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Let $k \subset K \subset L$ be finite purely inseparable extensions of characteristic $p >0$. I am interested in knowing whether the group $$ \frac{(kK^p)^\ast \cap (k^\ast {L^p}^\ast)}{k^\ast {K^p}^\ast} $$ is trivial ($kK^p$ is the subextension of $L$ generated by $k$ and $k^p$).

If $k \subset K^p$ or $K^p \subset k$ it is obviously trivial.

Also,if $k$ and $L^p$ are linearly independent over $k^p$, then we have $(kK^p) \cap L^p = K^p$, so the group above is trivial. Indeed, one can take a $k$-basis $(e_1,\dotsc,e_m)$ of $K/k$ and complete it into a basis $(e_1,\dotsc,e_n)$ of $L/k$. Then $(e_1^p,\dotsc,e_m^p)$ is a $k^p$-basis of $K^p$ and $(e_1^p,\dotsc,e_n^p)$ is a $k^p$-basis of $L^p$. Now because of the hypothesis, the family $(e_1^p,\dotsc,e_m^p)$ is still linearly independent over $k$. So, given an element $x \in (kK^p) \cap L^p$ it can be written as \begin{align*} x & = \lambda_1 e_1^p + \dotsb \lambda_m e_m^p \in kK^p \\ & = \alpha_1 e_1^p +\dotsb+ \alpha_n e_n^p \in L^p \end{align*} for scalars $\lambda_i$, $\alpha_j \in k$, and since $(e_1^p,\dotsc,e_n^p)$ is $k$-linearly independent, one finds that $\alpha_{m+1}=\dotsb=\alpha_n =0$, thus $x \in K^p$.

How can one go further ? If you have any ideas or references that can help, you're welcome !

Let $k \subset K \subset L$ be finite purely inseparable extensions of characteristic $p >0$. I am interested in knowing whether the group $$ \frac{(kK^p)^\ast \cap (k^\ast {L^p}^\ast)}{k^\ast {K^p}^\ast} $$ is trivial ($kK^p$ is the subextension of $L$ generated by $k$ and $K^p$).

If $k \subset K^p$ or $K^p \subset k$ it is obviously trivial.

Also,if $k$ and $L^p$ are linearly independent over $k^p$, then we have $(kK^p) \cap L^p = K^p$, so the group above is trivial. Indeed, one can take a $k$-basis $(e_1,\dotsc,e_m)$ of $K/k$ and complete it into a basis $(e_1,\dotsc,e_n)$ of $L/k$. Then $(e_1^p,\dotsc,e_m^p)$ is a $k^p$-basis of $K^p$ and $(e_1^p,\dotsc,e_n^p)$ is a $k^p$-basis of $L^p$. Now because of the hypothesis, the family $(e_1^p,\dotsc,e_m^p)$ is still linearly independent over $k$. So, given an element $x \in (kK^p) \cap L^p$ it can be written as \begin{align*} x & = \lambda_1 e_1^p + \dotsb +\lambda_m e_m^p \in kK^p \\ & = \alpha_1^p e_1^p +\dotsb+ \alpha_m^p e_m^p + \dotsb + \alpha_n^p e_n^p \in L^p \end{align*} for scalars $\lambda_i$, $\alpha_j \in k$, and since $(e_1^p,\dotsc,e_n^p)$ is $k$-linearly independent, one finds that $\alpha_{m+1}=\dotsb=\alpha_n =0$, thus $x \in K^p$.

How can one go further ? If you have any ideas or references that can help, you're welcome !

Let $k \subset K \subset L$ be finite purely inseparable extensions of characteristic $p >0$. I am interested in knowing whether the group $$ \frac{(kK^p)^\ast \cap (k^\ast {L^p}^\ast)}{k^\ast {K^p}^\ast} $$ is trivial ($kK^p$ is the subextension of $L$ generated by $k$ and $k^p$).

If $k \subset K^p$ or $K^p \subset k$ it is obviously trivial.

Also,if $k$ and $L^p$ are linearly independent over $k^p$, then we have $(kK^p) \cap L^p = K^p$, so the group above is trivial. Indeed, one can take a $k$-basis $(e_1,\dotsc,e_m)$ of $K/k$ and complete it into a basis $(e_1,\dotsc,e_n)$ of $L/k$. Then $(e_1^p,\dotsc,e_m^p)$ is a $k^p$-basis of $K^p$ and $(e_1^p,\dotsc,e_n^p)$ is a $k^p$-basis of $L^p$. Now because of the hypothesis, the family $(e_1,^p,\dotsc,e_m^p)$ is still linearly independent over $k$. So, given an element $x \in (kK^p) \cap L^p$ it can be written as \begin{align*} x & = \lambda_1 e_1^p + \dotsb \lambda_m e_m^p \in kK^p \\ & = \alpha_1 e_1^p +\dotsb+ \alpha_n e_n^p \in L^p \end{align*} for scalars $\lambda_i$, $\alpha_j \in k$, and since $(e_1^p,\dotsc,e_n^p)$ is $k$-linearly independent, one finds that $\alpha_{m+1}=\dotsb=\alpha_n =0$, thus $x \in K^p$.

