q}_2$ ?

We have given the basis $\{2 \pi, 4\}$ of $\log_2(U)$, and therefore $\log_2(\alpha_1), \log_2(\alpha_2)$ all can be obtained by linear combination of the basis $\{2\pi, 4 \}$, where the uniformizer is $\pi=-1+\sqrt{-1}$. So we can write \begin{align} &\log_2(\alpha_1)=2 \pi a_1+4b_1, \log_2(\alpha_2)=2 \pi a_2+4b_2 \\ \Rightarrow &\log_2(\alpha_1)=(-2a_1+4b_1)+\sqrt{-1} \cdot 2a_1, \cdots, \log_2(\alpha_n2)=(-2a_2+4b_2)+ \sqrt{-1} \cdot 2a_2 \end{align} for scalars $a_i,b_i \in \mathbb{Z}_2$.

Since $\sqrt{-1} \notin \mathbb{Q}_2$, I think $\log_2(\alpha_1), \log_2(\alpha_2)$ are linearly independent over the algebraic closure $A$ of $\mathbb{Q}$ in $\mathbb{Q}_2$ because for scalars (not all zero) $\beta_1, \beta_2, \cdots, \beta_n \in \mathbb{Q}$ the linear combination \begin{align*} &\beta_1 \log_2(\alpha_1)+\cdots+\beta_2 \log_2(\alpha_2) \neq 0 \\ \Rightarrow & \beta_1((-2a_1+4b_1)+\sqrt{-1} \cdot 2a_1)+\cdots+2 \beta_2((-2a_2+4b_2)+\sqrt{-1} \cdot 2a_2) \neq 0, \end{align*} where $\beta_i \in \mathbb{Q}$ and $a_i,b_i \in \mathbb{Z}_2$.

The logarithms $\log_2(\alpha_1), \log_2(\alpha_2)$ will be "linearly independent" over $A=\bar{Q} \cap \mathbb Q_2$. The question remains

Can we talk about "algebraic independence" of the logarithms $\log_2(\alpha_1), \log_2(\alpha_2)$ ?

So it seems this example works. I appreciate your comments/answers.

Let $p$ be a prime number, $\mathbb{Q}_p$ the field of $p$-adic numbers, and $\mathbb{C}_p$ the completion of the algebraic closure of $\mathbb{Q}_p$. Let $U_p$ be the units $(1+\mathfrak{m})$ of $\mathbb{C}_p$, the $p$-adic version of Baker theorem is:

Theorem (Brumer): Let $\alpha_1, \cdots, \alpha_n$ be elements of principal units $U_p$ which are algebraic over the rationals $\mathbb{Q}$ and their $p$-adic logarithms $\log_p(\alpha_1), \cdots, \log_p(\alpha_n)$ are linearly independent over $\mathbb{Q}$. Then $\log_p(\alpha_1), \cdots, \log_p(\alpha_n)$ are then linearly independent over the algebraic closure $A$ of $\mathbb{Q}$ in $\mathbb{Q}_p$.

I am confused if the above result is the true the $p$-adic version of Baker theorem or not because wikipedia says the Baker's theorem in $p$-adic setting is an open problem.

Anyway, I want to construct an example where the above theorem(Brumer) holds:

Fix $p=2$, assume, $K=\mathbb Q_2(\sqrt{-1})$ with uniformizer $\pi=-1+\sqrt{-1}$ and "restrict" ourselves on the principal units $U=1+\mathfrak{m}_K$, where $\mathfrak{m}_K$ is the maximal ideal of $K$. Further, assume we know a basis $\{2\pi, 4\}$ of $\log_2(U)$, where $\log_2$ is the $2$-adic logarithm. Take $\alpha_1, \alpha_2, \cdots, \alpha_n$ in the principal units $U_1=1+\mathfrak{m}_K$ of the quadratic extension $K$ satisfying the Brumer theorem above i.e., are algebraic over the rationals $\mathbb{Q}$ and their $2$-adic logarithm $\log_2(\alpha_1), \cdots, \log_2(\alpha_n)$ are linearly independent over $\mathbb{Q}$. My question:

Does the above theorem holds in the given situation? i.e., Are $\log_1(\alpha_1), \cdots, \log_2(\alpha_n)$ linearly independent over the algebraic closure $A$ of $\mathbb{Q}$ in $\mathbb{Q}_2$ ?

We have given the basis $\{2 \pi, 4\}$ of $\log_2(U)$, and therefore $\log_2(\alpha_1), \cdots, \log_2(\alpha_n)$ all can be obtained by linear combination of the basis $\{2\pi, 4 \}$, where the uniformizer is $\pi=-1+\sqrt{-1}$. So we can write \begin{align} &\log_2(\alpha_1)=2 \pi a_1+4b_1, \cdots, \log_2(\alpha_n)=2 \pi a_n+4b_n \\ \Rightarrow &\log_2(\alpha_1)=(-2a_1+4b_1)+\sqrt{-1} \cdot 2a_1, \cdots, \log_2(\alpha_n)=(-2a_n+4b_n)+ \sqrt{-1} \cdot 2a_n \end{align} for scalars $a_i,b_i \in \mathbb{Z}_2$.

