Mixing
find all algebraic elements of $mathbb{Q}(pi)$ over $mathbb{Q}(pi^2 -2pi +5)$
$begingroup$
Well, it is pretty clear that $mathbb{Q}(pi^2-2pi+5)$ are all algebraic. I don't have any idea how to justify it, but I think those are the only algebraic elements of $mathbb{Q}(pi)$ over $mathbb{Q}(pi^2-2pi+5)$, yet I don't succeed to show it or recall a theorem which might help.
I would also like an approach for the general case, when relating $mathbb{Q}(pi)$ as $mathbb{Q}(x)$ and asking the same question regarding a general polynomial $g(x)$.
abstract-algebra field-theory
asked Jan 13 at 13:17
dandan
547613
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add a comment |
$begingroup$
Well, it is pretty clear that $mathbb{Q}(pi^2-2pi+5)$ are all algebraic. I don't have any idea how to justify it, but I think those are the only algebraic elements of $mathbb{Q}(pi)$ over $mathbb{Q}(pi^2-2pi+5)$, yet I don't succeed to show it or recall a theorem which might help.
I would also like an approach for the general case, when relating $mathbb{Q}(pi)$ as $mathbb{Q}(x)$ and asking the same question regarding a general polynomial $g(x)$.
abstract-algebra field-theory
asked Jan 13 at 13:17
dandan
547613
$endgroup$
add a comment |
$begingroup$
Well, it is pretty clear that $mathbb{Q}(pi^2-2pi+5)$ are all algebraic. I don't have any idea how to justify it, but I think those are the only algebraic elements of $mathbb{Q}(pi)$ over $mathbb{Q}(pi^2-2pi+5)$, yet I don't succeed to show it or recall a theorem which might help.
I would also like an approach for the general case, when relating $mathbb{Q}(pi)$ as $mathbb{Q}(x)$ and asking the same question regarding a general polynomial $g(x)$.
abstract-algebra field-theory
asked Jan 13 at 13:17
dandan
547613
$endgroup$
Well, it is pretty clear that $mathbb{Q}(pi^2-2pi+5)$ are all algebraic. I don't have any idea how to justify it, but I think those are the only algebraic elements of $mathbb{Q}(pi)$ over $mathbb{Q}(pi^2-2pi+5)$, yet I don't succeed to show it or recall a theorem which might help.
I would also like an approach for the general case, when relating $mathbb{Q}(pi)$ as $mathbb{Q}(x)$ and asking the same question regarding a general polynomial $g(x)$.
abstract-algebra field-theory
abstract-algebra field-theory
asked Jan 13 at 13:17
dandan
547613
asked Jan 13 at 13:17
dandan
547613
asked Jan 13 at 13:17
dandan
547613
asked Jan 13 at 13:17
dandan
547613
asked Jan 13 at 13:17
dandan
547613
547613
add a comment |
add a comment |
1 Answer
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$begingroup$
$pi$ is algebraic, as it is the root of $X^2-2X+5 - (pi^2-2pi + 5)$.
Then all polynomials in $pi$ are algebraic, because algebraic numbers are stable under product and sum. Moreover, nonzero algebraic numbers are stable under inverses, so all rational fractions in $pi$ are algebraic : $mathbb{Q}(pi)/mathbb{Q}(pi^2-2pi+5)$ is an algebraic extension.
answered Jan 13 at 14:07
MaxMax
14.3k11142
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1 Answer
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1 Answer
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oldest
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active
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active
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votes
$begingroup$
$pi$ is algebraic, as it is the root of $X^2-2X+5 - (pi^2-2pi + 5)$.
Then all polynomials in $pi$ are algebraic, because algebraic numbers are stable under product and sum. Moreover, nonzero algebraic numbers are stable under inverses, so all rational fractions in $pi$ are algebraic : $mathbb{Q}(pi)/mathbb{Q}(pi^2-2pi+5)$ is an algebraic extension.
answered Jan 13 at 14:07
MaxMax
14.3k11142
$endgroup$
add a comment |
$begingroup$
$pi$ is algebraic, as it is the root of $X^2-2X+5 - (pi^2-2pi + 5)$.
Then all polynomials in $pi$ are algebraic, because algebraic numbers are stable under product and sum. Moreover, nonzero algebraic numbers are stable under inverses, so all rational fractions in $pi$ are algebraic : $mathbb{Q}(pi)/mathbb{Q}(pi^2-2pi+5)$ is an algebraic extension.
answered Jan 13 at 14:07
MaxMax
14.3k11142
$endgroup$
add a comment |
$begingroup$
$pi$ is algebraic, as it is the root of $X^2-2X+5 - (pi^2-2pi + 5)$.
Then all polynomials in $pi$ are algebraic, because algebraic numbers are stable under product and sum. Moreover, nonzero algebraic numbers are stable under inverses, so all rational fractions in $pi$ are algebraic : $mathbb{Q}(pi)/mathbb{Q}(pi^2-2pi+5)$ is an algebraic extension.
answered Jan 13 at 14:07
MaxMax
14.3k11142
$endgroup$
$pi$ is algebraic, as it is the root of $X^2-2X+5 - (pi^2-2pi + 5)$.
Then all polynomials in $pi$ are algebraic, because algebraic numbers are stable under product and sum. Moreover, nonzero algebraic numbers are stable under inverses, so all rational fractions in $pi$ are algebraic : $mathbb{Q}(pi)/mathbb{Q}(pi^2-2pi+5)$ is an algebraic extension.
answered Jan 13 at 14:07
MaxMax
14.3k11142
answered Jan 13 at 14:07
MaxMax
14.3k11142
answered Jan 13 at 14:07
MaxMax
14.3k11142
answered Jan 13 at 14:07
MaxMax
14.3k11142
14.3k11142
add a comment |
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