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Minimal polynomial for $sqrt[5]2$ over $mathbb Q(sqrt[]3)$
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Find the minimal polynomial for $sqrt[5]2$ over $mathbb
Q(sqrt3)$.
The natural candidate is $x^5-2$. But the problem is to prove it is irreducible. In theory, I can write down the roots explicitly and by squaring multiple times somehow show that no root lies in the field. But even then there may be a quadratic factor, and using the method of undetermined coefficients is way too long. Can I prove irreducibility easier?
abstract-algebra polynomials galois-theory extension-field irreducible-polynomials
asked Jul 15 at 20:44
user437309
556212
add a comment |Â
up vote
0
down vote
favorite
Find the minimal polynomial for $sqrt[5]2$ over $mathbb
Q(sqrt3)$.
The natural candidate is $x^5-2$. But the problem is to prove it is irreducible. In theory, I can write down the roots explicitly and by squaring multiple times somehow show that no root lies in the field. But even then there may be a quadratic factor, and using the method of undetermined coefficients is way too long. Can I prove irreducibility easier?
abstract-algebra polynomials galois-theory extension-field irreducible-polynomials
asked Jul 15 at 20:44
user437309
556212
3
If you already know that $X^5 - 2$ is irreducible in $mathbbQ[X]$, it suffices to note that $2$ and $5$ are coprime.
– Daniel Fischer♦
Jul 15 at 21:04
@DanielFischer Why is it sufficient?
– user437309
Jul 15 at 21:13
1
@user437307 We know that $[Q(sqrt3,sqrt[5]2):Q(sqrt3)]*[Q(sqrt3):Q]=[Q(sqrt3,sqrt[5]2):Q]$ and $[Q(sqrt3,sqrt[5]2):Q(sqrt[5]2)]*[Q(sqrt[5]2):Q]=[Q(sqrt3,sqrt[5]2):Q]$ Now use the fact that $2$ and $5$ are coprime
– Sorfosh
Jul 15 at 21:54
@Sorfosh Do you mean I should use it to conclude that since $5k=2n$, $2n$ is divisible by $5$, so $n$ is divisible by $5$. And since $nle 5$ and it's divisible by $5$, $n=5$? Here $n=[mathbb Q(sqrt[5]2,sqrt3):mathbb Q(sqrt 3)]$. But I'm not sure how this is related to irreducibility.
– user437309
Jul 15 at 22:28
1
@user437307 Yes, the idea is to use it to conclude $n = 5$. And since $n$ is the degree of the minimal polynomial of $sqrt[5]2$ over $mathbbQ(sqrt3)$ it follows that $X^5 - 2$ is that minimal polynomial, hence irreducible.
– Daniel Fischer♦
Jul 16 at 9:45
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Find the minimal polynomial for $sqrt[5]2$ over $mathbb
Q(sqrt3)$.
The natural candidate is $x^5-2$. But the problem is to prove it is irreducible. In theory, I can write down the roots explicitly and by squaring multiple times somehow show that no root lies in the field. But even then there may be a quadratic factor, and using the method of undetermined coefficients is way too long. Can I prove irreducibility easier?
abstract-algebra polynomials galois-theory extension-field irreducible-polynomials
asked Jul 15 at 20:44
user437309
556212
Find the minimal polynomial for $sqrt[5]2$ over $mathbb
Q(sqrt3)$.
The natural candidate is $x^5-2$. But the problem is to prove it is irreducible. In theory, I can write down the roots explicitly and by squaring multiple times somehow show that no root lies in the field. But even then there may be a quadratic factor, and using the method of undetermined coefficients is way too long. Can I prove irreducibility easier?
abstract-algebra polynomials galois-theory extension-field irreducible-polynomials
asked Jul 15 at 20:44
user437309
556212
asked Jul 15 at 20:44
user437309
556212
asked Jul 15 at 20:44
user437309
556212
asked Jul 15 at 20:44
user437309
556212
556212
3
If you already know that $X^5 - 2$ is irreducible in $mathbbQ[X]$, it suffices to note that $2$ and $5$ are coprime.
– Daniel Fischer♦
Jul 15 at 21:04
@DanielFischer Why is it sufficient?
