What is $mathbbQ(sqrt[4]3i)$? [closed]

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Is it $mathbbQ(sqrt[4]3i) = a+bsqrt[4]3i+csqrt[4]9i + dsqrt[4]27 : a,b,c,d in mathbbQ $? Is it isomorphic to $mathbbQ(sqrt[4]3)$?

field-theory

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edited Jul 19 at 0:20

asked Jul 19 at 0:08

math4everyone

38918

closed as off-topic by amWhy, user223391, Parcly Taxel, Isaac Browne, Mostafa Ayaz Jul 19 at 7:36

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Community, Parcly Taxel, Isaac Browne, Mostafa Ayaz

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Unless I am mistaken, $sqrt[3]2i$ is not killed by a polynomial of degree $3$ over $mathbbQ$. The degree $3$ term, after valuation, is a rational multiple of $i$ whereas the degree $1$ term necessarily is an irrational multiple of $i$, so that the imaginary parts can not possibly add up to zero.
  – Suzet
  Jul 19 at 0:13
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Note if $alpha = sqrt[3] 2,i$ then $alpha^2= -sqrt[3] 4$ and $alpha^3=-2i$, thus both $i$ and $sqrt[3] 2$ are in your field, which therefore has degree at least $6$. In particular, $mathbb Q(sqrt[3] 2)$ is a proper subfield of your field.
  – lulu
  Jul 19 at 0:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Suzet The original question asked me to show that $mathbbQ(sqrt[4]3)$ and$mathbbQ(sqrt[4]3i)$ are isomorphic, but first I wanted to understand simpler examples.
  – math4everyone
  Jul 19 at 0:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oddly, that example is significantly easier than the one you posed, as both values are roots of the irreducible polynomial $x^4-3$.
  – lulu
  Jul 19 at 0:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @lulu Then I should modify the question
  – math4everyone
  Jul 19 at 0:19
  

  
  
  
  

 | 
show 3 more comments

up vote
0
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Is it $mathbbQ(sqrt[4]3i) = a+bsqrt[4]3i+csqrt[4]9i + dsqrt[4]27 : a,b,c,d in mathbbQ $? Is it isomorphic to $mathbbQ(sqrt[4]3)$?

field-theory

share|cite|improve this question

edited Jul 19 at 0:20

asked Jul 19 at 0:08

math4everyone

38918

closed as off-topic by amWhy, user223391, Parcly Taxel, Isaac Browne, Mostafa Ayaz Jul 19 at 7:36

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Community, Parcly Taxel, Isaac Browne, Mostafa Ayaz

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Unless I am mistaken, $sqrt[3]2i$ is not killed by a polynomial of degree $3$ over $mathbbQ$. The degree $3$ term, after valuation, is a rational multiple of $i$ whereas the degree $1$ term necessarily is an irrational multiple of $i$, so that the imaginary parts can not possibly add up to zero.
  – Suzet
  Jul 19 at 0:13
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Note if $alpha = sqrt[3] 2,i$ then $alpha^2= -sqrt[3] 4$ and $alpha^3=-2i$, thus both $i$ and $sqrt[3] 2$ are in your field, which therefore has degree at least $6$. In particular, $mathbb Q(sqrt[3] 2)$ is a proper subfield of your field.
  – lulu
  Jul 19 at 0:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Suzet The original question asked me to show that $mathbbQ(sqrt[4]3)$ and$mathbbQ(sqrt[4]3i)$ are isomorphic, but first I wanted to understand simpler examples.
  – math4everyone
  Jul 19 at 0:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oddly, that example is significantly easier than the one you posed, as both values are roots of the irreducible polynomial $x^4-3$.
  – lulu
  Jul 19 at 0:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @lulu Then I should modify the question
  – math4everyone
  Jul 19 at 0:19
  

  
  
