Why is $x^4 - 4x + 2$ irreducible over $mathbb Q(i)$?
$begingroup$
This is an exercise from Garling's A Course in Galois Theory Ch. 5 on irreducible polynomials.
The other part of the question was to show that $x^5 - 4x + 2$ was irreducible over $mathbb Q(i)$. I got that one. Clearly the polynomial is irreducible over $mathbb Q$ by Eisenstein and Gauss' lemma, and, since 5 and 2 are relatively prime, it follows from a previous exercise that a root of the polynomial must have degree 5 over $mathbb Q(i)$. So the given polynomial must be the root's minimal polynomial over $mathbb Q(i)$ (and not just over $mathbb Q$). So in particular the polynomial must be irreducible over $mathbb Q(i)$.
But that trick doesn't work for $x^4 - 4x + 2$ anymore, and I'm stumped. I convinced myself that it can't have a linear factor over $mathbb Z[i]$ because that would mean a root that was a factor of $2$, and none of the factors of $2$ (namely, $pm 2, pm 1, pm 2i, pm i, pm (1 - i), pm (1 + i)$) solve the polynomial, but I don't see why it can't have a quadratic factor. Help!
abstract-algebra irreducible-polynomials
asked Dec 11 '18 at 1:48
user3339517user3339517
315
$endgroup$
add a comment |
$begingroup$
This is an exercise from Garling's A Course in Galois Theory Ch. 5 on irreducible polynomials.
The other part of the question was to show that $x^5 - 4x + 2$ was irreducible over $mathbb Q(i)$. I got that one. Clearly the polynomial is irreducible over $mathbb Q$ by Eisenstein and Gauss' lemma, and, since 5 and 2 are relatively prime, it follows from a previous exercise that a root of the polynomial must have degree 5 over $mathbb Q(i)$. So the given polynomial must be the root's minimal polynomial over $mathbb Q(i)$ (and not just over $mathbb Q$). So in particular the polynomial must be irreducible over $mathbb Q(i)$.
But that trick doesn't work for $x^4 - 4x + 2$ anymore, and I'm stumped. I convinced myself that it can't have a linear factor over $mathbb Z[i]$ because that would mean a root that was a factor of $2$, and none of the factors of $2$ (namely, $pm 2, pm 1, pm 2i, pm i, pm (1 - i), pm (1 + i)$) solve the polynomial, but I don't see why it can't have a quadratic factor. Help!
abstract-algebra irreducible-polynomials
asked Dec 11 '18 at 1:48
user3339517user3339517
315
$endgroup$
3
$begingroup$
Hmm... in $mathbb{F}_5[x]$ the polynomial $x^4 - 4x + 2$ appears to have exactly one root $x = 2$, and therefore it would be a product of a linear factor $x-2$ and an irreducible cubic factor. And also, there's a ring homomorphism $mathbb{Z}[i] to mathbb{F}_5$ which sends $i mapsto 2$. If I'm not mistaken, those facts could be put together into a proof.
$endgroup$
– Daniel Schepler
Dec 11 '18 at 2:01
$begingroup$
I will have to think about that but thank you for the idea.
$endgroup$
– user3339517
Dec 11 '18 at 2:08
2
$begingroup$
I think that’s right, @DanielSchepler. Since this quartic has no roots in $Bbb Q(i)$, the only possible factorization is into two quadratics. But since $(x-2)(x^3+2x^2+4x+4)$ is the only factorization into irreducibles over $Bbb F_5$, you get a contradiction to the existence of a pair of irreducible quadratic factors over $Bbb Q(i)$. Why don’t you write it up as an answer?
$endgroup$
– Lubin
Dec 11 '18 at 3:53
add a comment |
$begingroup$
This is an exercise from Garling's A Course in Galois Theory Ch. 5 on irreducible polynomials.
The other part of the question was to show that $x^5 - 4x + 2$ was irreducible over $mathbb Q(i)$. I got that one. Clearly the polynomial is irreducible over $mathbb Q$ by Eisenstein and Gauss' lemma, and, since 5 and 2 are relatively prime, it follows from a previous exercise that a root of the polynomial must have degree 5 over $mathbb Q(i)$. So the given polynomial must be the root's minimal polynomial over $mathbb Q(i)$ (and not just over $mathbb Q$). So in particular the polynomial must be irreducible over $mathbb Q(i)$.
But that trick doesn't work for $x^4 - 4x + 2$ anymore, and I'm stumped. I convinced myself that it can't have a linear factor over $mathbb Z[i]$ because that would mean a root that was a factor of $2$, and none of the factors of $2$ (namely, $pm 2, pm 1, pm 2i, pm i, pm (1 - i), pm (1 + i)$) solve the polynomial, but I don't see why it can't have a quadratic factor. Help!
abstract-algebra irreducible-polynomials
asked Dec 11 '18 at 1:48
user3339517user3339517
315
$endgroup$
This is an exercise from Garling's A Course in Galois Theory Ch. 5 on irreducible polynomials.
