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?







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marked as duplicate by uniquesolution, Arthur, Lord Shark the Unknown abstract-algebra
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  • 3




    What if $betain K$?
    – Bernard
    Aug 24 at 10:12










  • My problem says only this @Bernard
    – 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 $K$."
    – Gerry Myerson
    Aug 24 at 10:14










  • @Arthur Have you looked at the other one? It is identical.
    – 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.
    – Arthur
    Aug 24 at 10:16















up vote
1
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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?







share|cite|improve this question












marked as duplicate by uniquesolution, Arthur, Lord Shark the Unknown abstract-algebra
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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 $betain K$?
    – Bernard
    Aug 24 at 10:12










  • My problem says only this @Bernard
    – 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 $K$."
    – Gerry Myerson
    Aug 24 at 10:14










  • @Arthur Have you looked at the other one? It is identical.
    – 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.
    – Arthur
    Aug 24 at 10:16













up vote
1
down vote

favorite









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?







share|cite|improve this question













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









share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 24 at 10:10









mandella

713420




713420




marked as duplicate by uniquesolution, Arthur, Lord Shark the Unknown abstract-algebra
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marked as duplicate by uniquesolution, Arthur, Lord Shark the Unknown abstract-algebra
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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 $betain K$?
    – Bernard
    Aug 24 at 10:12










  • My problem says only this @Bernard
    – 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 $K$."
    – Gerry Myerson
    Aug 24 at 10:14










  • @Arthur Have you looked at the other one? It is identical.
    – 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.
    – Arthur
    Aug 24 at 10:16













  • 3




    What if $betain K$?
    – Bernard
    Aug 24 at 10:12










  • My problem says only this @Bernard
    – 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 $K$."
    – Gerry Myerson
    Aug 24 at 10:14










  • @Arthur Have you looked at the other one? It is identical.
    – 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.
    – Arthur
    Aug 24 at 10:16








3




3




What if $betain K$?
– Bernard
Aug 24 at 10:12




What if $betain K$?
– Bernard
Aug 24 at 10:12












My problem says only this @Bernard
– mandella
Aug 24 at 10:14





My problem says only this @Bernard
– 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 $K$."
– 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 $K$."
– Gerry Myerson
Aug 24 at 10:14












@Arthur Have you looked at the other one? It is identical.
– uniquesolution
Aug 24 at 10:15




@Arthur Have you looked at the other one? It is identical.
– 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.
– 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.
– Arthur
Aug 24 at 10:16











1 Answer
1






active

oldest

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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 $$[K:mathbbQ]=4neq 2=deg(p).$$






share|cite|improve this answer




















  • Isn 't $[K:mathbbQ]=3$ and $1,i,sqrt2$ a basis for the $mathbbQ$-vector space $K$?
    – Peter Melech
    Aug 24 at 10:33











  • If it is $3$ then "not equal still holds? @PeterMelech
    – mandella
    Aug 24 at 10:37






  • 1




    Yes sure the argument is fine!
    – 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 $lbrace 1, i, sqrt2, isqrt2rbrace$.
    – YumekuiMath
    Aug 24 at 10:37







  • 1




    @YumekuiMath Ah sure! Thanks a lot!
    – Peter Melech
    Aug 24 at 10:38

















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 $$[K:mathbbQ]=4neq 2=deg(p).$$






share|cite|improve this answer




















  • Isn 't $[K:mathbbQ]=3$ and $1,i,sqrt2$ a basis for the $mathbbQ$-vector space $K$?
    – Peter Melech
    Aug 24 at 10:33











  • If it is $3$ then "not equal still holds? @PeterMelech
    – mandella
    Aug 24 at 10:37






  • 1




    Yes sure the argument is fine!
    – 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 $lbrace 1, i, sqrt2, isqrt2rbrace$.
    – YumekuiMath
    Aug 24 at 10:37







  • 1




    @YumekuiMath Ah sure! Thanks a lot!
    – Peter Melech
    Aug 24 at 10:38














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 $$[K:mathbbQ]=4neq 2=deg(p).$$






share|cite|improve this answer




















  • Isn 't $[K:mathbbQ]=3$ and $1,i,sqrt2$ a basis for the $mathbbQ$-vector space $K$?
    – Peter Melech
    Aug 24 at 10:33











  • If it is $3$ then "not equal still holds? @PeterMelech
    – mandella
    Aug 24 at 10:37






  • 1




    Yes sure the argument is fine!
    – 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 $lbrace 1, i, sqrt2, isqrt2rbrace$.
    – YumekuiMath
    Aug 24 at 10:37







  • 1




    @YumekuiMath Ah sure! Thanks a lot!
    – Peter Melech
    Aug 24 at 10:38












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 $$[K:mathbbQ]=4neq 2=deg(p).$$






share|cite|improve this answer












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 $$[K:mathbbQ]=4neq 2=deg(p).$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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 $K$?
    – Peter Melech
    Aug 24 at 10:33











  • If it is $3$ then "not equal still holds? @PeterMelech
    – mandella
    Aug 24 at 10:37






  • 1




    Yes sure the argument is fine!
    – 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 $lbrace 1, i, sqrt2, isqrt2rbrace$.
    – YumekuiMath
    Aug 24 at 10:37







  • 1




    @YumekuiMath Ah sure! Thanks a lot!
    – Peter Melech
    Aug 24 at 10:38
















  • Isn 't $[K:mathbbQ]=3$ and $1,i,sqrt2$ a basis for the $mathbbQ$-vector space $K$?
    – Peter Melech
    Aug 24 at 10:33











  • If it is $3$ then "not equal still holds? @PeterMelech
    – mandella
    Aug 24 at 10:37






  • 1




    Yes sure the argument is fine!
    – 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 $lbrace 1, i, sqrt2, isqrt2rbrace$.
    – YumekuiMath
    Aug 24 at 10:37







  • 1




    @YumekuiMath Ah sure! Thanks a lot!
    – Peter Melech
    Aug 24 at 10:38















Isn 't $[K:mathbbQ]=3$ and $1,i,sqrt2$ a basis for the $mathbbQ$-vector space $K$?
– Peter Melech
Aug 24 at 10:33





Isn 't $[K:mathbbQ]=3$ and $1,i,sqrt2$ a basis for the $mathbbQ$-vector space $K$?
– Peter Melech
Aug 24 at 10:33













If it is $3$ then "not equal still holds? @PeterMelech
– mandella
Aug 24 at 10:37




If it is $3$ then "not equal still holds? @PeterMelech
– mandella
Aug 24 at 10:37




1




1




Yes sure the argument is fine!
– Peter Melech
Aug 24 at 10:37




Yes sure the argument is fine!
– 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 $lbrace 1, i, sqrt2, isqrt2rbrace$.
– 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 $lbrace 1, i, sqrt2, isqrt2rbrace$.
– YumekuiMath
Aug 24 at 10:37





1




1




@YumekuiMath Ah sure! Thanks a lot!
– Peter Melech
Aug 24 at 10:38




@YumekuiMath Ah sure! Thanks a lot!
– Peter Melech
Aug 24 at 10:38


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