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A question about non-archimedesarchimedean binomial expansion

Suppose we work in the number field $Q(\omega_{pq})$ where$\mathbb Q(\omega_{pq})$ with $\omega_{pq}$ is thea $p*q$$p q$th

  root of unit andunity, where $p*q$ are the product of two distinctive prime$p$, $p,q$$q$ are distinct primes. Similarly $\omega_{p}$ and $\omega_{q}$ are the $p$th and $q$th rootroots of unit respectively. Now we note the binomial expansion

$(1-(1-\omega_{p}))^{\frac{p}{q}}=1-\frac{p}{q}*(1-\omega_{p})+ C_{\frac{p}{q}}^{2}*(1-\omega_{p})^{2} + ...$ $$ (1-(1-\omega_{p}))^{\frac{p}{q}}=1-\frac{p}{q}(1-\omega_{p})+ \binom{p/q}2(1-\omega_{p})^{2} + \dotsb. $$

By simple analysis, we know it should converge byin the non-archimedesarchimedean topology to one of the $q$th rootroots of unitunity $\omega_{q}^{k}$ at a certain place $\beta$ over $1-\omega_{p}$, or equivalently over $p$. And denote this expansion asDenote the limit by $s$, again. Again by simple analysis of the expansion of $s-1$, we know $s\neq1$.

So, by the property of product of converge series in non-archimedes topology we know $1, s, s^{2}, ... s^{q-1}$$1, s, s^{2}, \dotsc, s^{q-1}$ will go once through all the $q$th rootroots of unitunity while all of them belong to $1+\pi*Z_{\beta}$$1+\pi\mathbb Z_{\beta}$ where $\pi$ is a primal element of place $\beta$  . So $1+s+s^{2}+...+s^{q-1}=q+\pi*Z_{\beta}\neq0$$1+s+s^{2}+\dotsb+s^{q-1}\in q+\pi\mathbb Z_{\beta}$, so $\neq0$, which is a contradiction withof the property of a $q$th root of unitunity. So what

What is wrong in the above argument?

A question about non-archimedes binomial expansion

Suppose we work in number field $Q(\omega_{pq})$ where $\omega_{pq}$ is the $p*q$th

  root of unit and $p*q$ are the product of two distinctive prime $p,q$. Similarly $\omega_{p}$ and $\omega_{q}$ are the $p$th and $q$th root of unit respectively. Now we note the binomial expansion

$(1-(1-\omega_{p}))^{\frac{p}{q}}=1-\frac{p}{q}*(1-\omega_{p})+ C_{\frac{p}{q}}^{2}*(1-\omega_{p})^{2} + ...$

By simple analysis, we know it should converge by non-archimedes topology to one of the $q$th root of unit $\omega_{q}^{k}$ at certain place $\beta$ over $1-\omega_{p}$, or equivalently over $p$. And denote this expansion as $s$, again by simple analysis of the expansion $s-1$, we know $s\neq1$.

So, by the property of product of converge series in non-archimedes topology we know $1, s, s^{2}, ... s^{q-1}$ will go through all the $q$th root of unit while all of them belong to $1+\pi*Z_{\beta}$ where $\pi$ is a primal element of place $\beta$  . So $1+s+s^{2}+...+s^{q-1}=q+\pi*Z_{\beta}\neq0$ which is a contradiction with the property of $q$th root of unit. So what is wrong in the above argument?

A question about non-archimedean binomial expansion

Suppose we work in the number field $\mathbb Q(\omega_{pq})$ with $\omega_{pq}$ a $p q$th root of unity, where $p$, $q$ are distinct primes. Similarly $\omega_{p}$ and $\omega_{q}$ are $p$th and $q$th roots of unit respectively. Now we note the binomial expansion $$ (1-(1-\omega_{p}))^{\frac{p}{q}}=1-\frac{p}{q}(1-\omega_{p})+ \binom{p/q}2(1-\omega_{p})^{2} + \dotsb. $$

By simple analysis, we know it should converge in the non-archimedean topology to one of the $q$th roots of unity $\omega_{q}^{k}$ at a certain place $\beta$ over $1-\omega_{p}$, or equivalently over $p$. Denote the limit by $s$. Again by simple analysis of the expansion of $s-1$, we know $s\neq1$.

So we know $1, s, s^{2}, \dotsc, s^{q-1}$ will go once through all the $q$th roots of unity while all of them belong to $1+\pi\mathbb Z_{\beta}$ where $\pi$ is a primal element of place $\beta$. So $1+s+s^{2}+\dotsb+s^{q-1}\in q+\pi\mathbb Z_{\beta}$, so $\neq0$, which is a contradiction of the property of a $q$th root of unity.

What is wrong in the above argument?

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A question about non-archimedes binomial expansion

Suppose we work in number field $Q(\omega_{pq})$ where $\omega_{pq}$ is the $p*q$th

root of unit and $p*q$ are the product of two distinctive prime $p,q$. Similarly $\omega_{p}$ and $\omega_{q}$ are the $p$th and $q$th root of unit respectively. Now we note the binomial expansion

$(1-(1-\omega_{p}))^{\frac{p}{q}}=1-\frac{p}{q}*(1-\omega_{p})+ C_{\frac{p}{q}}^{2}*(1-\omega_{p})^{2} + ...$

By simple analysis, we know it should converge by non-archimedes topology to one of the $q$th root of unit $\omega_{q}^{k}$ at certain place $\beta$ over $1-\omega_{p}$, or equivalently over $p$. And denote this expansion as $s$, again by simple analysis of the expansion $s-1$, we know $s\neq1$.

So, by the property of product of converge series in non-archimedes topology we know $1, s, s^{2}, ... s^{q-1}$ will go through all the $q$th root of unit while all of them belong to $1+\pi*Z_{\beta}$ where $\pi$ is a primal element of place $\beta$ . So $1+s+s^{2}+...+s^{q-1}=q+\pi*Z_{\beta}\neq0$ which is a contradiction with the property of $q$th root of unit. So what is wrong in the above argument?