The other part of the question was to show that $x^5 - 4x + 2$ was irreducible over $mathbb Q(i)$. I got that one. Clearly the polynomial is irreducible over $mathbb Q$ by Eisenstein and Gauss' lemma, and, since 5 and 2 are relatively prime, it follows from a previous exercise that a root of the polynomial must have degree 5 over $mathbb Q(i)$. So the given polynomial must be the root's minimal polynomial over $mathbb Q(i)$ (and not just over $mathbb Q$). So in particular the polynomial must be irreducible over $mathbb Q(i)$.
But that trick doesn't work for $x^4 - 4x + 2$ anymore, and I'm stumped. I convinced myself that it can't have a linear factor over $mathbb Z[i]$ because that would mean a root that was a factor of $2$, and none of the factors of $2$ (namely, $pm 2, pm 1, pm 2i, pm i, pm (1 - i), pm (1 + i)$) solve the polynomial, but I don't see why it can't have a quadratic factor. Help!
abstract-algebra irreducible-polynomials
abstract-algebra irreducible-polynomials
asked Dec 11 '18 at 1:48
user3339517user3339517
315
asked Dec 11 '18 at 1:48
user3339517user3339517
315
asked Dec 11 '18 at 1:48
user3339517user3339517
315
asked Dec 11 '18 at 1:48
user3339517user3339517
315
asked Dec 11 '18 at 1:48
user3339517user3339517
315
315
3
$begingroup$
Hmm... in $mathbb{F}_5[x]$ the polynomial $x^4 - 4x + 2$ appears to have exactly one root $x = 2$, and therefore it would be a product of a linear factor $x-2$ and an irreducible cubic factor. And also, there's a ring homomorphism $mathbb{Z}[i] to mathbb{F}_5$ which sends $i mapsto 2$. If I'm not mistaken, those facts could be put together into a proof.
$endgroup$
– Daniel Schepler
Dec 11 '18 at 2:01
$begingroup$
I will have to think about that but thank you for the idea.
$endgroup$
– user3339517
Dec 11 '18 at 2:08
2
$begingroup$
I think that’s right, @DanielSchepler. Since this quartic has no roots in $Bbb Q(i)$, the only possible factorization is into two quadratics. But since $(x-2)(x^3+2x^2+4x+4)$ is the only factorization into irreducibles over $Bbb F_5$, you get a contradiction to the existence of a pair of irreducible quadratic factors over $Bbb Q(i)$. Why don’t you write it up as an answer?
$endgroup$
– Lubin
Dec 11 '18 at 3:53
add a comment |
3
$begingroup$
Hmm... in $mathbb{F}_5[x]$ the polynomial $x^4 - 4x + 2$ appears to have exactly one root $x = 2$, and therefore it would be a product of a linear factor $x-2$ and an irreducible cubic factor. And also, there's a ring homomorphism $mathbb{Z}[i] to mathbb{F}_5$ which sends $i mapsto 2$. If I'm not mistaken, those facts could be put together into a proof.
$endgroup$
– Daniel Schepler
Dec 11 '18 at 2:01
$begingroup$
I will have to think about that but thank you for the idea.
$endgroup$
– user3339517
Dec 11 '18 at 2:08
2
$begingroup$
I think that’s right, @DanielSchepler. Since this quartic has no roots in $Bbb Q(i)$, the only possible factorization is into two quadratics. But since $(x-2)(x^3+2x^2+4x+4)$ is the only factorization into irreducibles over $Bbb F_5$, you get a contradiction to the existence of a pair of irreducible quadratic factors over $Bbb Q(i)$. Why don’t you write it up as an answer?
$endgroup$
– Lubin
Dec 11 '18 at 3:53
3
3
$begingroup$
Hmm... in $mathbb{F}_5[x]$ the polynomial $x^4 - 4x + 2$ appears to have exactly one root $x = 2$, and therefore it would be a product of a linear factor $x-2$ and an irreducible cubic factor. And also, there's a ring homomorphism $mathbb{Z}[i] to mathbb{F}_5$ which sends $i mapsto 2$. If I'm not mistaken, those facts could be put together into a proof.
$endgroup$
– Daniel Schepler
Dec 11 '18 at 2:01
$begingroup$
Hmm... in $mathbb{F}_5[x]$ the polynomial $x^4 - 4x + 2$ appears to have exactly one root $x = 2$, and therefore it would be a product of a linear factor $x-2$ and an irreducible cubic factor. And also, there's a ring homomorphism $mathbb{Z}[i] to mathbb{F}_5$ which sends $i mapsto 2$. If I'm not mistaken, those facts could be put together into a proof.
