Proving that $mathbb{Q}$ adjoin the square root of every prime is an infinite extension
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How would one show that $[mathbb{Q}(sqrt2, sqrt3,...,sqrt{p_n},...)]=infty$? I know that we want to show there is no finite basis over the rationals, but I'm not sure how one would determine that such a basis does not exist.
galois-theory algebraic-number-theory
asked Nov 7 '18 at 15:48
J.Jones5552J.Jones5552
624
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add a comment |
$begingroup$
How would one show that $[mathbb{Q}(sqrt2, sqrt3,...,sqrt{p_n},...)]=infty$? I know that we want to show there is no finite basis over the rationals, but I'm not sure how one would determine that such a basis does not exist.
galois-theory algebraic-number-theory
asked Nov 7 '18 at 15:48
J.Jones5552J.Jones5552
624
$endgroup$
$begingroup$
Isn't it enough that there are infinitely many primes? Maybe I've misunderstood this question.
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– David R.
Nov 7 '18 at 22:27
1
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See e.g. math.stackexchange.com/a/1609061/300700
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– nguyen quang do
Nov 9 '18 at 16:35
add a comment |
$begingroup$
How would one show that $[mathbb{Q}(sqrt2, sqrt3,...,sqrt{p_n},...)]=infty$? I know that we want to show there is no finite basis over the rationals, but I'm not sure how one would determine that such a basis does not exist.
galois-theory algebraic-number-theory
asked Nov 7 '18 at 15:48
J.Jones5552J.Jones5552
624
$endgroup$
How would one show that $[mathbb{Q}(sqrt2, sqrt3,...,sqrt{p_n},...)]=infty$? I know that we want to show there is no finite basis over the rationals, but I'm not sure how one would determine that such a basis does not exist.
galois-theory algebraic-number-theory
galois-theory algebraic-number-theory
asked Nov 7 '18 at 15:48
J.Jones5552J.Jones5552
624
asked Nov 7 '18 at 15:48
J.Jones5552J.Jones5552
624
asked Nov 7 '18 at 15:48
J.Jones5552J.Jones5552
624
asked Nov 7 '18 at 15:48
J.Jones5552J.Jones5552
624
asked Nov 7 '18 at 15:48
J.Jones5552J.Jones5552
624
624
$begingroup$
Isn't it enough that there are infinitely many primes? Maybe I've misunderstood this question.
$endgroup$
– David R.
Nov 7 '18 at 22:27
1
$begingroup$
See e.g. math.stackexchange.com/a/1609061/300700
$endgroup$
– nguyen quang do
Nov 9 '18 at 16:35
add a comment |
$begingroup$
Isn't it enough that there are infinitely many primes? Maybe I've misunderstood this question.
$endgroup$
– David R.
Nov 7 '18 at 22:27
1
$begingroup$
See e.g. math.stackexchange.com/a/1609061/300700
$endgroup$
– nguyen quang do
Nov 9 '18 at 16:35
$begingroup$
Isn't it enough that there are infinitely many primes? Maybe I've misunderstood this question.
$endgroup$
– David R.
Nov 7 '18 at 22:27
$begingroup$
Isn't it enough that there are infinitely many primes? Maybe I've misunderstood this question.
$endgroup$
– David R.
Nov 7 '18 at 22:27
1
1
$begingroup$
See e.g. math.stackexchange.com/a/1609061/300700
$endgroup$
– nguyen quang do
Nov 9 '18 at 16:35
$begingroup$
See e.g. math.stackexchange.com/a/1609061/300700
$endgroup$
– nguyen quang do
Nov 9 '18 at 16:35
add a comment |
1 Answer
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$begingroup$
Kummer theory tells us that as $ 2$, $3,ldots,{p_n}$
generate a subgroup of order $2^n$ in $Bbb Q^*/(Bbb Q^*)^2$ then the
extension field $K_n=Bbb Q(sqrt2,ldots,sqrt {p_n})$ has degree $2^n$
over $Bbb Q$. So your field contains subfields of arbitrarily large degree
over $Bbb Q$.
answered Nov 7 '18 at 16:48
Lord Shark the UnknownLord Shark the Unknown
107k1162135
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1 Answer
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1 Answer
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$begingroup$
Kummer theory tells us that as $ 2$, $3,ldots,{p_n}$
generate a subgroup of order $2^n$ in $Bbb Q^*/(Bbb Q^*)^2$ then the
extension field $K_n=Bbb Q(sqrt2,ldots,sqrt {p_n})$ has degree $2^n$
over $Bbb Q$. So your field contains subfields of arbitrarily large degree
over $Bbb Q$.
answered Nov 7 '18 at 16:48
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
add a comment |
$begingroup$
Kummer theory tells us that as $ 2$, $3,ldots,{p_n}$
generate a subgroup of order $2^n$ in $Bbb Q^*/(Bbb Q^*)^2$ then the
extension field $K_n=Bbb Q(sqrt2,ldots,sqrt {p_n})$ has degree $2^n$
over $Bbb Q$. So your field contains subfields of arbitrarily large degree
over $Bbb Q$.
answered Nov 7 '18 at 16:48
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
add a comment |
$begingroup$
Kummer theory tells us that as $ 2$, $3,ldots,{p_n}$
generate a subgroup of order $2^n$ in $Bbb Q^*/(Bbb Q^*)^2$ then the
extension field $K_n=Bbb Q(sqrt2,ldots,sqrt {p_n})$ has degree $2^n$
over $Bbb Q$. So your field contains subfields of arbitrarily large degree
over $Bbb Q$.
answered Nov 7 '18 at 16:48
Lord Shark the UnknownLord Shark the Unknown
107k1162135
$endgroup$
Kummer theory tells us that as $ 2$, $3,ldots,{p_n}$
generate a subgroup of order $2^n$ in $Bbb Q^*/(Bbb Q^*)^2$ then the
extension field $K_n=Bbb Q(sqrt2,ldots,sqrt {p_n})$ has degree $2^n$
over $Bbb Q$. So your field contains subfields of arbitrarily large degree
over $Bbb Q$.
answered Nov 7 '18 at 16:48
Lord Shark the UnknownLord Shark the Unknown
107k1162135
edited Jan 2 at 2:47
answered Nov 7 '18 at 16:48
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered Nov 7 '18 at 16:48
Lord Shark the UnknownLord Shark the Unknown
107k1162135
answered Nov 7 '18 at 16:48
Lord Shark the UnknownLord Shark the Unknown
107k1162135
107k1162135
add a comment |
add a comment |
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$begingroup$
Isn't it enough that there are infinitely many primes? Maybe I've misunderstood this question.
$endgroup$
– David R.
Nov 7 '18 at 22:27
1
$begingroup$
See e.g. math.stackexchange.com/a/1609061/300700
$endgroup$
– nguyen quang do
Nov 9 '18 at 16:35