Proving an isomorphism between field extensions
$begingroup$
I am trying to prove the following:
Two field extensions $mathbb{Q}[sqrt{a}]$ and $mathbb{Q}[sqrt{b}],$ where $a,b in mathbb{Q}^times,$ are isomorphic if and only if $sqrt{a/b} in mathbb{Q}.$
I started backwards, that is I first assumed the fields were isomorphic. So there exists $phi$ such that
begin{gather}
phi(sqrt{a}) = s + tsqrt{b}.
end{gather}
This means that this must satisfy the following
begin{gather}
a = s^2 + 2stsqrt{b} + t^2b.
end{gather}
But why is it true that one of $s,t$ must be 0. However, I know that if I can show that then it will be clear that $sqrt{a/b} in mathbb{Q}.$ However, the other direction is not at all obvious, does anyone have any hints?
EDIT: So I started with an element of $u + vsqrt{a}$ and I tried mapping this to $u + vsqrt{b/a}.$ But this doesn't satisfy the multiplicative property of the isomorphism.
abstract-algebra polynomials field-theory irreducible-polynomials
asked Dec 5 '18 at 20:20
JosabanksJosabanks
936
$endgroup$
add a comment |
$begingroup$
I am trying to prove the following:
Two field extensions $mathbb{Q}[sqrt{a}]$ and $mathbb{Q}[sqrt{b}],$ where $a,b in mathbb{Q}^times,$ are isomorphic if and only if $sqrt{a/b} in mathbb{Q}.$
I started backwards, that is I first assumed the fields were isomorphic. So there exists $phi$ such that
begin{gather}
phi(sqrt{a}) = s + tsqrt{b}.
end{gather}
This means that this must satisfy the following
begin{gather}
a = s^2 + 2stsqrt{b} + t^2b.
end{gather}
But why is it true that one of $s,t$ must be 0. However, I know that if I can show that then it will be clear that $sqrt{a/b} in mathbb{Q}.$ However, the other direction is not at all obvious, does anyone have any hints?
EDIT: So I started with an element of $u + vsqrt{a}$ and I tried mapping this to $u + vsqrt{b/a}.$ But this doesn't satisfy the multiplicative property of the isomorphism.
abstract-algebra polynomials field-theory irreducible-polynomials
asked Dec 5 '18 at 20:20
JosabanksJosabanks
936
$endgroup$
4
$begingroup$
That implies $stsqrt b$ is rational.
$endgroup$
– Lord Shark the Unknown
Dec 5 '18 at 20:36
add a comment |
$begingroup$
I am trying to prove the following:
Two field extensions $mathbb{Q}[sqrt{a}]$ and $mathbb{Q}[sqrt{b}],$ where $a,b in mathbb{Q}^times,$ are isomorphic if and only if $sqrt{a/b} in mathbb{Q}.$
I started backwards, that is I first assumed the fields were isomorphic. So there exists $phi$ such that
begin{gather}
phi(sqrt{a}) = s + tsqrt{b}.
end{gather}
This means that this must satisfy the following
begin{gather}
a = s^2 + 2stsqrt{b} + t^2b.
end{gather}
But why is it true that one of $s,t$ must be 0. However, I know that if I can show that then it will be clear that $sqrt{a/b} in mathbb{Q}.$ However, the other direction is not at all obvious, does anyone have any hints?
EDIT: So I started with an element of $u + vsqrt{a}$ and I tried mapping this to $u + vsqrt{b/a}.$ But this doesn't satisfy the multiplicative property of the isomorphism.
abstract-algebra polynomials field-theory irreducible-polynomials
asked Dec 5 '18 at 20:20
JosabanksJosabanks
936
$endgroup$
I am trying to prove the following:
Two field extensions $mathbb{Q}[sqrt{a}]$ and $mathbb{Q}[sqrt{b}],$ where $a,b in mathbb{Q}^times,$ are isomorphic if and only if $sqrt{a/b} in mathbb{Q}.$
I started backwards, that is I first assumed the fields were isomorphic. So there exists $phi$ such that
begin{gather}
phi(sqrt{a}) = s + tsqrt{b}.
end{gather}
This means that this must satisfy the following
begin{gather}
a = s^2 + 2stsqrt{b} + t^2b.
end{gather}
But why is it true that one of $s,t$ must be 0. However, I know that if I can show that then it will be clear that $sqrt{a/b} in mathbb{Q}.$ However, the other direction is not at all obvious, does anyone have any hints?
EDIT: So I started with an element of $u + vsqrt{a}$ and I tried mapping this to $u + vsqrt{b/a}.$ But this doesn't satisfy the multiplicative property of the isomorphism.
abstract-algebra polynomials field-theory irreducible-polynomials
abstract-algebra polynomials field-theory irreducible-polynomials
asked Dec 5 '18 at 20:20
JosabanksJosabanks
936
asked Dec 5 '18 at 20:20
JosabanksJosabanks
936
asked Dec 5 '18 at 20:20
JosabanksJosabanks
936
asked Dec 5 '18 at 20:20
JosabanksJosabanks
936
asked Dec 5 '18 at 20:20
JosabanksJosabanks
936
936
4
$begingroup$
That implies $stsqrt b$ is rational.
$endgroup$
– Lord Shark the Unknown
Dec 5 '18 at 20:36
add a comment |
4
$begingroup$
That implies $stsqrt b$ is rational.
$endgroup$
– Lord Shark the Unknown
Dec 5 '18 at 20:36
4
4
$begingroup$
That implies $stsqrt b$ is rational.
$endgroup$
– Lord Shark the Unknown
Dec 5 '18 at 20:36
$begingroup$
That implies $stsqrt b$ is rational.
$endgroup$
– Lord Shark the Unknown
Dec 5 '18 at 20:36
add a comment |
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$begingroup$
That implies $stsqrt b$ is rational.
$endgroup$
– Lord Shark the Unknown
Dec 5 '18 at 20:36