Finding the norm map on number fields

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Let $K$ be a number field and $L$ be a finite extension of $K$ of degree $n$. I want to define a norm map on $L$. So that, for some $alpha in L$, I want to find the matrix $mu_alpha$, the map of the transformation $x mapsto alpha x$ for for any $xin K$.

For example let $L = mathbbQ(sqrt2)$ and $K=mathbbQ$. $L$ has basis $(1,sqrt2)$. Let $alpha = a +bsqrt2$. If we check for the bais elements,we see that $1 mapsto a + bsqrt2$ and $sqrt2 mapsto 2b+asqrt2$. Thus, I assume that the transformation matrix for $mu_alpha$ is given by:

$
M=
left[ beginarraycc
a & 2b \
b & a\
endarray right]
$

However, $1 mapsto a+bsqrt2$ but we have $[1,0] left[ beginarraycc
a & 2b \
b & a\
endarray right] = [a , 2b] = a + 2bsqrt2 neq a+bsqrt2 .$

Where is the problem and how can we find the desired transformation matrix?

linear-algebra field-theory algebraic-number-theory extension-field

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asked Jul 28 at 19:40

Ninja

1,074520

add a comment | 

up vote
1
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1

Let $K$ be a number field and $L$ be a finite extension of $K$ of degree $n$. I want to define a norm map on $L$. So that, for some $alpha in L$, I want to find the matrix $mu_alpha$, the map of the transformation $x mapsto alpha x$ for for any $xin K$.

For example let $L = mathbbQ(sqrt2)$ and $K=mathbbQ$. $L$ has basis $(1,sqrt2)$. Let $alpha = a +bsqrt2$. If we check for the bais elements,we see that $1 mapsto a + bsqrt2$ and $sqrt2 mapsto 2b+asqrt2$. Thus, I assume that the transformation matrix for $mu_alpha$ is given by:

$
M=
left[ beginarraycc
a & 2b \
b & a\
endarray right]
$

However, $1 mapsto a+bsqrt2$ but we have $[1,0] left[ beginarraycc
a & 2b \
b & a\
endarray right] = [a , 2b] = a + 2bsqrt2 neq a+bsqrt2 .$

Where is the problem and how can we find the desired transformation matrix?

linear-algebra field-theory algebraic-number-theory extension-field

share|cite|improve this question

asked Jul 28 at 19:40

Ninja

1,074520

add a comment | 

up vote
1
down vote

favorite

1

up vote
1
down vote

favorite

1

1

Let $K$ be a number field and $L$ be a finite extension of $K$ of degree $n$. I want to define a norm map on $L$. So that, for some $alpha in L$, I want to find the matrix $mu_alpha$, the map of the transformation $x mapsto alpha x$ for for any $xin K$.

For example let $L = mathbbQ(sqrt2)$ and $K=mathbbQ$. $L$ has basis $(1,sqrt2)$. Let $alpha = a +bsqrt2$. If we check for the bais elements,we see that $1 mapsto a + bsqrt2$ and $sqrt2 mapsto 2b+asqrt2$. Thus, I assume that the transformation matrix for $mu_alpha$ is given by:

$
M=
left[ beginarraycc
a & 2b \
b & a\
endarray right]
$

However, $1 mapsto a+bsqrt2$ but we have $[1,0] left[ beginarraycc
a & 2b \
b & a\
endarray right] = [a , 2b] = a + 2bsqrt2 neq a+bsqrt2 .$

Where is the problem and how can we find the desired transformation matrix?

linear-algebra field-theory algebraic-number-theory extension-field

share|cite|improve this question

asked Jul 28 at 19:40

Ninja

1,074520

Let $K$ be a number field and $L$ be a finite extension of $K$ of degree $n$. I want to define a norm map on $L$. So that, for some $alpha in L$, I want to find the matrix $mu_alpha$, the map of the transformation $x mapsto alpha x$ for for any $xin K$.

