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$x^4+x^3+x^2+x+1$ irreducible over $mathbb F_7$
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The answer claims that $x^4+x^3+x^2+x+1$ is irreducible over $mathbb F_7$. I can check that it has no roots in the field, but why can't it be written as a product of two quadratics? I tried the method of undetermined coefficients: $$x^4+x^3+x^2+x+1=(x^2+ax+b)(x^2+cx+d)=\ x^4+(a+c)x^3+(ac+b+d)x^2+(bc+ad)x+bd$$ so $$a+c=1\ac+b+d=1\bc+ad=1\bd=1$$ but I wasn't able to solve this. Maybe there is a way to see irreducibility by only using general finite fields theory?
abstract-algebra polynomials field-theory finite-fields irreducible-polynomials
asked Aug 6 at 19:31
user437309
556212
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up vote
7
down vote
favorite
This question came from the answer here.
The answer claims that $x^4+x^3+x^2+x+1$ is irreducible over $mathbb F_7$. I can check that it has no roots in the field, but why can't it be written as a product of two quadratics? I tried the method of undetermined coefficients: $$x^4+x^3+x^2+x+1=(x^2+ax+b)(x^2+cx+d)=\ x^4+(a+c)x^3+(ac+b+d)x^2+(bc+ad)x+bd$$ so $$a+c=1\ac+b+d=1\bc+ad=1\bd=1$$ but I wasn't able to solve this. Maybe there is a way to see irreducibility by only using general finite fields theory?
abstract-algebra polynomials field-theory finite-fields irreducible-polynomials
asked Aug 6 at 19:31
user437309
556212
Worst case scenario, check the different choices for $b$ and $a$. $c$ and $d$ are dictated at that point. I'd like to see a slicker way.
– Randall
Aug 6 at 19:38
this question may help.
– lulu
Aug 6 at 19:43
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
This question came from the answer here.
The answer claims that $x^4+x^3+x^2+x+1$ is irreducible over $mathbb F_7$. I can check that it has no roots in the field, but why can't it be written as a product of two quadratics? I tried the method of undetermined coefficients: $$x^4+x^3+x^2+x+1=(x^2+ax+b)(x^2+cx+d)=\ x^4+(a+c)x^3+(ac+b+d)x^2+(bc+ad)x+bd$$ so $$a+c=1\ac+b+d=1\bc+ad=1\bd=1$$ but I wasn't able to solve this. Maybe there is a way to see irreducibility by only using general finite fields theory?
abstract-algebra polynomials field-theory finite-fields irreducible-polynomials
asked Aug 6 at 19:31
user437309
556212
This question came from the answer here.
The answer claims that $x^4+x^3+x^2+x+1$ is irreducible over $mathbb F_7$. I can check that it has no roots in the field, but why can't it be written as a product of two quadratics? I tried the method of undetermined coefficients: $$x^4+x^3+x^2+x+1=(x^2+ax+b)(x^2+cx+d)=\ x^4+(a+c)x^3+(ac+b+d)x^2+(bc+ad)x+bd$$ so $$a+c=1\ac+b+d=1\bc+ad=1\bd=1$$ but I wasn't able to solve this. Maybe there is a way to see irreducibility by only using general finite fields theory?
abstract-algebra polynomials field-theory finite-fields irreducible-polynomials
asked Aug 6 at 19:31
user437309
556212
asked Aug 6 at 19:31
user437309
556212
asked Aug 6 at 19:31
user437309
556212
asked Aug 6 at 19:31
user437309
556212
556212
Worst case scenario, check the different choices for $b$ and $a$. $c$ and $d$ are dictated at that point. I'd like to see a slicker way.
– Randall
Aug 6 at 19:38
this question may help.
– lulu
Aug 6 at 19:43
add a comment |Â
Worst case scenario, check the different choices for $b$ and $a$. $c$ and $d$ are dictated at that point. I'd like to see a slicker way.
– Randall
Aug 6 at 19:38
this question may help.
– lulu
Aug 6 at 19:43
Worst case scenario, check the different choices for $b$ and $a$. $c$ and $d$ are dictated at that point. I'd like to see a slicker way.
– Randall
Aug 6 at 19:38
Worst case scenario, check the different choices for $b$ and $a$. $c$ and $d$ are dictated at that point. I'd like to see a slicker way.
– Randall
Aug 6 at 19:38
this question may help.
– lulu
Aug 6 at 19:43
this question may help.
– lulu
Aug 6 at 19:43
add a comment |Â
4 Answers
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11
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Any element in any field satisfying $x^4+x^3+x^2+x+1=0$ also satisfies $x^5=1$. In other words, it has order $5$ in the multiplicative group of that field.
The field with $p^k$ elements has a multiplicative group with $p^k-1$ elements, so it has elements of order $5$ if and only if $5$ divides $p^k-1$. This is not the case for $p=7$ and $k=1,2,3$, but it is the case for $p=7,k=4$.
