If $p, q$ are prime integers, then $mathbb{Q}(sqrt{p})$ is not isomorphic (as a field) to...
This question already has an answer here:
$mathbb{Q}(sqrt{n}) cong mathbb{Q}(sqrt{m})$ iff $n=m$ [duplicate]
2 answers
My strategy is something like this: suppose $phi: mathbb{Q}(sqrt{p}) to mathbb{Q}(sqrt{q})$ is a isomophism such that $phi(x) = x$ for all $x in mathbb{Q}$ and let $phi(sqrt{p}) = a + b sqrt{q}$. Then $p = phi(p) = phi(sqrt{p}^2) = phi(sqrt{p})^2 = (a + b sqrt{q})^2 = a^2 + 2ab sqrt{q} + b^2 p$.
I feel like '$phi(x) = x$ for all $x in mathbb{Q}$' can be proved from the supposition that $phi: mathbb{Q}(sqrt{p}) to mathbb{Q}(sqrt{q})$ is an isomorphism, but I am not sure how to prove it. Can anyone help me?
field-theory extension-field
asked Nov 26 '18 at 6:28
Nuntractatuses Amável
61812
marked as duplicate by Watson, KReiser, ancientmathematician, Paul Frost, Did Nov 26 '18 at 11:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
$mathbb{Q}(sqrt{n}) cong mathbb{Q}(sqrt{m})$ iff $n=m$ [duplicate]
2 answers
My strategy is something like this: suppose $phi: mathbb{Q}(sqrt{p}) to mathbb{Q}(sqrt{q})$ is a isomophism such that $phi(x) = x$ for all $x in mathbb{Q}$ and let $phi(sqrt{p}) = a + b sqrt{q}$. Then $p = phi(p) = phi(sqrt{p}^2) = phi(sqrt{p})^2 = (a + b sqrt{q})^2 = a^2 + 2ab sqrt{q} + b^2 p$.
I feel like '$phi(x) = x$ for all $x in mathbb{Q}$' can be proved from the supposition that $phi: mathbb{Q}(sqrt{p}) to mathbb{Q}(sqrt{q})$ is an isomorphism, but I am not sure how to prove it. Can anyone help me?
field-theory extension-field
asked Nov 26 '18 at 6:28
Nuntractatuses Amável
61812
marked as duplicate by Watson, KReiser, ancientmathematician, Paul Frost, Did Nov 26 '18 at 11:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Even more general: math.stackexchange.com/questions/353337
– Watson
Nov 26 '18 at 7:44
add a comment |
This question already has an answer here:
$mathbb{Q}(sqrt{n}) cong mathbb{Q}(sqrt{m})$ iff $n=m$ [duplicate]
2 answers
My strategy is something like this: suppose $phi: mathbb{Q}(sqrt{p}) to mathbb{Q}(sqrt{q})$ is a isomophism such that $phi(x) = x$ for all $x in mathbb{Q}$ and let $phi(sqrt{p}) = a + b sqrt{q}$. Then $p = phi(p) = phi(sqrt{p}^2) = phi(sqrt{p})^2 = (a + b sqrt{q})^2 = a^2 + 2ab sqrt{q} + b^2 p$.
I feel like '$phi(x) = x$ for all $x in mathbb{Q}$' can be proved from the supposition that $phi: mathbb{Q}(sqrt{p}) to mathbb{Q}(sqrt{q})$ is an isomorphism, but I am not sure how to prove it. Can anyone help me?
field-theory extension-field
asked Nov 26 '18 at 6:28
Nuntractatuses Amável
61812
This question already has an answer here:
$mathbb{Q}(sqrt{n}) cong mathbb{Q}(sqrt{m})$ iff $n=m$ [duplicate]
2 answers
My strategy is something like this: suppose $phi: mathbb{Q}(sqrt{p}) to mathbb{Q}(sqrt{q})$ is a isomophism such that $phi(x) = x$ for all $x in mathbb{Q}$ and let $phi(sqrt{p}) = a + b sqrt{q}$. Then $p = phi(p) = phi(sqrt{p}^2) = phi(sqrt{p})^2 = (a + b sqrt{q})^2 = a^2 + 2ab sqrt{q} + b^2 p$.