For example let $L = mathbbQ(sqrt2)$ and $K=mathbbQ$. $L$ has basis $(1,sqrt2)$. Let $alpha = a +bsqrt2$. If we check for the bais elements,we see that $1 mapsto a + bsqrt2$ and $sqrt2 mapsto 2b+asqrt2$. Thus, I assume that the transformation matrix for $mu_alpha$ is given by:

$
M=
left[ beginarraycc
a & 2b \
b & a\
endarray right]
$

However, $1 mapsto a+bsqrt2$ but we have $[1,0] left[ beginarraycc
a & 2b \
b & a\
endarray right] = [a , 2b] = a + 2bsqrt2 neq a+bsqrt2 .$

Where is the problem and how can we find the desired transformation matrix?

linear-algebra field-theory algebraic-number-theory extension-field

share|cite|improve this question

asked Jul 28 at 19:40

Ninja

1,074520

share|cite|improve this question

share|cite|improve this question

asked Jul 28 at 19:40

Ninja

1,074520

asked Jul 28 at 19:40

Ninja

1,074520

asked Jul 28 at 19:40

Ninja

1,074520

1,074520

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote



accepted

You should multiply the matrix with the coordinate vector from the right side. Indeed you would have:

$$left[ beginarraycc
a & 2b \
b & a\
endarray right]left[ beginarrayc
1 \
0 \
endarray right]=[a enspace b]$$

On the other side if you want to keep left multiplication the transformation matrix should be:

$$M = left[ beginarraycc
a & b \
2b & a\
endarray right]$$

In other words, instead of writing the values for the basis elements in the columns you are doing it in a row.

---

Let's determine the transformation matrix under left multiplication. We know that it's of the form: $M= left[ beginarraycc
x & y \
z & t\
endarray right]$, where $x,y,z,t$ are yet to be determined.

Now we know that $[1,0]M = [a,b]$ by the definition of the function. Now just perform the multiplication on the left-hand side and comapre coefficients to get: $x=a$ and $y=b$.

Similarly from $[0,1]M = [2b,a]$ we get that $z=2b$ and $t=a$. Therefore the transformation matrix becomes $M = left[ beginarraycc
a & b \
2b & a\
endarray right]$

share|cite|improve this answer

edited Jul 29 at 11:50

Ninja

1,074520

answered Jul 28 at 22:27

Stefan4024

28k52974

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you explain the reason please?
  – Ninja
  Jul 28 at 22:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ninja I added some further explanation. I hope it answers your question
  – Stefan4024
  Jul 28 at 22:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  One last question: In the notes www1.spms.ntu.edu.sg/~frederique/antchap1.pdf, how does the author write the matrix? It multiplies from left and different from the one that we say.
  – Ninja
  Jul 29 at 10:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ninja I'm not sure how he computes it, but something seems fishy. I mean the author has $(1,sqrt2)$ on the left, which can't be written like this in a basis.
  – Stefan4024
  Jul 29 at 11:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For the general case (not only quadratic, see math.stackexchange.com/a/2710883/300700
  – nguyen quang do
  Jul 30 at 12:33
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

You should multiply the matrix with the coordinate vector from the right side. Indeed you would have:

$$left[ beginarraycc
a & 2b \
b & a\
endarray right]left[ beginarrayc
1 \
0 \
endarray right]=[a enspace b]$$

On the other side if you want to keep left multiplication the transformation matrix should be:

$$M = left[ beginarraycc
a & b \
2b & a\
endarray right]$$

In other words, instead of writing the values for the basis elements in the columns you are doing it in a row.

---

Let's determine the transformation matrix under left multiplication. We know that it's of the form: $M= left[ beginarraycc
x & y \
z & t\
endarray right]$, where $x,y,z,t$ are yet to be determined.

Now we know that $[1,0]M = [a,b]$ by the definition of the function. Now just perform the multiplication on the left-hand side and comapre coefficients to get: $x=a$ and $y=b$.

