Mixing
$mathbbQ(sqrt d)$ is the smallest subfield of $mathbbC$ containing $mathbbQ$ and $sqrt d$ using field norm
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
From Aluffi's "Algebra chapter 0" problem 4.10 from chapter 3. Prove that $mathbbQ(sqrt d)$ is a field and in fact the smallest subfield of $mathbbC$ containing both $mathbbQ$ and $sqrt d$.
I know how to show it directly by a standard argument (which has been reproduced many times on this site). But the author hints to use field norm. I am a little confused by how exactly are we supposed to use it here.
The only thing I can think of is this:
Let $zinmathbbQ(sqrt d)$ and write $z=a+bsqrt d$. We claim that $fraca-bsqrt da^2-b^2d$ is the multiplicative inverse of $z$. We only need to show $a^2-b^2dne0$. For this we note that $a^2-b^2d=N(z)$ and it is nonzero if $zne0$ (which has been proved in the same exercise).
I do not see how we can use norm to show the minimality of $mathbbQ(sqrt d)$.
abstract-algebra ring-theory field-theory
asked Jul 31 at 19:37
lanskey
544213
add a comment |Â
up vote
0
down vote
favorite
From Aluffi's "Algebra chapter 0" problem 4.10 from chapter 3. Prove that $mathbbQ(sqrt d)$ is a field and in fact the smallest subfield of $mathbbC$ containing both $mathbbQ$ and $sqrt d$.
I know how to show it directly by a standard argument (which has been reproduced many times on this site). But the author hints to use field norm. I am a little confused by how exactly are we supposed to use it here.
The only thing I can think of is this:
Let $zinmathbbQ(sqrt d)$ and write $z=a+bsqrt d$. We claim that $fraca-bsqrt da^2-b^2d$ is the multiplicative inverse of $z$. We only need to show $a^2-b^2dne0$. For this we note that $a^2-b^2d=N(z)$ and it is nonzero if $zne0$ (which has been proved in the same exercise).
I do not see how we can use norm to show the minimality of $mathbbQ(sqrt d)$.
abstract-algebra ring-theory field-theory
asked Jul 31 at 19:37
lanskey
544213
1
You use it to prove $a+bsqrt d:a,binBbb Q$ is a field.
– Lord Shark the Unknown
Jul 31 at 19:38
How do you define $Bbb Q(sqrt d)$? For me, it is defined as the smallest subfield of $Bbb C$ containing $Bbb Q$ and $sqrt d$...
– Kenny Lau
Jul 31 at 19:39
Yes, of course. But in the exercise it is defined as $a+bsqrt d mid a,binmathbbQ$.
– lanskey
Jul 31 at 19:40
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
From Aluffi's "Algebra chapter 0" problem 4.10 from chapter 3. Prove that $mathbbQ(sqrt d)$ is a field and in fact the smallest subfield of $mathbbC$ containing both $mathbbQ$ and $sqrt d$.
I know how to show it directly by a standard argument (which has been reproduced many times on this site). But the author hints to use field norm. I am a little confused by how exactly are we supposed to use it here.
The only thing I can think of is this:
Let $zinmathbbQ(sqrt d)$ and write $z=a+bsqrt d$. We claim that $fraca-bsqrt da^2-b^2d$ is the multiplicative inverse of $z$. We only need to show $a^2-b^2dne0$. For this we note that $a^2-b^2d=N(z)$ and it is nonzero if $zne0$ (which has been proved in the same exercise).
I do not see how we can use norm to show the minimality of $mathbbQ(sqrt d)$.
abstract-algebra ring-theory field-theory
asked Jul 31 at 19:37
lanskey
544213
From Aluffi's "Algebra chapter 0" problem 4.10 from chapter 3. Prove that $mathbbQ(sqrt d)$ is a field and in fact the smallest subfield of $mathbbC$ containing both $mathbbQ$ and $sqrt d$.
I know how to show it directly by a standard argument (which has been reproduced many times on this site). But the author hints to use field norm. I am a little confused by how exactly are we supposed to use it here.
The only thing I can think of is this:
Let $zinmathbbQ(sqrt d)$ and write $z=a+bsqrt d$. We claim that $fraca-bsqrt da^2-b^2d$ is the multiplicative inverse of $z$. We only need to show $a^2-b^2dne0$. For this we note that $a^2-b^2d=N(z)$ and it is nonzero if $zne0$ (which has been proved in the same exercise).