  
  

 | 
show 3 more comments

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Is it $mathbbQ(sqrt[4]3i) = a+bsqrt[4]3i+csqrt[4]9i + dsqrt[4]27 : a,b,c,d in mathbbQ $? Is it isomorphic to $mathbbQ(sqrt[4]3)$?

field-theory

share|cite|improve this question

edited Jul 19 at 0:20

asked Jul 19 at 0:08

math4everyone

38918

Is it $mathbbQ(sqrt[4]3i) = a+bsqrt[4]3i+csqrt[4]9i + dsqrt[4]27 : a,b,c,d in mathbbQ $? Is it isomorphic to $mathbbQ(sqrt[4]3)$?

field-theory

share|cite|improve this question

edited Jul 19 at 0:20

asked Jul 19 at 0:08

math4everyone

38918

share|cite|improve this question

share|cite|improve this question

edited Jul 19 at 0:20

edited Jul 19 at 0:20

edited Jul 19 at 0:20

asked Jul 19 at 0:08

math4everyone

38918

asked Jul 19 at 0:08

math4everyone

38918

asked Jul 19 at 0:08

math4everyone

38918

38918

closed as off-topic by amWhy, user223391, Parcly Taxel, Isaac Browne, Mostafa Ayaz Jul 19 at 7:36

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Community, Parcly Taxel, Isaac Browne, Mostafa Ayaz

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by amWhy, user223391, Parcly Taxel, Isaac Browne, Mostafa Ayaz Jul 19 at 7:36

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Community, Parcly Taxel, Isaac Browne, Mostafa Ayaz

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Unless I am mistaken, $sqrt[3]2i$ is not killed by a polynomial of degree $3$ over $mathbbQ$. The degree $3$ term, after valuation, is a rational multiple of $i$ whereas the degree $1$ term necessarily is an irrational multiple of $i$, so that the imaginary parts can not possibly add up to zero.
  – Suzet
  Jul 19 at 0:13
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Note if $alpha = sqrt[3] 2,i$ then $alpha^2= -sqrt[3] 4$ and $alpha^3=-2i$, thus both $i$ and $sqrt[3] 2$ are in your field, which therefore has degree at least $6$. In particular, $mathbb Q(sqrt[3] 2)$ is a proper subfield of your field.
  – lulu
  Jul 19 at 0:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Suzet The original question asked me to show that $mathbbQ(sqrt[4]3)$ and$mathbbQ(sqrt[4]3i)$ are isomorphic, but first I wanted to understand simpler examples.
  – math4everyone
  Jul 19 at 0:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oddly, that example is significantly easier than the one you posed, as both values are roots of the irreducible polynomial $x^4-3$.
  – lulu
  Jul 19 at 0:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @lulu Then I should modify the question
  – math4everyone
  Jul 19 at 0:19
  

  
  
  
  

 | 
show 3 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Unless I am mistaken, $sqrt[3]2i$ is not killed by a polynomial of degree $3$ over $mathbbQ$. The degree $3$ term, after valuation, is a rational multiple of $i$ whereas the degree $1$ term necessarily is an irrational multiple of $i$, so that the imaginary parts can not possibly add up to zero.
  – Suzet
  Jul 19 at 0:13
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Note if $alpha = sqrt[3] 2,i$ then $alpha^2= -sqrt[3] 4$ and $alpha^3=-2i$, thus both $i$ and $sqrt[3] 2$ are in your field, which therefore has degree at least $6$. In particular, $mathbb Q(sqrt[3] 2)$ is a proper subfield of your field.
  – lulu
  Jul 19 at 0:14
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Suzet The original question asked me to show that $mathbbQ(sqrt[4]3)$ and$mathbbQ(sqrt[4]3i)$ are isomorphic, but first I wanted to understand simpler examples.
  – math4everyone
  Jul 19 at 0:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oddly, that example is significantly easier than the one you posed, as both values are roots of the irreducible polynomial $x^4-3$.
  – lulu
  Jul 19 at 0:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @lulu Then I should modify the question
  – math4everyone
  Jul 19 at 0:19
  

