Find the quotient field of $mathbb{Z}[sqrt {2}]$
$begingroup$
My idea: how to make a connection between this with isomorphism.
abstract-algebra quotient-spaces
edited May 17 '14 at 2:12
Benjamin Dickman
10.3k22968
asked May 17 '14 at 1:44
Danial JeoDanial Jeo
508
$endgroup$
add a comment |
$begingroup$
My idea: how to make a connection between this with isomorphism.
abstract-algebra quotient-spaces
edited May 17 '14 at 2:12
Benjamin Dickman
10.3k22968
asked May 17 '14 at 1:44
Danial JeoDanial Jeo
508
$endgroup$
1
$begingroup$
Given a quotient of elements in $Z[sqrt{2}]$ you get an element of $Q[sqrt 2]$ by multiplying top and bottom by the conjugate of the denominator.
$endgroup$
– Dane
May 17 '14 at 1:46
$begingroup$
can we replace root of 2 with any number outside Q. @Dane
$endgroup$
– Danial Jeo
May 17 '14 at 1:47
add a comment |
$begingroup$
My idea: how to make a connection between this with isomorphism.
abstract-algebra quotient-spaces
edited May 17 '14 at 2:12
Benjamin Dickman
10.3k22968
asked May 17 '14 at 1:44
Danial JeoDanial Jeo
508
$endgroup$
My idea: how to make a connection between this with isomorphism.
abstract-algebra quotient-spaces
abstract-algebra quotient-spaces
edited May 17 '14 at 2:12
Benjamin Dickman
10.3k22968
asked May 17 '14 at 1:44
Danial JeoDanial Jeo
508
edited May 17 '14 at 2:12
Benjamin Dickman
10.3k22968
asked May 17 '14 at 1:44
Danial JeoDanial Jeo
508
edited May 17 '14 at 2:12
Benjamin Dickman
10.3k22968
edited May 17 '14 at 2:12
Benjamin Dickman
10.3k22968
edited May 17 '14 at 2:12
Benjamin Dickman
10.3k22968
10.3k22968
asked May 17 '14 at 1:44
Danial JeoDanial Jeo
508
asked May 17 '14 at 1:44
Danial JeoDanial Jeo
508
asked May 17 '14 at 1:44
Danial JeoDanial Jeo
508
508
1
$begingroup$
Given a quotient of elements in $Z[sqrt{2}]$ you get an element of $Q[sqrt 2]$ by multiplying top and bottom by the conjugate of the denominator.
$endgroup$
– Dane
May 17 '14 at 1:46
$begingroup$
can we replace root of 2 with any number outside Q. @Dane
$endgroup$
– Danial Jeo
May 17 '14 at 1:47
add a comment |
1
$begingroup$
Given a quotient of elements in $Z[sqrt{2}]$ you get an element of $Q[sqrt 2]$ by multiplying top and bottom by the conjugate of the denominator.
$endgroup$
– Dane
May 17 '14 at 1:46
$begingroup$
can we replace root of 2 with any number outside Q. @Dane
$endgroup$
– Danial Jeo
May 17 '14 at 1:47
1
1
$begingroup$
Given a quotient of elements in $Z[sqrt{2}]$ you get an element of $Q[sqrt 2]$ by multiplying top and bottom by the conjugate of the denominator.
$endgroup$
– Dane
May 17 '14 at 1:46
$begingroup$
Given a quotient of elements in $Z[sqrt{2}]$ you get an element of $Q[sqrt 2]$ by multiplying top and bottom by the conjugate of the denominator.
$endgroup$
– Dane
May 17 '14 at 1:46
$begingroup$
can we replace root of 2 with any number outside Q. @Dane
$endgroup$
– Danial Jeo
May 17 '14 at 1:47
$begingroup$
can we replace root of 2 with any number outside Q. @Dane
$endgroup$
– Danial Jeo
May 17 '14 at 1:47
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
$mathbb Q[sqrt 2]=mathbb Q(sqrt 2)$ because $sqrt 2$ is algebraic.
$mathbb Q(sqrt 2)$ is the smallest subfield of $mathbb C$ that contains $mathbb Q$ and $sqrt 2$.
$mathbb Q(sqrt 2)$ is thus the smallest subfield of $mathbb C$ that contains $mathbb Z$ and $sqrt 2$.