– user437309
Jul 15 at 21:13
1
@user437307 We know that $[Q(sqrt3,sqrt[5]2):Q(sqrt3)]*[Q(sqrt3):Q]=[Q(sqrt3,sqrt[5]2):Q]$ and $[Q(sqrt3,sqrt[5]2):Q(sqrt[5]2)]*[Q(sqrt[5]2):Q]=[Q(sqrt3,sqrt[5]2):Q]$ Now use the fact that $2$ and $5$ are coprime
– Sorfosh
Jul 15 at 21:54
@Sorfosh Do you mean I should use it to conclude that since $5k=2n$, $2n$ is divisible by $5$, so $n$ is divisible by $5$. And since $nle 5$ and it's divisible by $5$, $n=5$? Here $n=[mathbb Q(sqrt[5]2,sqrt3):mathbb Q(sqrt 3)]$. But I'm not sure how this is related to irreducibility.
– user437309
Jul 15 at 22:28
1
@user437307 Yes, the idea is to use it to conclude $n = 5$. And since $n$ is the degree of the minimal polynomial of $sqrt[5]2$ over $mathbbQ(sqrt3)$ it follows that $X^5 - 2$ is that minimal polynomial, hence irreducible.
– Daniel Fischer♦
Jul 16 at 9:45
add a comment |Â
3
If you already know that $X^5 - 2$ is irreducible in $mathbbQ[X]$, it suffices to note that $2$ and $5$ are coprime.
– Daniel Fischer♦
Jul 15 at 21:04
@DanielFischer Why is it sufficient?
– user437309
Jul 15 at 21:13
1
@user437307 We know that $[Q(sqrt3,sqrt[5]2):Q(sqrt3)]*[Q(sqrt3):Q]=[Q(sqrt3,sqrt[5]2):Q]$ and $[Q(sqrt3,sqrt[5]2):Q(sqrt[5]2)]*[Q(sqrt[5]2):Q]=[Q(sqrt3,sqrt[5]2):Q]$ Now use the fact that $2$ and $5$ are coprime
– Sorfosh
Jul 15 at 21:54
@Sorfosh Do you mean I should use it to conclude that since $5k=2n$, $2n$ is divisible by $5$, so $n$ is divisible by $5$. And since $nle 5$ and it's divisible by $5$, $n=5$? Here $n=[mathbb Q(sqrt[5]2,sqrt3):mathbb Q(sqrt 3)]$. But I'm not sure how this is related to irreducibility.
– user437309
Jul 15 at 22:28
1
@user437307 Yes, the idea is to use it to conclude $n = 5$. And since $n$ is the degree of the minimal polynomial of $sqrt[5]2$ over $mathbbQ(sqrt3)$ it follows that $X^5 - 2$ is that minimal polynomial, hence irreducible.
– Daniel Fischer♦
Jul 16 at 9:45
3
3
If you already know that $X^5 - 2$ is irreducible in $mathbbQ[X]$, it suffices to note that $2$ and $5$ are coprime.
– Daniel Fischer♦
Jul 15 at 21:04
If you already know that $X^5 - 2$ is irreducible in $mathbbQ[X]$, it suffices to note that $2$ and $5$ are coprime.
– Daniel Fischer♦
Jul 15 at 21:04
@DanielFischer Why is it sufficient?
– user437309
Jul 15 at 21:13
@DanielFischer Why is it sufficient?
– user437309
Jul 15 at 21:13
1
1
@user437307 We know that $[Q(sqrt3,sqrt[5]2):Q(sqrt3)]*[Q(sqrt3):Q]=[Q(sqrt3,sqrt[5]2):Q]$ and $[Q(sqrt3,sqrt[5]2):Q(sqrt[5]2)]*[Q(sqrt[5]2):Q]=[Q(sqrt3,sqrt[5]2):Q]$ Now use the fact that $2$ and $5$ are coprime
– Sorfosh
Jul 15 at 21:54
@user437307 We know that $[Q(sqrt3,sqrt[5]2):Q(sqrt3)]*[Q(sqrt3):Q]=[Q(sqrt3,sqrt[5]2):Q]$ and $[Q(sqrt3,sqrt[5]2):Q(sqrt[5]2)]*[Q(sqrt[5]2):Q]=[Q(sqrt3,sqrt[5]2):Q]$ Now use the fact that $2$ and $5$ are coprime
– Sorfosh
Jul 15 at 21:54
@Sorfosh Do you mean I should use it to conclude that since $5k=2n$, $2n$ is divisible by $5$, so $n$ is divisible by $5$. And since $nle 5$ and it's divisible by $5$, $n=5$? Here $n=[mathbb Q(sqrt[5]2,sqrt3):mathbb Q(sqrt 3)]$. But I'm not sure how this is related to irreducibility.