Since $\sqrt{-1} \notin \mathbb{Q}_2$, I think $\log_2(\alpha_1), \cdots, \log_2(\alpha_n)$ are linearly independent over the algebraic closure $A$ of $\mathbb{Q}$ in $\mathbb{Q}_2$ because for scalars (not all zero) $\beta_1, \beta_2, \cdots, \beta_n \in \mathbb{Q}$ the linear combination \begin{align*} &\beta_1 \log_2(\alpha_1)+\cdots+\beta_n \log_2(\alpha_n) \neq 0 \\ \Rightarrow & \beta_1((-2a_1+4b_1)+\sqrt{-1} \cdot 2a_1)+\cdots+2 \beta_n((-2a_n+4b_n)+\sqrt{-1} \cdot 2a_n) \neq 0, \end{align*} where $\beta_i \in \mathbb{Q}$ and $a_i,b_i \in \mathbb{Z}_2$.

Can we talk about linear independence of the logarithms $\log_2(\alpha_1), \cdots, \log_2(\alpha_n)$ ?

Can we talk about algebraic independence of the logarithms $\log_2(\alpha_1), \cdots, \log_2(\alpha_n)$ ?

So it seems this example works. I appreciate your comments/answers.

Let $p$ be a prime number, $\mathbb{Q}_p$ the field of $p$-adic numbers, and $\mathbb{C}_p$ the completion of the algebraic closure of $\mathbb{Q}_p$. Let $U_p$ be the units $(1+\mathfrak{m})$ of $\mathbb{C}_p$, the $p$-adic version of Baker theorem is:

Theorem (Brumer): Let $\alpha_1, \cdots, \alpha_n$ be elements of principal units $U_p$ which are algebraic over the rationals $\mathbb{Q}$ and their $p$-adic logarithms $\log_p(\alpha_1), \cdots, \log_p(\alpha_n)$ are linearly independent over $\mathbb{Q}$. Then $\log_p(\alpha_1), \cdots, \log_p(\alpha_n)$ are then linearly independent over the algebraic closure $A$ of $\mathbb{Q}$ in $\mathbb{Q}_p$.

I am confused if the above result is the true the $p$-adic version of Baker theorem or not because wikipedia says the Baker's theorem in $p$-adic setting is an open problem.

Anyway, I want to construct an example where the above theorem(Brumer) holds:

Fix $p=2$, and assume, $K=\mathbb Q_2(\sqrt{-1})$ with uniformizer $\pi=-1+\sqrt{-1}$ and "restrict" ourselves on the principal units $U=1+\mathfrak{m}_K$, where $\mathfrak{m}_K$ is the maximal ideal of $K$. Further, assume we know a basis $\{2\pi, 4\}$ of $\log_2(U)$, where $\log_2$ is the $2$-adic logarithm. Take $\alpha_1, \alpha_2, \cdots, \alpha_n$ in the principal units $U_1=1+\mathfrak{m}_K$ of the quadratic extension $K$ satisfying the Brumer theorem above i.e., are algebraic over the rationals $\mathbb{Q}$ and their $2$-adic logarithm $\log_2(\alpha_1), \cdots, \log_2(\alpha_n)$ are linearly independent over $\mathbb{Q}$ for $n=2$. My question:

Does the above theorem holds in the given situation? i.e., Are $\log_1(\alpha_1), \log_2(\alpha_2)$ linearly independent over the algebraic closure $A$ of $\mathbb{Q}$ in $\mathbb{Q}_2$ ?

We have given the basis $\{2 \pi, 4\}$ of $\log_2(U)$, and therefore $\log_2(\alpha_1), \log_2(\alpha_2)$ all can be obtained by linear combination of the basis $\{2\pi, 4 \}$, where the uniformizer is $\pi=-1+\sqrt{-1}$. So we can write \begin{align} &\log_2(\alpha_1)=2 \pi a_1+4b_1, \log_2(\alpha_2)=2 \pi a_2+4b_2 \\ \Rightarrow &\log_2(\alpha_1)=(-2a_1+4b_1)+\sqrt{-1} \cdot 2a_1, \cdots, \log_2(\alpha_n2)=(-2a_2+4b_2)+ \sqrt{-1} \cdot 2a_2 \end{align} for scalars $a_i,b_i \in \mathbb{Z}_2$.