Therefore, the smallest field containing $mathbbF_7$ and roots of $x^4+x^3+x^2+x+1$ has $7^4$ elements. If that polynomial were reducible, there would be a smaller splitting field with such roots, but there isn’t.
answered Aug 6 at 19:43
Alon Amit
10.3k3765
Why would there be a smaller splitting field (than that with $7^4$ elements) with the roots of $x^4+x^3+x^2+x+1$ if that polynomial were reducible?
– user437309
Aug 6 at 20:00
Take a factor of that polynomial and consider the splitting field of that factor.
– Alon Amit
Aug 6 at 20:01
1
I have lost count of the questions I have settled here using this argument. Welcome to the club :-)
– Jyrki Lahtonen
Aug 6 at 21:12
@JyrkiLahtonen proud to be a member!
– Alon Amit
Aug 6 at 22:08
1
@user437309 yes, that’s right.
– Alon Amit
Aug 7 at 19:45
 |Â
show 10 more comments
up vote
3
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If you want to continue your approach in case you cannot come up with some clever argument like that in Alon Amit's answer, then you can do it as follows. First, note that $c=1-a$ and $d=b^-1$. Thus, the equation $ac+b+d=1$ becomes
$$a(1-a)+b+b^-1=1,.tag*$$
The equation $bc+ad=1$ is now
$$b(1-a)+ab^-1=1,.$$
Thus,
$$a^-1big(1-b(1-a)big)=b^-1=1-b-a(1-a),.$$
That is,
$$1-b(1-a)=a-ab-a^2(1-a),.$$
Consequently,
$$(2a-1)b=-1+a-a^2+a^3=(a-1)(a^2+1),.tag#$$
Thus, $anotin0,1,4$ for the equation above to be possible (noting that $aneq 0$ and $bneq 0$). You are left with four choices of $a$, namely, $ain2,3,5,6$.
If $a=2$, then, using (#), we get $b=3^-1cdot 5=5cdot 5=4$, so $b^-1=2$, but then $$a(1-a)+b+b^-1=2cdot(-1)+4+2=4neq 1,.$$
If $a=3$, then we have by (#) that $b=5^-1cdot20=-3=4$, whence
$$a(1-a)+b+b^-1=3cdot(-2)+4+2=0neq 1,.$$
If $a=5$, then $b=3^-1cdot (-1)=-5=2$, so $b^-1=4$, but then
$$a(1-a)+b+b^-1=5cdot(-4)+2+4=0neq 1,.$$
Finally, if $a=6$, then $b=4^-1cdot3=2cdot 3=-1$, whence $b^-1=-1$, and
$$a(1-a)+b+b^-1=6cdot(-5)+(-1)+(-1)=-4neq 1,.$$
Thus, (*) cannot be satisfied.
answered Aug 6 at 20:01
Batominovski
23.4k22779
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Yet an other solution. It is based on the fact that
$$ f = x^4+x^3+x^2+x+1inBbb F_7[x] $$
is reciprocal, and tries to use the idea in the OP. (There should be no splitting in two polynomials of degree two.)
Assume there is a factorization
$$
f=(x^2+ax+b)(x^2+cx+d) ,qquad a,b,c,dinBbb F_7 .$$
Let $une 0$ be a root of the first factor in some extension (of degree two). Then $1/u$ is also a root. We distinguish two cases.
First case: $1/u$ is also a root of the first factor. Then $b=1$, then $d=1$, $c+a=1$, the product
$$(x^2+ax+1)(x^2+(1-a)x+1)$$
is reciprocal,
and we have to look for a match of the coefficient in $x^2$, for a few values of $a$ (modulo the action $aleftrightarrow (1-a)$). There is no such $a$ with $1+a(1-a)+1=1$.
Second case: $1/u$ is not a root of the first factor. Let $u$, $v$ be the two roots of $x^2+ax+b$. Then $1/u$, $1/v$ are the roots of the second factor, which is the reciprocal polynomial, made monic, so we expect an equality of reciprocal polynomials
$$
(x^2+ax+b)(1+ax+bx^2) = b(x^4+x^3+x^2+x+1) .
$$
The two equations obtained, $ab+a=b$ and $1+a^2+b^2=b$, have no solution in $Bbb F_7$. (The substitution of $a=b/(b+1)$ in the second equation leads to... $(b^4+b^3+b^2+b+1)/(b+1)^2=0$.)
Note: This is arguably shorter then taking all possible polynomials $x^2+ax+b$, $bne 0$, associating $c=1-a$, $d=1/b$, and checking if the other conditions among $a,b,c,d$ are satisfied.