I do not see how we can use norm to show the minimality of $mathbbQ(sqrt d)$.
abstract-algebra ring-theory field-theory
asked Jul 31 at 19:37
lanskey
544213
edited Jul 31 at 19:38
asked Jul 31 at 19:37
lanskey
544213
asked Jul 31 at 19:37
lanskey
544213
asked Jul 31 at 19:37
lanskey
544213
544213
1
You use it to prove $a+bsqrt d:a,binBbb Q$ is a field.
– Lord Shark the Unknown
Jul 31 at 19:38
How do you define $Bbb Q(sqrt d)$? For me, it is defined as the smallest subfield of $Bbb C$ containing $Bbb Q$ and $sqrt d$...
– Kenny Lau
Jul 31 at 19:39
Yes, of course. But in the exercise it is defined as $a+bsqrt d mid a,binmathbbQ$.
– lanskey
Jul 31 at 19:40
add a comment |Â
1
You use it to prove $a+bsqrt d:a,binBbb Q$ is a field.
– Lord Shark the Unknown
Jul 31 at 19:38
How do you define $Bbb Q(sqrt d)$? For me, it is defined as the smallest subfield of $Bbb C$ containing $Bbb Q$ and $sqrt d$...
– Kenny Lau
Jul 31 at 19:39
Yes, of course. But in the exercise it is defined as $a+bsqrt d mid a,binmathbbQ$.
– lanskey
Jul 31 at 19:40
1
1
You use it to prove $a+bsqrt d:a,binBbb Q$ is a field.
– Lord Shark the Unknown
Jul 31 at 19:38
You use it to prove $a+bsqrt d:a,binBbb Q$ is a field.
– Lord Shark the Unknown
Jul 31 at 19:38
How do you define $Bbb Q(sqrt d)$? For me, it is defined as the smallest subfield of $Bbb C$ containing $Bbb Q$ and $sqrt d$...
– Kenny Lau
Jul 31 at 19:39
How do you define $Bbb Q(sqrt d)$? For me, it is defined as the smallest subfield of $Bbb C$ containing $Bbb Q$ and $sqrt d$...
– Kenny Lau
Jul 31 at 19:39
Yes, of course. But in the exercise it is defined as $a+bsqrt d mid a,binmathbbQ$.
– lanskey
Jul 31 at 19:40
Yes, of course. But in the exercise it is defined as $a+bsqrt d mid a,binmathbbQ$.
– lanskey
Jul 31 at 19:40
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
5
down vote
accepted
Prove that $mathbbQ(sqrtd)$ is a field and in fact the smallest
subfield of $mathbbC$ containing both $mathbbQ$ and $sqrtd$.
This is confusing, since $mathbbQ(sqrtd)$ may well be defined to be the smallest subfield of $mathbbC$ containing both $mathbbQ$ and $sqrtd$.
One reasonable interpretation of the question is to suppose that we're looking at the set $a+bsqrtd,mid, a,b in mathbbQ$, and calling that $mathbbQ(sqrtd)$. Our goal then is to show that this is the smallest subfield containing both $mathbbQ$ and $sqrtd$.
Now, clearly any subfield of $mathbbC$ containing both $mathbbQ$ and $sqrtd$ must contain all numbers of the form $a+bsqrtd$, so the only thing left is to show that this is actually a field. If it is, it has to be the smallest subfield with the required property.
To show that it's a field we need to show that every $a+bsqrtd$ has a multiplicative inverse of the same form, and this is where the norm comes in. Once you show that, you're done.
answered Jul 31 at 19:44
Alon Amit
10.2k3765
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Prove that $mathbbQ(sqrtd)$ is a field and in fact the smallest
subfield of $mathbbC$ containing both $mathbbQ$ and $sqrtd$.
This is confusing, since $mathbbQ(sqrtd)$ may well be defined to be the smallest subfield of $mathbbC$ containing both $mathbbQ$ and $sqrtd$.
One reasonable interpretation of the question is to suppose that we're looking at the set $a+bsqrtd,mid, a,b in mathbbQ$, and calling that $mathbbQ(sqrtd)$. Our goal then is to show that this is the smallest subfield containing both $mathbbQ$ and $sqrtd$.
Now, clearly any subfield of $mathbbC$ containing both $mathbbQ$ and $sqrtd$ must contain all numbers of the form $a+bsqrtd$, so the only thing left is to show that this is actually a field. If it is, it has to be the smallest subfield with the required property.