$endgroup$
– Daniel Schepler
Dec 11 '18 at 2:01
$begingroup$
I will have to think about that but thank you for the idea.
$endgroup$
– user3339517
Dec 11 '18 at 2:08
$begingroup$
I will have to think about that but thank you for the idea.
$endgroup$
– user3339517
Dec 11 '18 at 2:08
2
2
$begingroup$
I think that’s right, @DanielSchepler. Since this quartic has no roots in $Bbb Q(i)$, the only possible factorization is into two quadratics. But since $(x-2)(x^3+2x^2+4x+4)$ is the only factorization into irreducibles over $Bbb F_5$, you get a contradiction to the existence of a pair of irreducible quadratic factors over $Bbb Q(i)$. Why don’t you write it up as an answer?
$endgroup$
– Lubin
Dec 11 '18 at 3:53
$begingroup$
I think that’s right, @DanielSchepler. Since this quartic has no roots in $Bbb Q(i)$, the only possible factorization is into two quadratics. But since $(x-2)(x^3+2x^2+4x+4)$ is the only factorization into irreducibles over $Bbb F_5$, you get a contradiction to the existence of a pair of irreducible quadratic factors over $Bbb Q(i)$. Why don’t you write it up as an answer?
$endgroup$
– Lubin
Dec 11 '18 at 3:53
add a comment |
1 Answer
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$begingroup$
By Gauss's lemma, since $mathbb{Z}[i]$ is a UFD, we see that to prove $x^4 - 4x + 2$ is irreducible over $mathbb{Q}(i)$ it will suffice to prove it is irreducible over $mathbb{Z}[i]$. As you checked, this polynomial does not have any linear factors; therefore, if it were reducible, the irreducible factors would have to be quadratic.
On the other hand, over $mathbb{F}_5$, the polynomial has exactly one root $x=2$ (which is not a double root). Therefore, here the irreducible factorization would have to be $x^4 - 4x + 2 = (x-2) (x^3 + 2x^2 + 4x + 4)$.
However, we have a ring homomorphism $mathbb{Z}[i] to mathbb{F}_5$ defined by $a + bi mapsto a + 2b$ (which induces an isomorphism $mathbb{Z}[i] / langle 2 - i rangle simeq mathbb{F}_5$). Therefore, any factorization of $x^4 - 4x + 2$ into quadratic factors over $mathbb{Z}[i]$ would induce a factorization into quadratic factors over $mathbb{F}_5$, which is a contradiction since $mathbb{F}_5[x]$ is a UFD.
answered Dec 11 '18 at 17:37
Daniel ScheplerDaniel Schepler
8,8691620
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add a comment |
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$begingroup$
By Gauss's lemma, since $mathbb{Z}[i]$ is a UFD, we see that to prove $x^4 - 4x + 2$ is irreducible over $mathbb{Q}(i)$ it will suffice to prove it is irreducible over $mathbb{Z}[i]$. As you checked, this polynomial does not have any linear factors; therefore, if it were reducible, the irreducible factors would have to be quadratic.
On the other hand, over $mathbb{F}_5$, the polynomial has exactly one root $x=2$ (which is not a double root). Therefore, here the irreducible factorization would have to be $x^4 - 4x + 2 = (x-2) (x^3 + 2x^2 + 4x + 4)$.
However, we have a ring homomorphism $mathbb{Z}[i] to mathbb{F}_5$ defined by $a + bi mapsto a + 2b$ (which induces an isomorphism $mathbb{Z}[i] / langle 2 - i rangle simeq mathbb{F}_5$). Therefore, any factorization of $x^4 - 4x + 2$ into quadratic factors over $mathbb{Z}[i]$ would induce a factorization into quadratic factors over $mathbb{F}_5$, which is a contradiction since $mathbb{F}_5[x]$ is a UFD.
answered Dec 11 '18 at 17:37
Daniel ScheplerDaniel Schepler
8,8691620
$endgroup$
add a comment |
$begingroup$
By Gauss's lemma, since $mathbb{Z}[i]$ is a UFD, we see that to prove $x^4 - 4x + 2$ is irreducible over $mathbb{Q}(i)$ it will suffice to prove it is irreducible over $mathbb{Z}[i]$. As you checked, this polynomial does not have any linear factors; therefore, if it were reducible, the irreducible factors would have to be quadratic.
On the other hand, over $mathbb{F}_5$, the polynomial has exactly one root $x=2$ (which is not a double root). Therefore, here the irreducible factorization would have to be $x^4 - 4x + 2 = (x-2) (x^3 + 2x^2 + 4x + 4)$.
