Prove or disprove that $[L:K] = deg(p)$ [duplicate]
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How is the degree of the minimal polynomial related to the degree of a field extension?
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Let $L/K$ be a finite extension and let $beta in L$. If $p$ is the minimal polynomial of $beta$ then is it true that $[L:K] = deg(p)$? If not, give a counterexample.
Can someone help me figure out if this is true or not?
abstract-algebra extension-field minimal-polynomials
marked as duplicate by uniquesolution, Arthur, Lord Shark the Unknown
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Aug 24 at 11:34
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
How is the degree of the minimal polynomial related to the degree of a field extension?
1 answer
Let $L/K$ be a finite extension and let $beta in L$. If $p$ is the minimal polynomial of $beta$ then is it true that $[L:K] = deg(p)$? If not, give a counterexample.
Can someone help me figure out if this is true or not?
abstract-algebra extension-field minimal-polynomials
marked as duplicate by uniquesolution, Arthur, Lord Shark the Unknown
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Aug 24 at 11:34
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
What if $betain K$?
â Bernard
Aug 24 at 10:12
My problem says only this @Bernard
â mandella
Aug 24 at 10:14
1
You write "the minimal polynomial" but I think you have to specify it as "the minimal polynomial of $beta$ over $K$."
â Gerry Myerson
Aug 24 at 10:14
@Arthur Have you looked at the other one? It is identical.
â uniquesolution
Aug 24 at 10:15
@uniquesolution I only looked at the title first, and misread it to boot. Then I actually looked at the question, and that's why I've deleted my comment.
â Arthur
Aug 24 at 10:16
 |Â
show 4 more comments
up vote
1
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up vote
1
down vote
favorite
This question already has an answer here:
How is the degree of the minimal polynomial related to the degree of a field extension?
1 answer
Let $L/K$ be a finite extension and let $beta in L$. If $p$ is the minimal polynomial of $beta$ then is it true that $[L:K] = deg(p)$? If not, give a counterexample.
Can someone help me figure out if this is true or not?
abstract-algebra extension-field minimal-polynomials
This question already has an answer here:
How is the degree of the minimal polynomial related to the degree of a field extension?
1 answer
Let $L/K$ be a finite extension and let $beta in L$. If $p$ is the minimal polynomial of $beta$ then is it true that $[L:K] = deg(p)$? If not, give a counterexample.
Can someone help me figure out if this is true or not?
This question already has an answer here:
How is the degree of the minimal polynomial related to the degree of a field extension?
1 answer
abstract-algebra extension-field minimal-polynomials
asked Aug 24 at 10:10
mandella
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713420
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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by uniquesolution, Arthur, Lord Shark the Unknown
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Aug 24 at 11:34
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
3
What if $betain K$?
â Bernard
Aug 24 at 10:12
My problem says only this @Bernard
â mandella
Aug 24 at 10:14
1
You write "the minimal polynomial" but I think you have to specify it as "the minimal polynomial of $beta$ over $K$."
â Gerry Myerson
Aug 24 at 10:14
@Arthur Have you looked at the other one? It is identical.
â uniquesolution
Aug 24 at 10:15
@uniquesolution I only looked at the title first, and misread it to boot. Then I actually looked at the question, and that's why I've deleted my comment.
â Arthur
Aug 24 at 10:16
 |Â
show 4 more comments
3
What if $betain K$?
â Bernard
Aug 24 at 10:12
My problem says only this @Bernard
â mandella
Aug 24 at 10:14
1
You write "the minimal polynomial" but I think you have to specify it as "the minimal polynomial of $beta$ over $K$."
â Gerry Myerson
Aug 24 at 10:14
@Arthur Have you looked at the other one? It is identical.
â uniquesolution
Aug 24 at 10:15
@uniquesolution I only looked at the title first, and misread it to boot. Then I actually looked at the question, and that's why I've deleted my comment.
â Arthur
Aug 24 at 10:16
3
3
What if $betain K$?
â Bernard
Aug 24 at 10:12
What if $betain K$?
â Bernard
Aug 24 at 10:12
My problem says only this @Bernard
â mandella
Aug 24 at 10:14
My problem says only this @Bernard
â mandella
Aug 24 at 10:14
1
1
You write "the minimal polynomial" but I think you have to specify it as "the minimal polynomial of $beta$ over $K$."
â Gerry Myerson
Aug 24 at 10:14
You write "the minimal polynomial" but I think you have to specify it as "the minimal polynomial of $beta$ over $K$."