To show that it's a field we need to show that every $a+bsqrtd$ has a multiplicative inverse of the same form, and this is where the norm comes in. Once you show that, you're done.
answered Jul 31 at 19:44
Alon Amit
10.2k3765
add a comment |Â
up vote
5
down vote
accepted
Prove that $mathbbQ(sqrtd)$ is a field and in fact the smallest
subfield of $mathbbC$ containing both $mathbbQ$ and $sqrtd$.
This is confusing, since $mathbbQ(sqrtd)$ may well be defined to be the smallest subfield of $mathbbC$ containing both $mathbbQ$ and $sqrtd$.
One reasonable interpretation of the question is to suppose that we're looking at the set $a+bsqrtd,mid, a,b in mathbbQ$, and calling that $mathbbQ(sqrtd)$. Our goal then is to show that this is the smallest subfield containing both $mathbbQ$ and $sqrtd$.
Now, clearly any subfield of $mathbbC$ containing both $mathbbQ$ and $sqrtd$ must contain all numbers of the form $a+bsqrtd$, so the only thing left is to show that this is actually a field. If it is, it has to be the smallest subfield with the required property.
To show that it's a field we need to show that every $a+bsqrtd$ has a multiplicative inverse of the same form, and this is where the norm comes in. Once you show that, you're done.
answered Jul 31 at 19:44
Alon Amit
10.2k3765
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Prove that $mathbbQ(sqrtd)$ is a field and in fact the smallest
subfield of $mathbbC$ containing both $mathbbQ$ and $sqrtd$.
This is confusing, since $mathbbQ(sqrtd)$ may well be defined to be the smallest subfield of $mathbbC$ containing both $mathbbQ$ and $sqrtd$.
One reasonable interpretation of the question is to suppose that we're looking at the set $a+bsqrtd,mid, a,b in mathbbQ$, and calling that $mathbbQ(sqrtd)$. Our goal then is to show that this is the smallest subfield containing both $mathbbQ$ and $sqrtd$.
Now, clearly any subfield of $mathbbC$ containing both $mathbbQ$ and $sqrtd$ must contain all numbers of the form $a+bsqrtd$, so the only thing left is to show that this is actually a field. If it is, it has to be the smallest subfield with the required property.
To show that it's a field we need to show that every $a+bsqrtd$ has a multiplicative inverse of the same form, and this is where the norm comes in. Once you show that, you're done.
answered Jul 31 at 19:44
Alon Amit
10.2k3765
Prove that $mathbbQ(sqrtd)$ is a field and in fact the smallest
subfield of $mathbbC$ containing both $mathbbQ$ and $sqrtd$.
This is confusing, since $mathbbQ(sqrtd)$ may well be defined to be the smallest subfield of $mathbbC$ containing both $mathbbQ$ and $sqrtd$.
One reasonable interpretation of the question is to suppose that we're looking at the set $a+bsqrtd,mid, a,b in mathbbQ$, and calling that $mathbbQ(sqrtd)$. Our goal then is to show that this is the smallest subfield containing both $mathbbQ$ and $sqrtd$.
Now, clearly any subfield of $mathbbC$ containing both $mathbbQ$ and $sqrtd$ must contain all numbers of the form $a+bsqrtd$, so the only thing left is to show that this is actually a field. If it is, it has to be the smallest subfield with the required property.
To show that it's a field we need to show that every $a+bsqrtd$ has a multiplicative inverse of the same form, and this is where the norm comes in. Once you show that, you're done.
answered Jul 31 at 19:44
Alon Amit
10.2k3765
answered Jul 31 at 19:44
Alon Amit
10.2k3765
answered Jul 31 at 19:44
Alon Amit
10.2k3765
answered Jul 31 at 19:44
Alon Amit
10.2k3765
10.2k3765
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2868401%2fmathbbq-sqrt-d-is-the-smallest-subfield-of-mathbbc-containing-math%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
You use it to prove $a+bsqrt d:a,binBbb Q$ is a field.
– Lord Shark the Unknown
Jul 31 at 19:38
How do you define $Bbb Q(sqrt d)$? For me, it is defined as the smallest subfield of $Bbb C$ containing $Bbb Q$ and $sqrt d$...
– Kenny Lau
Jul 31 at 19:39
Yes, of course. But in the exercise it is defined as $a+bsqrt d mid a,binmathbbQ$.
– lanskey
Jul 31 at 19:40