However, we have a ring homomorphism $mathbb{Z}[i] to mathbb{F}_5$ defined by $a + bi mapsto a + 2b$ (which induces an isomorphism $mathbb{Z}[i] / langle 2 - i rangle simeq mathbb{F}_5$). Therefore, any factorization of $x^4 - 4x + 2$ into quadratic factors over $mathbb{Z}[i]$ would induce a factorization into quadratic factors over $mathbb{F}_5$, which is a contradiction since $mathbb{F}_5[x]$ is a UFD.
answered Dec 11 '18 at 17:37
Daniel ScheplerDaniel Schepler
8,8691620
$endgroup$
add a comment |
$begingroup$
By Gauss's lemma, since $mathbb{Z}[i]$ is a UFD, we see that to prove $x^4 - 4x + 2$ is irreducible over $mathbb{Q}(i)$ it will suffice to prove it is irreducible over $mathbb{Z}[i]$. As you checked, this polynomial does not have any linear factors; therefore, if it were reducible, the irreducible factors would have to be quadratic.
On the other hand, over $mathbb{F}_5$, the polynomial has exactly one root $x=2$ (which is not a double root). Therefore, here the irreducible factorization would have to be $x^4 - 4x + 2 = (x-2) (x^3 + 2x^2 + 4x + 4)$.
However, we have a ring homomorphism $mathbb{Z}[i] to mathbb{F}_5$ defined by $a + bi mapsto a + 2b$ (which induces an isomorphism $mathbb{Z}[i] / langle 2 - i rangle simeq mathbb{F}_5$). Therefore, any factorization of $x^4 - 4x + 2$ into quadratic factors over $mathbb{Z}[i]$ would induce a factorization into quadratic factors over $mathbb{F}_5$, which is a contradiction since $mathbb{F}_5[x]$ is a UFD.
answered Dec 11 '18 at 17:37
Daniel ScheplerDaniel Schepler
8,8691620
$endgroup$
By Gauss's lemma, since $mathbb{Z}[i]$ is a UFD, we see that to prove $x^4 - 4x + 2$ is irreducible over $mathbb{Q}(i)$ it will suffice to prove it is irreducible over $mathbb{Z}[i]$. As you checked, this polynomial does not have any linear factors; therefore, if it were reducible, the irreducible factors would have to be quadratic.
On the other hand, over $mathbb{F}_5$, the polynomial has exactly one root $x=2$ (which is not a double root). Therefore, here the irreducible factorization would have to be $x^4 - 4x + 2 = (x-2) (x^3 + 2x^2 + 4x + 4)$.
However, we have a ring homomorphism $mathbb{Z}[i] to mathbb{F}_5$ defined by $a + bi mapsto a + 2b$ (which induces an isomorphism $mathbb{Z}[i] / langle 2 - i rangle simeq mathbb{F}_5$). Therefore, any factorization of $x^4 - 4x + 2$ into quadratic factors over $mathbb{Z}[i]$ would induce a factorization into quadratic factors over $mathbb{F}_5$, which is a contradiction since $mathbb{F}_5[x]$ is a UFD.
answered Dec 11 '18 at 17:37
Daniel ScheplerDaniel Schepler
8,8691620
answered Dec 11 '18 at 17:37
Daniel ScheplerDaniel Schepler
8,8691620
answered Dec 11 '18 at 17:37
Daniel ScheplerDaniel Schepler
8,8691620
answered Dec 11 '18 at 17:37
Daniel ScheplerDaniel Schepler
8,8691620
8,8691620
add a comment |
add a comment |
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$begingroup$
Hmm... in $mathbb{F}_5[x]$ the polynomial $x^4 - 4x + 2$ appears to have exactly one root $x = 2$, and therefore it would be a product of a linear factor $x-2$ and an irreducible cubic factor. And also, there's a ring homomorphism $mathbb{Z}[i] to mathbb{F}_5$ which sends $i mapsto 2$. If I'm not mistaken, those facts could be put together into a proof.
$endgroup$
– Daniel Schepler
Dec 11 '18 at 2:01
$begingroup$
I will have to think about that but thank you for the idea.
$endgroup$
– user3339517
Dec 11 '18 at 2:08
2
$begingroup$
I think that’s right, @DanielSchepler. Since this quartic has no roots in $Bbb Q(i)$, the only possible factorization is into two quadratics. But since $(x-2)(x^3+2x^2+4x+4)$ is the only factorization into irreducibles over $Bbb F_5$, you get a contradiction to the existence of a pair of irreducible quadratic factors over $Bbb Q(i)$. Why don’t you write it up as an answer?
$endgroup$
– Lubin
Dec 11 '18 at 3:53