How can one go further ? If you have any ideas or references that can help, you're welcome !

Let $k \subset K \subset L$ be finite purely inseparable extensions of characteristic $p >0$. I am interested in knowing whether the group $$ \frac{(kK^p)^\ast \cap (k^\ast {L^p}^\ast)}{k^\ast {K^p}^\ast} $$ is trivial ($kK^p$ is the subextension of $L$ generated by $k$ and $k^p$).

If $k \subset K^p$ or $K^p \subset k$ it is obviously trivial.

Also,if $k$ and $L^p$ are linearly independent over $k^p$, then we have $(kK^p) \cap L^p = K^p$, so the group above is trivial. Indeed, one can take a $k$-basis $(e_1,\dotsc,e_m)$ of $K/k$ and complete it into a basis $(e_1,\dotsc,e_n)$ of $L/k$. Then $(e_1^p,\dotsc,e_m^p)$ is a $k^p$-basis of $K^p$ and $(e_1^p,\dotsc,e_n^p)$ is a $k^p$-basis of $L^p$. Now because of the hypothesis, the family $(e_1^p,\dotsc,e_m^p)$ is still linearly independent over $k$. So, given an element $x \in (kK^p) \cap L^p$ it can be written as \begin{align*} x & = \lambda_1 e_1^p + \dotsb \lambda_m e_m^p \in kK^p \\ & = \alpha_1 e_1^p +\dotsb+ \alpha_n e_n^p \in L^p \end{align*} for scalars $\lambda_i$, $\alpha_j \in k$, and since $(e_1^p,\dotsc,e_n^p)$ is $k$-linearly independent, one finds that $\alpha_{m+1}=\dotsb=\alpha_n =0$, thus $x \in K^p$.

How can one go further ? If you have any ideas or references that can help, you're welcome !

Let $k \subset K \subset L$ be finite purely inseparable extensions of characteristic $p >0$. I am interested in knowing whether the group $$ \frac{(kK^p)^\ast \cap (k^\ast {L^p}^\ast)}{k^\ast {K^p}^\ast} $$ is trivial ($kK^p$ is the subextension of $L$ generated by $k$ and $k^p$).

If $k \subset K^p$ or $K^p \subset k$ it is obviously trivial.

Also,if $k$ and $L^p$ are linearly independent over $k^p$, then we have $(kK^p) \cap L^p = K^p$, so the group above is trivial. Indeed, one can take a $k$-basis $e_1,\cdots,e_m)$ of $K/k$ and complete it into a basis $(e_1,\cdots,e_n)$ of $L/k$. Then $(e_1^p,\cdots,e_m^p)$ is a $k^p$-basis of $K^p$ and $(e_1^p,\cdots,e_n^p)$ is a $k^p$-basis of $L^p$. Now because of the hypothesis, the family $(e_1,^p,\cdots,e_m^p)$ is still linearly independent over $k$. So, given an element $x \in (kK^p) \cap L^p$ it can be written as \begin{align*} x & = \lambda_1 e_1^p + \cdots \lambda_m e_m^p \in kK^p \\ & = \alpha_1 e_1^p +\cdots+ \alpha_n e_n^p \in L^p \end{align*} for scalars $\lambda_i$, $\alpha_j \in k$, and since $(e_1^p,\cdots,e_n^p)$ is $k$-linearly independent, one finds that $\alpha_{m+1}=\cdots=\alpha_n =0$, thus $x \in K^p$.

How can one go further ? If you have any ideas or references that can help, you're welcome !

Let $k \subset K \subset L$ be finite purely inseparable extensions of characteristic $p >0$. I am interested in knowing whether the group $$ \frac{(kK^p)^\ast \cap (k^\ast {L^p}^\ast)}{k^\ast {K^p}^\ast} $$ is trivial ($kK^p$ is the subextension of $L$ generated by $k$ and $k^p$).

If $k \subset K^p$ or $K^p \subset k$ it is obviously trivial.

Also,if $k$ and $L^p$ are linearly independent over $k^p$, then we have $(kK^p) \cap L^p = K^p$, so the group above is trivial. Indeed, one can take a $k$-basis $(e_1,\dotsc,e_m)$ of $K/k$ and complete it into a basis $(e_1,\dotsc,e_n)$ of $L/k$. Then $(e_1^p,\dotsc,e_m^p)$ is a $k^p$-basis of $K^p$ and $(e_1^p,\dotsc,e_n^p)$ is a $k^p$-basis of $L^p$. Now because of the hypothesis, the family $(e_1,^p,\dotsc,e_m^p)$ is still linearly independent over $k$. So, given an element $x \in (kK^p) \cap L^p$ it can be written as \begin{align*} x & = \lambda_1 e_1^p + \dotsb \lambda_m e_m^p \in kK^p \\ & = \alpha_1 e_1^p +\dotsb+ \alpha_n e_n^p \in L^p \end{align*} for scalars $\lambda_i$, $\alpha_j \in k$, and since $(e_1^p,\dotsc,e_n^p)$ is $k$-linearly independent, one finds that $\alpha_{m+1}=\dotsb=\alpha_n =0$, thus $x \in K^p$.

How can one go further ? If you have any ideas or references that can help, you're welcome !