Mixing
all algebraic numbers over $mathbb{Q}(g(pi))$ are the elements in $mathbb{Q}$
$begingroup$
Let $g(x)$ be a polynomial in $mathbb{Q}[x]$ such the $deg(g)>1$. I wish to show that $mathbb{Q}$ are the only algebraic elements in $mathbb{Q(g(pi))}$.
I tried to show that $g(pi)$ is not algebraic, yet even if I assume it, I don't know how to explain why the only elements in $mathbb{Q}(alpha)$ when $alpha$ is not algebraic are $mathbb{Q}$.
abstract-algebra field-theory extension-field
asked Jan 31 at 17:18
dandan
623613
$endgroup$
add a comment |
$begingroup$
Let $g(x)$ be a polynomial in $mathbb{Q}[x]$ such the $deg(g)>1$. I wish to show that $mathbb{Q}$ are the only algebraic elements in $mathbb{Q(g(pi))}$.
I tried to show that $g(pi)$ is not algebraic, yet even if I assume it, I don't know how to explain why the only elements in $mathbb{Q}(alpha)$ when $alpha$ is not algebraic are $mathbb{Q}$.
abstract-algebra field-theory extension-field
asked Jan 31 at 17:18
dandan
623613
$endgroup$
1
$begingroup$
Suppose that $g(pi) $ is algebraic. Then there is a polynomial $f$ over the rationals from which is zero. Therefore $f(g(pi)) =0$, which contradicts the fact of that $pi$ is not algebraic.
$endgroup$
– Frank Murphy
Jan 31 at 17:35
add a comment |
$begingroup$
Let $g(x)$ be a polynomial in $mathbb{Q}[x]$ such the $deg(g)>1$. I wish to show that $mathbb{Q}$ are the only algebraic elements in $mathbb{Q(g(pi))}$.
I tried to show that $g(pi)$ is not algebraic, yet even if I assume it, I don't know how to explain why the only elements in $mathbb{Q}(alpha)$ when $alpha$ is not algebraic are $mathbb{Q}$.
abstract-algebra field-theory extension-field
asked Jan 31 at 17:18
dandan
623613
$endgroup$
Let $g(x)$ be a polynomial in $mathbb{Q}[x]$ such the $deg(g)>1$. I wish to show that $mathbb{Q}$ are the only algebraic elements in $mathbb{Q(g(pi))}$.
I tried to show that $g(pi)$ is not algebraic, yet even if I assume it, I don't know how to explain why the only elements in $mathbb{Q}(alpha)$ when $alpha$ is not algebraic are $mathbb{Q}$.
abstract-algebra field-theory extension-field
abstract-algebra field-theory extension-field
asked Jan 31 at 17:18
dandan
623613
asked Jan 31 at 17:18
dandan
623613
asked Jan 31 at 17:18
dandan
623613
asked Jan 31 at 17:18
dandan
623613
asked Jan 31 at 17:18
dandan
623613
623613
1
$begingroup$
Suppose that $g(pi) $ is algebraic. Then there is a polynomial $f$ over the rationals from which is zero. Therefore $f(g(pi)) =0$, which contradicts the fact of that $pi$ is not algebraic.
$endgroup$
– Frank Murphy
Jan 31 at 17:35
add a comment |
1
$begingroup$
Suppose that $g(pi) $ is algebraic. Then there is a polynomial $f$ over the rationals from which is zero. Therefore $f(g(pi)) =0$, which contradicts the fact of that $pi$ is not algebraic.
$endgroup$
– Frank Murphy
Jan 31 at 17:35
1
1
$begingroup$
Suppose that $g(pi) $ is algebraic. Then there is a polynomial $f$ over the rationals from which is zero. Therefore $f(g(pi)) =0$, which contradicts the fact of that $pi$ is not algebraic.
$endgroup$
– Frank Murphy
Jan 31 at 17:35
$begingroup$
Suppose that $g(pi) $ is algebraic. Then there is a polynomial $f$ over the rationals from which is zero. Therefore $f(g(pi)) =0$, which contradicts the fact of that $pi$ is not algebraic.
$endgroup$
– Frank Murphy
Jan 31 at 17:35
add a comment |
1 Answer
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$begingroup$
Hint: If $alpha$ is a complex number that is transcendental over $mathbb{Q}$ then $mathbb{Q}(alpha)$ is isomorphic to $mathbb{Q}(x)$ i.e., to the field of rational functions over $mathbb{Q}$.
Transcendental Extensions. $F(alpha)$ isomorphic to $F(x)$
answered Jan 31 at 17:30
Dietrich BurdeDietrich Burde
81.8k648106
$endgroup$
add a comment |
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$begingroup$
Hint: If $alpha$ is a complex number that is transcendental over $mathbb{Q}$ then $mathbb{Q}(alpha)$ is isomorphic to $mathbb{Q}(x)$ i.e., to the field of rational functions over $mathbb{Q}$.
Transcendental Extensions. $F(alpha)$ isomorphic to $F(x)$
answered Jan 31 at 17:30
Dietrich BurdeDietrich Burde
81.8k648106
$endgroup$
add a comment |
$begingroup$
Hint: If $alpha$ is a complex number that is transcendental over $mathbb{Q}$ then $mathbb{Q}(alpha)$ is isomorphic to $mathbb{Q}(x)$ i.e., to the field of rational functions over $mathbb{Q}$.
Transcendental Extensions. $F(alpha)$ isomorphic to $F(x)$
answered Jan 31 at 17:30
Dietrich BurdeDietrich Burde
81.8k648106
$endgroup$
add a comment |
$begingroup$
Hint: If $alpha$ is a complex number that is transcendental over $mathbb{Q}$ then $mathbb{Q}(alpha)$ is isomorphic to $mathbb{Q}(x)$ i.e., to the field of rational functions over $mathbb{Q}$.
Transcendental Extensions. $F(alpha)$ isomorphic to $F(x)$
answered Jan 31 at 17:30
Dietrich BurdeDietrich Burde
81.8k648106
$endgroup$
Hint: If $alpha$ is a complex number that is transcendental over $mathbb{Q}$ then $mathbb{Q}(alpha)$ is isomorphic to $mathbb{Q}(x)$ i.e., to the field of rational functions over $mathbb{Q}$.
Transcendental Extensions. $F(alpha)$ isomorphic to $F(x)$
answered Jan 31 at 17:30
Dietrich BurdeDietrich Burde
81.8k648106
edited Jan 31 at 19:32
answered Jan 31 at 17:30
Dietrich BurdeDietrich Burde
81.8k648106
answered Jan 31 at 17:30
Dietrich BurdeDietrich Burde
81.8k648106
answered Jan 31 at 17:30
Dietrich BurdeDietrich Burde
81.8k648106
81.8k648106
add a comment |
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Suppose that $g(pi) $ is algebraic. Then there is a polynomial $f$ over the rationals from which is zero. Therefore $f(g(pi)) =0$, which contradicts the fact of that $pi$ is not algebraic.
$endgroup$
– Frank Murphy
Jan 31 at 17:35