Similarly from $[0,1]M = [2b,a]$ we get that $z=2b$ and $t=a$. Therefore the transformation matrix becomes $M = left[ beginarraycc
a & b \
2b & a\
endarray right]$

share|cite|improve this answer

edited Jul 29 at 11:50

Ninja

1,074520

answered Jul 28 at 22:27

Stefan4024

28k52974

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you explain the reason please?
  – Ninja
  Jul 28 at 22:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ninja I added some further explanation. I hope it answers your question
  – Stefan4024
  Jul 28 at 22:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  One last question: In the notes www1.spms.ntu.edu.sg/~frederique/antchap1.pdf, how does the author write the matrix? It multiplies from left and different from the one that we say.
  – Ninja
  Jul 29 at 10:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ninja I'm not sure how he computes it, but something seems fishy. I mean the author has $(1,sqrt2)$ on the left, which can't be written like this in a basis.
  – Stefan4024
  Jul 29 at 11:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For the general case (not only quadratic, see math.stackexchange.com/a/2710883/300700
  – nguyen quang do
  Jul 30 at 12:33
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

You should multiply the matrix with the coordinate vector from the right side. Indeed you would have:

$$left[ beginarraycc
a & 2b \
b & a\
endarray right]left[ beginarrayc
1 \
0 \
endarray right]=[a enspace b]$$

On the other side if you want to keep left multiplication the transformation matrix should be:

$$M = left[ beginarraycc
a & b \
2b & a\
endarray right]$$

In other words, instead of writing the values for the basis elements in the columns you are doing it in a row.

---

Let's determine the transformation matrix under left multiplication. We know that it's of the form: $M= left[ beginarraycc
x & y \
z & t\
endarray right]$, where $x,y,z,t$ are yet to be determined.

Now we know that $[1,0]M = [a,b]$ by the definition of the function. Now just perform the multiplication on the left-hand side and comapre coefficients to get: $x=a$ and $y=b$.

Similarly from $[0,1]M = [2b,a]$ we get that $z=2b$ and $t=a$. Therefore the transformation matrix becomes $M = left[ beginarraycc
a & b \
2b & a\
endarray right]$

share|cite|improve this answer

edited Jul 29 at 11:50

Ninja

1,074520

answered Jul 28 at 22:27

Stefan4024

28k52974

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you explain the reason please?
  – Ninja
  Jul 28 at 22:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ninja I added some further explanation. I hope it answers your question
  – Stefan4024
  Jul 28 at 22:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  One last question: In the notes www1.spms.ntu.edu.sg/~frederique/antchap1.pdf, how does the author write the matrix? It multiplies from left and different from the one that we say.
  – Ninja
  Jul 29 at 10:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ninja I'm not sure how he computes it, but something seems fishy. I mean the author has $(1,sqrt2)$ on the left, which can't be written like this in a basis.
  – Stefan4024
  Jul 29 at 11:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For the general case (not only quadratic, see math.stackexchange.com/a/2710883/300700
  – nguyen quang do
  Jul 30 at 12:33
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

You should multiply the matrix with the coordinate vector from the right side. Indeed you would have:

$$left[ beginarraycc
a & 2b \
b & a\
endarray right]left[ beginarrayc
1 \
0 \
endarray right]=[a enspace b]$$

On the other side if you want to keep left multiplication the transformation matrix should be:

$$M = left[ beginarraycc
a & b \
2b & a\
endarray right]$$

In other words, instead of writing the values for the basis elements in the columns you are doing it in a row.

---

Let's determine the transformation matrix under left multiplication. We know that it's of the form: $M= left[ beginarraycc
x & y \
z & t\
endarray right]$, where $x,y,z,t$ are yet to be determined.

Now we know that $[1,0]M = [a,b]$ by the definition of the function. Now just perform the multiplication on the left-hand side and comapre coefficients to get: $x=a$ and $y=b$.

Similarly from $[0,1]M = [2b,a]$ we get that $z=2b$ and $t=a$. Therefore the transformation matrix becomes $M = left[ beginarraycc
a & b \
2b & a\
endarray right]$

share|cite|improve this answer

edited Jul 29 at 11:50

Ninja

1,074520

answered Jul 28 at 22:27

Stefan4024

28k52974

You should multiply the matrix with the coordinate vector from the right side. Indeed you would have:

$$left[ beginarraycc
a & 2b \
b & a\
endarray right]left[ beginarrayc
1 \
0 \
endarray right]=[a enspace b]$$

On the other side if you want to keep left multiplication the transformation matrix should be:

$$M = left[ beginarraycc
a & b \
2b & a\
endarray right]$$

In other words, instead of writing the values for the basis elements in the columns you are doing it in a row.