  
  
  
  

Unless I am mistaken, $sqrt[3]2i$ is not killed by a polynomial of degree $3$ over $mathbbQ$. The degree $3$ term, after valuation, is a rational multiple of $i$ whereas the degree $1$ term necessarily is an irrational multiple of $i$, so that the imaginary parts can not possibly add up to zero.
– Suzet
Jul 19 at 0:13

Unless I am mistaken, $sqrt[3]2i$ is not killed by a polynomial of degree $3$ over $mathbbQ$. The degree $3$ term, after valuation, is a rational multiple of $i$ whereas the degree $1$ term necessarily is an irrational multiple of $i$, so that the imaginary parts can not possibly add up to zero.
– Suzet
Jul 19 at 0:13

1

1

Note if $alpha = sqrt[3] 2,i$ then $alpha^2= -sqrt[3] 4$ and $alpha^3=-2i$, thus both $i$ and $sqrt[3] 2$ are in your field, which therefore has degree at least $6$. In particular, $mathbb Q(sqrt[3] 2)$ is a proper subfield of your field.
– lulu
Jul 19 at 0:14

Note if $alpha = sqrt[3] 2,i$ then $alpha^2= -sqrt[3] 4$ and $alpha^3=-2i$, thus both $i$ and $sqrt[3] 2$ are in your field, which therefore has degree at least $6$. In particular, $mathbb Q(sqrt[3] 2)$ is a proper subfield of your field.
– lulu
Jul 19 at 0:14

@Suzet The original question asked me to show that $mathbbQ(sqrt[4]3)$ and$mathbbQ(sqrt[4]3i)$ are isomorphic, but first I wanted to understand simpler examples.
– math4everyone
Jul 19 at 0:16

@Suzet The original question asked me to show that $mathbbQ(sqrt[4]3)$ and$mathbbQ(sqrt[4]3i)$ are isomorphic, but first I wanted to understand simpler examples.
– math4everyone
Jul 19 at 0:16

Oddly, that example is significantly easier than the one you posed, as both values are roots of the irreducible polynomial $x^4-3$.
– lulu
Jul 19 at 0:17

Oddly, that example is significantly easier than the one you posed, as both values are roots of the irreducible polynomial $x^4-3$.
– lulu
Jul 19 at 0:17

@lulu Then I should modify the question
– math4everyone
Jul 19 at 0:19

@lulu Then I should modify the question
– math4everyone
Jul 19 at 0:19

 | 
show 3 more comments

1 Answer
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0
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accepted

Abstractly, $mathbbQ(sqrt[4]3i)$ is obtained by adjoining a root of $x^4-3$ to $mathbbQ$.

The same description applies to $mathbbQ(sqrt[4]3)$.

Hence, $mathbbQ(sqrt[4]3i)$ is isomorphic to $mathbbQ(sqrt[4]3)$.

A concrete isomorphism sends $sqrt[4]3i$ to $sqrt[4]3$.

If $alpha$ is algebraic of degree $n$, then $mathbbQ(alpha)$ is the set of rational linear combinations of $alpha^k$ for $k=0,dots,n$.

Therefore,
$
mathbbQ(sqrt[4]3i)=
a+bsqrt[4]3i+csqrt[4]9 + dsqrt[4]27i : a,b,c,d in mathbbQ
$.

share|cite|improve this answer

edited Jul 19 at 2:36

answered Jul 19 at 1:13

lhf

156k9160367

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
0
down vote



accepted

Abstractly, $mathbbQ(sqrt[4]3i)$ is obtained by adjoining a root of $x^4-3$ to $mathbbQ$.

The same description applies to $mathbbQ(sqrt[4]3)$.

Hence, $mathbbQ(sqrt[4]3i)$ is isomorphic to $mathbbQ(sqrt[4]3)$.