Since $\sqrt{-1} \notin \mathbb{Q}_2$, I think $\log_2(\alpha_1), \log_2(\alpha_2)$ are linearly independent over the algebraic closure $A$ of $\mathbb{Q}$ in $\mathbb{Q}_2$ because for scalars (not all zero) $\beta_1, \beta_2, \cdots, \beta_n \in \mathbb{Q}$ the linear combination \begin{align*} &\beta_1 \log_2(\alpha_1)+\cdots+\beta_2 \log_2(\alpha_2) \neq 0 \\ \Rightarrow & \beta_1((-2a_1+4b_1)+\sqrt{-1} \cdot 2a_1)+\cdots+2 \beta_2((-2a_2+4b_2)+\sqrt{-1} \cdot 2a_2) \neq 0, \end{align*} where $\beta_i \in \mathbb{Q}$ and $a_i,b_i \in \mathbb{Z}_2$.

The logarithms $\log_2(\alpha_1), \log_2(\alpha_2)$ will be "linearly independent" over $A=\bar{Q} \cap \mathbb Q_2$. The question remains

Can we talk about "algebraic independence" of the logarithms $\log_2(\alpha_1), \log_2(\alpha_2)$ ?

So it seems this example works. I appreciate your comments/answers.

Let $p$ be a prime number, $\mathbb{Q}_p$ the field of $p$-adic numbers, and $\mathbb{C}_p$ the completion of the algebraic closure of $\mathbb{Q}_p$. Let $U_p$ be the units $(1+\mathfrak{m})$ of $\mathbb{C}_p$, the $p$-adic version of Baker theorem is:

Theorem (Brumer): Let $\alpha_1, \cdots, \alpha_n$ be elements of principal units $U_p$ which are algebraic over the rationals $\mathbb{Q}$ and their $p$-adic logarithms $\log_p(\alpha_1), \cdots, \log_p(\alpha_n)$ are linearly independent over $\mathbb{Q}$. Then $\log_p(\alpha_1), \cdots, \log_p(\alpha_n)$ are then linearly independent over the algebraic closure $A$ of $\mathbb{Q}$ in $\mathbb{Q}_p$.

I am confused if the above result is the true the $p$-adic version of Baker theorem or not because wikipedia says the Baker's theorem in $p$-adic setting is an open problem.

Anyway, I want to construct an example where the above theorem(Brumer) holds:

Fix $p=2$, assume, $K=\mathbb Q_2(\sqrt{-1})$ with uniformizer $\pi=-1+\sqrt{-1}$ and "restrict" ourselves on the principal units $U=1+\mathfrak{m}_K$, where $\mathfrak{m}_K$ is the maximal ideal of $K$. Further, assume we know a basis $\{2\pi, 4\}$ of $\log_2(U)$, where $\log_2$ is the $2$-adic logarithm. Take $\alpha_1, \alpha_2, \cdots, \alpha_n$ in the principal units $U_1=1+\mathfrak{m}_K$ of the quadratic extension $K$ satisfying the Brumer theorem above i.e., are algebraic over the rationals $\mathbb{Q}$ and their $2$-adic logarithm $\log_2(\alpha_1), \cdots, \log_2(\alpha_n)$ are linearly independent over $\mathbb{Q}$. My question:

Does the above theorem holds in the given situation? i.e., Are $\log_1(\alpha_1), \cdots, \log_2(\alpha_n)$ linearly independent over the algebraic closure $A$ of $\mathbb{Q}$ in $\mathbb{Q}_2$ ?

We have given the basis $\{2 \pi, 4\}$ of $\log_2(U)$, and therefore $\log_p(\alpha_1), \cdots, \log_p(\alpha_n)$ all can be obtained by linear combination of the basis $\{2\pi, 4 \}$, where the uniformizer is $\pi=-1+\sqrt{-1}$. So we can write \begin{align} &\log_2(\alpha_1)=2 \pi a_1+4b_1, \cdots, \log_2(\alpha_n)=2 \pi a_n+4b_n \\ \Rightarrow &\log_2(\alpha_1)=(-2a_1+4b_1)+\sqrt{-1} \cdot 2a_1, \cdots, \log_2(\alpha_n)=(-2a_n+4b_n)+ \sqrt{-1} \cdot 2a_n \end{align} for scalars $a_i,b_i \in \mathbb{Z}_2$.

Since $\sqrt{-1} \notin \mathbb{Q}_2$, I think $\log_2(\alpha_1), \cdots, \log_2(\alpha_n)$ are linearly independent over the algebraic closure $A$ of $\mathbb{Q}$ in $\mathbb{Q}_2$ because for scalars (not all zero) $\beta_1, \beta_2, \cdots, \beta_n \in \mathbb{Q}$ the linear combination \begin{align*} &\beta_1 \log_2(\alpha_1)+\cdots+\beta_n \log_2(\alpha_n) \neq 0 \\ \Rightarrow & \beta_1((-2a_1+4b_1)+\sqrt{-1} \cdot 2a_1)+\cdots+2 \beta_n((-2a_n+4b_n)+\sqrt{-1} \cdot 2a_n) \neq 0, \end{align*} where $\beta_i \in \mathbb{Q}$ and $a_i,b_i \in \mathbb{Z}_2$.