– user437309
Jul 15 at 22:28
@Sorfosh Do you mean I should use it to conclude that since $5k=2n$, $2n$ is divisible by $5$, so $n$ is divisible by $5$. And since $nle 5$ and it's divisible by $5$, $n=5$? Here $n=[mathbb Q(sqrt[5]2,sqrt3):mathbb Q(sqrt 3)]$. But I'm not sure how this is related to irreducibility.
– user437309
Jul 15 at 22:28
1
1
@user437307 Yes, the idea is to use it to conclude $n = 5$. And since $n$ is the degree of the minimal polynomial of $sqrt[5]2$ over $mathbbQ(sqrt3)$ it follows that $X^5 - 2$ is that minimal polynomial, hence irreducible.
– Daniel Fischer♦
Jul 16 at 9:45
@user437307 Yes, the idea is to use it to conclude $n = 5$. And since $n$ is the degree of the minimal polynomial of $sqrt[5]2$ over $mathbbQ(sqrt3)$ it follows that $X^5 - 2$ is that minimal polynomial, hence irreducible.
– Daniel Fischer♦
Jul 16 at 9:45
add a comment |Â
1 Answer
1
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$Bbb Z[sqrt3]$ is UFD, $a^2-3b^2=2$ has no solution (since it has no solution mod $3$), so $2$ is prime in $Bbb Z[sqrt3]$. Now use Eisenstein.
answered Jul 15 at 20:55
Kenny Lau
18.7k2157
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
$Bbb Z[sqrt3]$ is UFD, $a^2-3b^2=2$ has no solution (since it has no solution mod $3$), so $2$ is prime in $Bbb Z[sqrt3]$. Now use Eisenstein.
answered Jul 15 at 20:55
Kenny Lau
18.7k2157
add a comment |Â
up vote
2
down vote
$Bbb Z[sqrt3]$ is UFD, $a^2-3b^2=2$ has no solution (since it has no solution mod $3$), so $2$ is prime in $Bbb Z[sqrt3]$. Now use Eisenstein.
answered Jul 15 at 20:55
Kenny Lau
18.7k2157
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$Bbb Z[sqrt3]$ is UFD, $a^2-3b^2=2$ has no solution (since it has no solution mod $3$), so $2$ is prime in $Bbb Z[sqrt3]$. Now use Eisenstein.
answered Jul 15 at 20:55
Kenny Lau
18.7k2157
$Bbb Z[sqrt3]$ is UFD, $a^2-3b^2=2$ has no solution (since it has no solution mod $3$), so $2$ is prime in $Bbb Z[sqrt3]$. Now use Eisenstein.
answered Jul 15 at 20:55
Kenny Lau
18.7k2157
answered Jul 15 at 20:55
Kenny Lau
18.7k2157
answered Jul 15 at 20:55
Kenny Lau
18.7k2157
answered Jul 15 at 20:55
Kenny Lau
18.7k2157
18.7k2157
add a comment |Â
add a comment |Â
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3
If you already know that $X^5 - 2$ is irreducible in $mathbbQ[X]$, it suffices to note that $2$ and $5$ are coprime.
– Daniel Fischer♦
Jul 15 at 21:04
@DanielFischer Why is it sufficient?
– user437309
Jul 15 at 21:13
1
@user437307 We know that $[Q(sqrt3,sqrt[5]2):Q(sqrt3)]*[Q(sqrt3):Q]=[Q(sqrt3,sqrt[5]2):Q]$ and $[Q(sqrt3,sqrt[5]2):Q(sqrt[5]2)]*[Q(sqrt[5]2):Q]=[Q(sqrt3,sqrt[5]2):Q]$ Now use the fact that $2$ and $5$ are coprime
– Sorfosh
Jul 15 at 21:54
@Sorfosh Do you mean I should use it to conclude that since $5k=2n$, $2n$ is divisible by $5$, so $n$ is divisible by $5$. And since $nle 5$ and it's divisible by $5$, $n=5$? Here $n=[mathbb Q(sqrt[5]2,sqrt3):mathbb Q(sqrt 3)]$. But I'm not sure how this is related to irreducibility.
– user437309
Jul 15 at 22:28
1
@user437307 Yes, the idea is to use it to conclude $n = 5$. And since $n$ is the degree of the minimal polynomial of $sqrt[5]2$ over $mathbbQ(sqrt3)$ it follows that $X^5 - 2$ is that minimal polynomial, hence irreducible.
– Daniel Fischer♦
Jul 16 at 9:45