A concrete isomorphism sends $sqrt[4]3i$ to $sqrt[4]3$.

If $alpha$ is algebraic of degree $n$, then $mathbbQ(alpha)$ is the set of rational linear combinations of $alpha^k$ for $k=0,dots,n$.

Therefore,
$
mathbbQ(sqrt[4]3i)=
a+bsqrt[4]3i+csqrt[4]9 + dsqrt[4]27i : a,b,c,d in mathbbQ
$.

share|cite|improve this answer

edited Jul 19 at 2:36

answered Jul 19 at 1:13

lhf

156k9160367

add a comment | 

up vote
0
down vote



accepted

Abstractly, $mathbbQ(sqrt[4]3i)$ is obtained by adjoining a root of $x^4-3$ to $mathbbQ$.

The same description applies to $mathbbQ(sqrt[4]3)$.

Hence, $mathbbQ(sqrt[4]3i)$ is isomorphic to $mathbbQ(sqrt[4]3)$.

A concrete isomorphism sends $sqrt[4]3i$ to $sqrt[4]3$.

If $alpha$ is algebraic of degree $n$, then $mathbbQ(alpha)$ is the set of rational linear combinations of $alpha^k$ for $k=0,dots,n$.

Therefore,
$
mathbbQ(sqrt[4]3i)=
a+bsqrt[4]3i+csqrt[4]9 + dsqrt[4]27i : a,b,c,d in mathbbQ
$.

share|cite|improve this answer

edited Jul 19 at 2:36

answered Jul 19 at 1:13

lhf

156k9160367

add a comment | 

up vote
0
down vote



accepted

up vote
0
down vote



accepted

Abstractly, $mathbbQ(sqrt[4]3i)$ is obtained by adjoining a root of $x^4-3$ to $mathbbQ$.

The same description applies to $mathbbQ(sqrt[4]3)$.

Hence, $mathbbQ(sqrt[4]3i)$ is isomorphic to $mathbbQ(sqrt[4]3)$.

A concrete isomorphism sends $sqrt[4]3i$ to $sqrt[4]3$.

If $alpha$ is algebraic of degree $n$, then $mathbbQ(alpha)$ is the set of rational linear combinations of $alpha^k$ for $k=0,dots,n$.

Therefore,
$
mathbbQ(sqrt[4]3i)=
a+bsqrt[4]3i+csqrt[4]9 + dsqrt[4]27i : a,b,c,d in mathbbQ
$.

share|cite|improve this answer

edited Jul 19 at 2:36

answered Jul 19 at 1:13

lhf

156k9160367

Abstractly, $mathbbQ(sqrt[4]3i)$ is obtained by adjoining a root of $x^4-3$ to $mathbbQ$.

The same description applies to $mathbbQ(sqrt[4]3)$.

Hence, $mathbbQ(sqrt[4]3i)$ is isomorphic to $mathbbQ(sqrt[4]3)$.

A concrete isomorphism sends $sqrt[4]3i$ to $sqrt[4]3$.

If $alpha$ is algebraic of degree $n$, then $mathbbQ(alpha)$ is the set of rational linear combinations of $alpha^k$ for $k=0,dots,n$.

Therefore,
$
mathbbQ(sqrt[4]3i)=
a+bsqrt[4]3i+csqrt[4]9 + dsqrt[4]27i : a,b,c,d in mathbbQ
$.

share|cite|improve this answer

edited Jul 19 at 2:36

answered Jul 19 at 1:13

lhf

156k9160367

share|cite|improve this answer

share|cite|improve this answer

edited Jul 19 at 2:36

edited Jul 19 at 2:36

edited Jul 19 at 2:36

answered Jul 19 at 1:13

lhf

156k9160367

answered Jul 19 at 1:13

lhf

156k9160367

answered Jul 19 at 1:13

lhf

156k9160367

156k9160367

add a comment | 

add a comment |