â Gerry Myerson
Aug 24 at 10:14
@Arthur Have you looked at the other one? It is identical.
â uniquesolution
Aug 24 at 10:15
@Arthur Have you looked at the other one? It is identical.
â uniquesolution
Aug 24 at 10:15
@uniquesolution I only looked at the title first, and misread it to boot. Then I actually looked at the question, and that's why I've deleted my comment.
â Arthur
Aug 24 at 10:16
@uniquesolution I only looked at the title first, and misread it to boot. Then I actually looked at the question, and that's why I've deleted my comment.
â Arthur
Aug 24 at 10:16
 |Â
show 4 more comments
1 Answer
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3
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Let $K=mathbbQ(i,sqrt2)$ and consider the extension $K/mathbbQ$. Consider the minimal polynomial of $i$ over $mathbbQ$, which is $p(x)=x^2+1$, i.e. $deg(p)=2$. But $$[K:mathbbQ]=4neq 2=deg(p).$$
Isn 't $[K:mathbbQ]=3$ and $1,i,sqrt2$ a basis for the $mathbbQ$-vector space $K$?
â Peter Melech
Aug 24 at 10:33
If it is $3$ then "not equal still holds? @PeterMelech
â mandella
Aug 24 at 10:37
1
Yes sure the argument is fine!
â Peter Melech
Aug 24 at 10:37
2
@PeterMelech $[K:mathbbQ]$ cannot be equal to 3 since $mathbbQ(sqrt2)$ is a subfield of $K$, which has degree 2 (so 2 must divide the degree of $K$ over $mathbbQ$). So a basis is $lbrace 1, i, sqrt2, isqrt2rbrace$.
â YumekuiMath
Aug 24 at 10:37
1
@YumekuiMath Ah sure! Thanks a lot!
â Peter Melech
Aug 24 at 10:38
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Let $K=mathbbQ(i,sqrt2)$ and consider the extension $K/mathbbQ$. Consider the minimal polynomial of $i$ over $mathbbQ$, which is $p(x)=x^2+1$, i.e. $deg(p)=2$. But $$[K:mathbbQ]=4neq 2=deg(p).$$
Isn 't $[K:mathbbQ]=3$ and $1,i,sqrt2$ a basis for the $mathbbQ$-vector space $K$?
â Peter Melech
Aug 24 at 10:33
If it is $3$ then "not equal still holds? @PeterMelech
â mandella
Aug 24 at 10:37
1
Yes sure the argument is fine!
â Peter Melech
Aug 24 at 10:37
2
@PeterMelech $[K:mathbbQ]$ cannot be equal to 3 since $mathbbQ(sqrt2)$ is a subfield of $K$, which has degree 2 (so 2 must divide the degree of $K$ over $mathbbQ$). So a basis is $lbrace 1, i, sqrt2, isqrt2rbrace$.
â YumekuiMath
Aug 24 at 10:37
1
@YumekuiMath Ah sure! Thanks a lot!
â Peter Melech
Aug 24 at 10:38
add a comment |Â
up vote
3
down vote
accepted
Let $K=mathbbQ(i,sqrt2)$ and consider the extension $K/mathbbQ$. Consider the minimal polynomial of $i$ over $mathbbQ$, which is $p(x)=x^2+1$, i.e. $deg(p)=2$. But $$[K:mathbbQ]=4neq 2=deg(p).$$
Isn 't $[K:mathbbQ]=3$ and $1,i,sqrt2$ a basis for the $mathbbQ$-vector space $K$?
â Peter Melech
Aug 24 at 10:33
If it is $3$ then "not equal still holds? @PeterMelech
â mandella
Aug 24 at 10:37
1
Yes sure the argument is fine!
â Peter Melech
Aug 24 at 10:37
2
@PeterMelech $[K:mathbbQ]$ cannot be equal to 3 since $mathbbQ(sqrt2)$ is a subfield of $K$, which has degree 2 (so 2 must divide the degree of $K$ over $mathbbQ$). So a basis is $lbrace 1, i, sqrt2, isqrt2rbrace$.
â YumekuiMath
Aug 24 at 10:37
1
@YumekuiMath Ah sure! Thanks a lot!