I feel like '$phi(x) = x$ for all $x in mathbb{Q}$' can be proved from the supposition that $phi: mathbb{Q}(sqrt{p}) to mathbb{Q}(sqrt{q})$ is an isomorphism, but I am not sure how to prove it. Can anyone help me?
This question already has an answer here:
$mathbb{Q}(sqrt{n}) cong mathbb{Q}(sqrt{m})$ iff $n=m$ [duplicate]
2 answers
field-theory extension-field
field-theory extension-field
asked Nov 26 '18 at 6:28
Nuntractatuses Amável
61812
asked Nov 26 '18 at 6:28
Nuntractatuses Amável
61812
asked Nov 26 '18 at 6:28
Nuntractatuses Amável
61812
asked Nov 26 '18 at 6:28
Nuntractatuses Amável
61812
asked Nov 26 '18 at 6:28
Nuntractatuses Amável
61812
61812
marked as duplicate by Watson, KReiser, ancientmathematician, Paul Frost, Did Nov 26 '18 at 11:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Watson, KReiser, ancientmathematician, Paul Frost, Did Nov 26 '18 at 11:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Even more general: math.stackexchange.com/questions/353337
– Watson
Nov 26 '18 at 7:44
add a comment |
Even more general: math.stackexchange.com/questions/353337
– Watson
Nov 26 '18 at 7:44
Even more general: math.stackexchange.com/questions/353337
– Watson
Nov 26 '18 at 7:44
Even more general: math.stackexchange.com/questions/353337
– Watson
Nov 26 '18 at 7:44
add a comment |
1 Answer
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Hint $phi(1)=1$ implies $phi(n)=n$.. Now use $phi(mx)=m phi(x)$ to deduce that $phi(x)=x$ for all $x in mathbb Q$.
Also note that $phi(sqrt{p})$ has to be a root of $X^2-p=0$, and the roots of this are...
answered Nov 26 '18 at 6:32
N. S.
102k5110205
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint $phi(1)=1$ implies $phi(n)=n$.. Now use $phi(mx)=m phi(x)$ to deduce that $phi(x)=x$ for all $x in mathbb Q$.
Also note that $phi(sqrt{p})$ has to be a root of $X^2-p=0$, and the roots of this are...
answered Nov 26 '18 at 6:32
N. S.
102k5110205
add a comment |
Hint $phi(1)=1$ implies $phi(n)=n$.. Now use $phi(mx)=m phi(x)$ to deduce that $phi(x)=x$ for all $x in mathbb Q$.
Also note that $phi(sqrt{p})$ has to be a root of $X^2-p=0$, and the roots of this are...
answered Nov 26 '18 at 6:32
N. S.
102k5110205
add a comment |
Hint $phi(1)=1$ implies $phi(n)=n$.. Now use $phi(mx)=m phi(x)$ to deduce that $phi(x)=x$ for all $x in mathbb Q$.
Also note that $phi(sqrt{p})$ has to be a root of $X^2-p=0$, and the roots of this are...
answered Nov 26 '18 at 6:32
N. S.
102k5110205
Hint $phi(1)=1$ implies $phi(n)=n$.. Now use $phi(mx)=m phi(x)$ to deduce that $phi(x)=x$ for all $x in mathbb Q$.
Also note that $phi(sqrt{p})$ has to be a root of $X^2-p=0$, and the roots of this are...
answered Nov 26 '18 at 6:32
N. S.
102k5110205
answered Nov 26 '18 at 6:32
N. S.
102k5110205
answered Nov 26 '18 at 6:32
N. S.
102k5110205
answered Nov 26 '18 at 6:32
N. S.
102k5110205
102k5110205
add a comment |
add a comment |
Even more general: math.stackexchange.com/questions/353337
– Watson
Nov 26 '18 at 7:44