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0

$begingroup$

Fix an algebraically closed field $F$.

Let $alpha_1,dotsc,alpha_nin F$ be roots of $1$.
Let $x=alpha_1+dotsc+alpha_n$ and $y=alpha_1^{-1}+dotsc+alpha_n^{-1}$.

I was thinking: Given $n$ and $x$, can we compute $y$?

If $F=mathbb{C}$ the answer is positive: $y$ is the complex conjugate of $x$.

If $F$ is the algebraic closure of the field with $5$ elements, the answer is negative. Both $alpha_1=1,alpha_2=1$ and $alpha_1=3,alpha_2=4$ give $x=2$, but the former gives $y=2$ while the latter gives $y=1$.

So the answer depends on $F$.

Question: What are the algebraically closed fields $F$ where we can always compute $y$ as a function of $n$ and $x$?

Partial answers are welcome too.

field-theory roots-of-unity

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asked Mar 11 at 11:42

Chris AChris A

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New contributor

Chris A is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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0

$begingroup$

Fix an algebraically closed field $F$.

Let $alpha_1,dotsc,alpha_nin F$ be roots of $1$.
Let $x=alpha_1+dotsc+alpha_n$ and $y=alpha_1^{-1}+dotsc+alpha_n^{-1}$.

I was thinking: Given $n$ and $x$, can we compute $y$?

If $F=mathbb{C}$ the answer is positive: $y$ is the complex conjugate of $x$.

If $F$ is the algebraic closure of the field with $5$ elements, the answer is negative. Both $alpha_1=1,alpha_2=1$ and $alpha_1=3,alpha_2=4$ give $x=2$, but the former gives $y=2$ while the latter gives $y=1$.

So the answer depends on $F$.

Question: What are the algebraically closed fields $F$ where we can always compute $y$ as a function of $n$ and $x$?

Partial answers are welcome too.

field-theory roots-of-unity

share|cite|improve this question

asked Mar 11 at 11:42

Chris AChris A

1

New contributor

Chris A is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

Fix an algebraically closed field $F$.

Let $alpha_1,dotsc,alpha_nin F$ be roots of $1$.
Let $x=alpha_1+dotsc+alpha_n$ and $y=alpha_1^{-1}+dotsc+alpha_n^{-1}$.

I was thinking: Given $n$ and $x$, can we compute $y$?

If $F=mathbb{C}$ the answer is positive: $y$ is the complex conjugate of $x$.

If $F$ is the algebraic closure of the field with $5$ elements, the answer is negative. Both $alpha_1=1,alpha_2=1$ and $alpha_1=3,alpha_2=4$ give $x=2$, but the former gives $y=2$ while the latter gives $y=1$.

So the answer depends on $F$.

Question: What are the algebraically closed fields $F$ where we can always compute $y$ as a function of $n$ and $x$?

Partial answers are welcome too.

field-theory roots-of-unity

share|cite|improve this question

asked Mar 11 at 11:42

Chris AChris A

1

New contributor

Chris A is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked Mar 11 at 11:42

Chris AChris A

1

New contributor

Chris A is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked Mar 11 at 11:42

Chris AChris A

1

New contributor

Chris A is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 11 at 11:42

Chris AChris A

1

New contributor

Chris A is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 11 at 11:42

Chris AChris A

1