â Peter Melech
Aug 24 at 10:38
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Let $K=mathbbQ(i,sqrt2)$ and consider the extension $K/mathbbQ$. Consider the minimal polynomial of $i$ over $mathbbQ$, which is $p(x)=x^2+1$, i.e. $deg(p)=2$. But $$[K:mathbbQ]=4neq 2=deg(p).$$
Let $K=mathbbQ(i,sqrt2)$ and consider the extension $K/mathbbQ$. Consider the minimal polynomial of $i$ over $mathbbQ$, which is $p(x)=x^2+1$, i.e. $deg(p)=2$. But $$[K:mathbbQ]=4neq 2=deg(p).$$
answered Aug 24 at 10:14
YumekuiMath
911
911
Isn 't $[K:mathbbQ]=3$ and $1,i,sqrt2$ a basis for the $mathbbQ$-vector space $K$?
â Peter Melech
Aug 24 at 10:33
If it is $3$ then "not equal still holds? @PeterMelech
â mandella
Aug 24 at 10:37
1
Yes sure the argument is fine!
â Peter Melech
Aug 24 at 10:37
2
@PeterMelech $[K:mathbbQ]$ cannot be equal to 3 since $mathbbQ(sqrt2)$ is a subfield of $K$, which has degree 2 (so 2 must divide the degree of $K$ over $mathbbQ$). So a basis is $lbrace 1, i, sqrt2, isqrt2rbrace$.
â YumekuiMath
Aug 24 at 10:37
1
@YumekuiMath Ah sure! Thanks a lot!
â Peter Melech
Aug 24 at 10:38
add a comment |Â
Isn 't $[K:mathbbQ]=3$ and $1,i,sqrt2$ a basis for the $mathbbQ$-vector space $K$?
â Peter Melech
Aug 24 at 10:33
If it is $3$ then "not equal still holds? @PeterMelech
â mandella
Aug 24 at 10:37
1
Yes sure the argument is fine!
â Peter Melech
Aug 24 at 10:37
2
@PeterMelech $[K:mathbbQ]$ cannot be equal to 3 since $mathbbQ(sqrt2)$ is a subfield of $K$, which has degree 2 (so 2 must divide the degree of $K$ over $mathbbQ$). So a basis is $lbrace 1, i, sqrt2, isqrt2rbrace$.
â YumekuiMath
Aug 24 at 10:37
1
@YumekuiMath Ah sure! Thanks a lot!
â Peter Melech
Aug 24 at 10:38
Isn 't $[K:mathbbQ]=3$ and $1,i,sqrt2$ a basis for the $mathbbQ$-vector space $K$?
â Peter Melech
Aug 24 at 10:33
Isn 't $[K:mathbbQ]=3$ and $1,i,sqrt2$ a basis for the $mathbbQ$-vector space $K$?
â Peter Melech
Aug 24 at 10:33
If it is $3$ then "not equal still holds? @PeterMelech
â mandella
Aug 24 at 10:37
If it is $3$ then "not equal still holds? @PeterMelech
â mandella
Aug 24 at 10:37
1
1
Yes sure the argument is fine!
â Peter Melech
Aug 24 at 10:37
Yes sure the argument is fine!
â Peter Melech
Aug 24 at 10:37
2
2
@PeterMelech $[K:mathbbQ]$ cannot be equal to 3 since $mathbbQ(sqrt2)$ is a subfield of $K$, which has degree 2 (so 2 must divide the degree of $K$ over $mathbbQ$). So a basis is $lbrace 1, i, sqrt2, isqrt2rbrace$.
â YumekuiMath
Aug 24 at 10:37
@PeterMelech $[K:mathbbQ]$ cannot be equal to 3 since $mathbbQ(sqrt2)$ is a subfield of $K$, which has degree 2 (so 2 must divide the degree of $K$ over $mathbbQ$). So a basis is $lbrace 1, i, sqrt2, isqrt2rbrace$.
â YumekuiMath
Aug 24 at 10:37
1
1
@YumekuiMath Ah sure! Thanks a lot!
â Peter Melech
Aug 24 at 10:38
@YumekuiMath Ah sure! Thanks a lot!
â Peter Melech
Aug 24 at 10:38
add a comment |Â
3
What if $betain K$?
â Bernard
Aug 24 at 10:12
My problem says only this @Bernard
â mandella
Aug 24 at 10:14
1
You write "the minimal polynomial" but I think you have to specify it as "the minimal polynomial of $beta$ over $K$."
â Gerry Myerson
Aug 24 at 10:14
@Arthur Have you looked at the other one? It is identical.
â uniquesolution
Aug 24 at 10:15
@uniquesolution I only looked at the title first, and misread it to boot. Then I actually looked at the question, and that's why I've deleted my comment.
â Arthur
Aug 24 at 10:16