So it seems this example works. I appreciate your comments/answers.

Let $p$ be a prime number, $\mathbb{Q}_p$ the field of $p$-adic numbers, and $\mathbb{C}_p$ the completion of the algebraic closure of $\mathbb{Q}_p$. Let $U_p$ be the units $(1+\mathfrak{m})$ of $\mathbb{C}_p$, the $p$-adic version of Baker theorem is:

Theorem (Brumer): Let $\alpha_1, \cdots, \alpha_n$ be elements of principal units $U_p$ which are algebraic over the rationals $\mathbb{Q}$ and their $p$-adic logarithms $\log_p(\alpha_1), \cdots, \log_p(\alpha_n)$ are linearly independent over $\mathbb{Q}$. Then $\log_p(\alpha_1), \cdots, \log_p(\alpha_n)$ are then linearly independent over the algebraic closure $A$ of $\mathbb{Q}$ in $\mathbb{Q}_p$.

I am confused if the above result is the true the $p$-adic version of Baker theorem or not because wikipedia says the Baker's theorem in $p$-adic setting is an open problem.

Anyway, I want to construct an example where the above theorem(Brumer) holds:

Fix $p=2$, assume, $K=\mathbb Q_2(\sqrt{-1})$ with uniformizer $\pi=-1+\sqrt{-1}$ and "restrict" ourselves on the principal units $U=1+\mathfrak{m}_K$, where $\mathfrak{m}_K$ is the maximal ideal of $K$. Further, assume we know a basis $\{2\pi, 4\}$ of $\log_2(U)$, where $\log_2$ is the $2$-adic logarithm. Take $\alpha_1, \alpha_2, \cdots, \alpha_n$ in the principal units $U_1=1+\mathfrak{m}_K$ of the quadratic extension $K$ satisfying the Brumer theorem above i.e., are algebraic over the rationals $\mathbb{Q}$ and their $2$-adic logarithm $\log_2(\alpha_1), \cdots, \log_2(\alpha_n)$ are linearly independent over $\mathbb{Q}$. My question:

Does the above theorem holds in the given situation? i.e., Are $\log_1(\alpha_1), \cdots, \log_2(\alpha_n)$ linearly independent over the algebraic closure $A$ of $\mathbb{Q}$ in $\mathbb{Q}_2$ ?

We have given the basis $\{2 \pi, 4\}$ of $\log_2(U)$, and therefore $\log_2(\alpha_1), \cdots, \log_2(\alpha_n)$ all can be obtained by linear combination of the basis $\{2\pi, 4 \}$, where the uniformizer is $\pi=-1+\sqrt{-1}$. So we can write \begin{align} &\log_2(\alpha_1)=2 \pi a_1+4b_1, \cdots, \log_2(\alpha_n)=2 \pi a_n+4b_n \\ \Rightarrow &\log_2(\alpha_1)=(-2a_1+4b_1)+\sqrt{-1} \cdot 2a_1, \cdots, \log_2(\alpha_n)=(-2a_n+4b_n)+ \sqrt{-1} \cdot 2a_n \end{align} for scalars $a_i,b_i \in \mathbb{Z}_2$.

Since $\sqrt{-1} \notin \mathbb{Q}_2$, I think $\log_2(\alpha_1), \cdots, \log_2(\alpha_n)$ are linearly independent over the algebraic closure $A$ of $\mathbb{Q}$ in $\mathbb{Q}_2$ because for scalars (not all zero) $\beta_1, \beta_2, \cdots, \beta_n \in \mathbb{Q}$ the linear combination \begin{align*} &\beta_1 \log_2(\alpha_1)+\cdots+\beta_n \log_2(\alpha_n) \neq 0 \\ \Rightarrow & \beta_1((-2a_1+4b_1)+\sqrt{-1} \cdot 2a_1)+\cdots+2 \beta_n((-2a_n+4b_n)+\sqrt{-1} \cdot 2a_n) \neq 0, \end{align*} where $\beta_i \in \mathbb{Q}$ and $a_i,b_i \in \mathbb{Z}_2$.

Can we talk about linear independence of the logarithms $\log_2(\alpha_1), \cdots, \log_2(\alpha_n)$ ?

Can we talk about algebraic independence of the logarithms $\log_2(\alpha_1), \cdots, \log_2(\alpha_n)$ ?

So it seems this example works. I appreciate your comments/answers.