---

Let's determine the transformation matrix under left multiplication. We know that it's of the form: $M= left[ beginarraycc
x & y \
z & t\
endarray right]$, where $x,y,z,t$ are yet to be determined.

Now we know that $[1,0]M = [a,b]$ by the definition of the function. Now just perform the multiplication on the left-hand side and comapre coefficients to get: $x=a$ and $y=b$.

Similarly from $[0,1]M = [2b,a]$ we get that $z=2b$ and $t=a$. Therefore the transformation matrix becomes $M = left[ beginarraycc
a & b \
2b & a\
endarray right]$

share|cite|improve this answer

edited Jul 29 at 11:50

Ninja

1,074520

answered Jul 28 at 22:27

Stefan4024

28k52974

share|cite|improve this answer

share|cite|improve this answer

edited Jul 29 at 11:50

Ninja

1,074520

edited Jul 29 at 11:50

Ninja

1,074520

edited Jul 29 at 11:50

Ninja

1,074520

1,074520

answered Jul 28 at 22:27

Stefan4024

28k52974

answered Jul 28 at 22:27

Stefan4024

28k52974

answered Jul 28 at 22:27

Stefan4024

28k52974

28k52974

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you explain the reason please?
  – Ninja
  Jul 28 at 22:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ninja I added some further explanation. I hope it answers your question
  – Stefan4024
  Jul 28 at 22:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  One last question: In the notes www1.spms.ntu.edu.sg/~frederique/antchap1.pdf, how does the author write the matrix? It multiplies from left and different from the one that we say.
  – Ninja
  Jul 29 at 10:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ninja I'm not sure how he computes it, but something seems fishy. I mean the author has $(1,sqrt2)$ on the left, which can't be written like this in a basis.
  – Stefan4024
  Jul 29 at 11:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For the general case (not only quadratic, see math.stackexchange.com/a/2710883/300700
  – nguyen quang do
  Jul 30 at 12:33
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can you explain the reason please?
  – Ninja
  Jul 28 at 22:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ninja I added some further explanation. I hope it answers your question
  – Stefan4024
  Jul 28 at 22:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  One last question: In the notes www1.spms.ntu.edu.sg/~frederique/antchap1.pdf, how does the author write the matrix? It multiplies from left and different from the one that we say.
  – Ninja
  Jul 29 at 10:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Ninja I'm not sure how he computes it, but something seems fishy. I mean the author has $(1,sqrt2)$ on the left, which can't be written like this in a basis.
  – Stefan4024
  Jul 29 at 11:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  For the general case (not only quadratic, see math.stackexchange.com/a/2710883/300700
  – nguyen quang do
  Jul 30 at 12:33
  

  
  
  
  

Can you explain the reason please?
– Ninja
Jul 28 at 22:28

Can you explain the reason please?
– Ninja
Jul 28 at 22:28

@Ninja I added some further explanation. I hope it answers your question
– Stefan4024
Jul 28 at 22:35

@Ninja I added some further explanation. I hope it answers your question
– Stefan4024
Jul 28 at 22:35

One last question: In the notes www1.spms.ntu.edu.sg/~frederique/antchap1.pdf, how does the author write the matrix? It multiplies from left and different from the one that we say.
– Ninja
Jul 29 at 10:53

One last question: In the notes www1.spms.ntu.edu.sg/~frederique/antchap1.pdf, how does the author write the matrix? It multiplies from left and different from the one that we say.
– Ninja
Jul 29 at 10:53

@Ninja I'm not sure how he computes it, but something seems fishy. I mean the author has $(1,sqrt2)$ on the left, which can't be written like this in a basis.
– Stefan4024
Jul 29 at 11:39

@Ninja I'm not sure how he computes it, but something seems fishy. I mean the author has $(1,sqrt2)$ on the left, which can't be written like this in a basis.
– Stefan4024
Jul 29 at 11:39

For the general case (not only quadratic, see math.stackexchange.com/a/2710883/300700
– nguyen quang do
Jul 30 at 12:33

For the general case (not only quadratic, see math.stackexchange.com/a/2710883/300700
– nguyen quang do
Jul 30 at 12:33

add a comment | 

 

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