Hence, $mathbb Q(sqrt 2)$ is the field of fractions of $mathbb Z[sqrt 2]$.
More generally, by the same argument, if $alpha$ is an algebraic number, then $mathbb Q[alpha]=mathbb Q(alpha)$ and is the field of fractions of $mathbb Z[alpha]$.
answered May 17 '14 at 2:05
lhflhf
165k10171396
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add a comment |
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$begingroup$
$mathbb Q[sqrt 2]=mathbb Q(sqrt 2)$ because $sqrt 2$ is algebraic.
$mathbb Q(sqrt 2)$ is the smallest subfield of $mathbb C$ that contains $mathbb Q$ and $sqrt 2$.
$mathbb Q(sqrt 2)$ is thus the smallest subfield of $mathbb C$ that contains $mathbb Z$ and $sqrt 2$.
Hence, $mathbb Q(sqrt 2)$ is the field of fractions of $mathbb Z[sqrt 2]$.
More generally, by the same argument, if $alpha$ is an algebraic number, then $mathbb Q[alpha]=mathbb Q(alpha)$ and is the field of fractions of $mathbb Z[alpha]$.
answered May 17 '14 at 2:05
lhflhf
165k10171396
$endgroup$
add a comment |
$begingroup$
$mathbb Q[sqrt 2]=mathbb Q(sqrt 2)$ because $sqrt 2$ is algebraic.
$mathbb Q(sqrt 2)$ is the smallest subfield of $mathbb C$ that contains $mathbb Q$ and $sqrt 2$.
$mathbb Q(sqrt 2)$ is thus the smallest subfield of $mathbb C$ that contains $mathbb Z$ and $sqrt 2$.
Hence, $mathbb Q(sqrt 2)$ is the field of fractions of $mathbb Z[sqrt 2]$.
More generally, by the same argument, if $alpha$ is an algebraic number, then $mathbb Q[alpha]=mathbb Q(alpha)$ and is the field of fractions of $mathbb Z[alpha]$.
answered May 17 '14 at 2:05
lhflhf
165k10171396
$endgroup$
add a comment |
$begingroup$
$mathbb Q[sqrt 2]=mathbb Q(sqrt 2)$ because $sqrt 2$ is algebraic.
$mathbb Q(sqrt 2)$ is the smallest subfield of $mathbb C$ that contains $mathbb Q$ and $sqrt 2$.
$mathbb Q(sqrt 2)$ is thus the smallest subfield of $mathbb C$ that contains $mathbb Z$ and $sqrt 2$.
Hence, $mathbb Q(sqrt 2)$ is the field of fractions of $mathbb Z[sqrt 2]$.
More generally, by the same argument, if $alpha$ is an algebraic number, then $mathbb Q[alpha]=mathbb Q(alpha)$ and is the field of fractions of $mathbb Z[alpha]$.
answered May 17 '14 at 2:05
lhflhf
165k10171396
$endgroup$
$mathbb Q[sqrt 2]=mathbb Q(sqrt 2)$ because $sqrt 2$ is algebraic.
$mathbb Q(sqrt 2)$ is the smallest subfield of $mathbb C$ that contains $mathbb Q$ and $sqrt 2$.
$mathbb Q(sqrt 2)$ is thus the smallest subfield of $mathbb C$ that contains $mathbb Z$ and $sqrt 2$.
Hence, $mathbb Q(sqrt 2)$ is the field of fractions of $mathbb Z[sqrt 2]$.
More generally, by the same argument, if $alpha$ is an algebraic number, then $mathbb Q[alpha]=mathbb Q(alpha)$ and is the field of fractions of $mathbb Z[alpha]$.
answered May 17 '14 at 2:05
lhflhf
165k10171396
edited May 17 '14 at 2:21
answered May 17 '14 at 2:05
lhflhf
165k10171396
answered May 17 '14 at 2:05
lhflhf
165k10171396
answered May 17 '14 at 2:05
lhflhf
165k10171396
165k10171396
add a comment |
add a comment |
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Given a quotient of elements in $Z[sqrt{2}]$ you get an element of $Q[sqrt 2]$ by multiplying top and bottom by the conjugate of the denominator.
$endgroup$
– Dane
May 17 '14 at 1:46
$begingroup$
can we replace root of 2 with any number outside Q. @Dane
$endgroup$
– Danial Jeo
May 17 '14 at 1:47