Note: Computer algebra programs show that we have an irreducible polynomial in our hands. For instance using sage:
sage: R.<x> = PolynomialRing(GF(7))
sage: f = x^4 + x^3 + x^2 + x + 1
sage: f.is_irreducible()
True
One can use then computer power also as follows to conclude. (With a slightly bigger effort than for the typing, one can also conclude humanly.) Assume $f$ splits in two (or more factors). One can check there is no root in $Bbb F_7$. Then there is a root in $Bbb F_7^2=Bbb F_49$, so the polynomials $f$ and $x^49-x$ have a common divisor. The gcd is but... we type
sage: gcd( x^49 - x, f )
1
sage: gcd( x^(7^4) - x, f )
x^4 + x^3 + x^2 + x + 1
For this, one can also compute easily $(x^49-x,f)$ as a human by noting that $f$ divides $x^5-1$, so $(x^49-x,f)=(x^49-x^4+x^4-x,f)=(x^4-x,f)=(x^3-1,f)=(x^2+x+1,f)=(x^2+x+1,x+1)=(1,x+1)=1$.
answered Aug 6 at 21:32
dan_fulea
4,2171211
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HINT.-If $x^4+x^3+x^2+x+1$ were reducible then we have either a linear factor with a cubic one or two quadratic factors.
We have $$x^4+x^3+x^2+x+1=-dfrac 14(-2x^2+(sqrt5-1)x-2)(2x^2+(sqrt5+1)x+2)$$
Since $mathbb F_7^2=1,2,4,0$ we see that $5$ is not a square modulo $7$. It follows $x^4+x^3+x^2+x+1$ is not a factor of two quadratics modulo $7$.
Besides that a linear factor is not possible is checked easily.
answered Aug 6 at 21:33
Piquito
17.4k31234
Explain the reason of your downvote, please. If there exists a factorization in two quadratics, then it would have two distinct factorizations in two quadratics over an extension of $mathbb F_7$. This is not possible.
– Piquito
Aug 6 at 22:38
I don't know who downvoted this answer, I just upvoted it.
– user437309
Aug 6 at 23:05
@user437309: I did not say, dear friend, that it was you who put the downvote. It should be mandatory in S.E. that a downvote is always with justification exposed. That the person that put it refuses to respond shows its (mathematically speaking) ilk.
– Piquito
Aug 7 at 0:07
Hmm. Over $BbbC$ the roots of the polynomial $x^4+x^3+x^2+x+1$ are $zeta^k$ with $zeta=e^2pi i/5$, $k=1,2,3,4$. Your argument shows that the quadratic factors with real coefficients, $(x-zeta)(x-zeta^4)$ and $(x-zeta^2)(x-zeta^3)$, don't exist in $BbbF_7$ because they involve $sqrt5$. But how do you rule out a factor like $(x-zeta)(x-zeta^2)$ showing up? True, Galois theory shows that $BbbQ(sqrt5)$ is the only quadratic subfield of $BbbQ(zeta)$. That does lead to an argument, but the connection to Galois theory of finite fields runs a bit deep.
– Jyrki Lahtonen
Aug 7 at 5:34
(cont'd) In if you really want to use Galois theory here, a simpler way is to use the Frobenius automorphism $F$ of all extensions of $BbbF_7$: If $zeta$ is a root of some polynomial with coefficients in $BbbF_7$ then so is $F(zeta)=zeta^7=zeta^2$, and therefore also $F(zeta^2)=zeta^14=zeta^4$, and therefore also $F(zeta^4)=zeta^3$. That's four zeros, so any polynomial with $zeta$ as a root has degree four.
– Jyrki Lahtonen
Aug 7 at 5:37
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
accepted
Any element in any field satisfying $x^4+x^3+x^2+x+1=0$ also satisfies $x^5=1$. In other words, it has order $5$ in the multiplicative group of that field.
The field with $p^k$ elements has a multiplicative group with $p^k-1$ elements, so it has elements of order $5$ if and only if $5$ divides $p^k-1$. This is not the case for $p=7$ and $k=1,2,3$, but it is the case for $p=7,k=4$.
Therefore, the smallest field containing $mathbbF_7$ and roots of $x^4+x^3+x^2+x+1$ has $7^4$ elements. If that polynomial were reducible, there would be a smaller splitting field with such roots, but there isn’t.
answered Aug 6 at 19:43
Alon Amit
10.3k3765
Why would there be a smaller splitting field (than that with $7^4$ elements) with the roots of $x^4+x^3+x^2+x+1$ if that polynomial were reducible?
– user437309
Aug 6 at 20:00
Take a factor of that polynomial and consider the splitting field of that factor.
– Alon Amit
Aug 6 at 20:01
1
I have lost count of the questions I have settled here using this argument. Welcome to the club :-)
– Jyrki Lahtonen
Aug 6 at 21:12
@JyrkiLahtonen proud to be a member!
– Alon Amit
Aug 6 at 22:08
1
@user437309 yes, that’s right.
– Alon Amit
Aug 7 at 19:45
 |Â
show 10 more comments
up vote
11
down vote
accepted
Any element in any field satisfying $x^4+x^3+x^2+x+1=0$ also satisfies $x^5=1$. In other words, it has order $5$ in the multiplicative group of that field.
The field with $p^k$ elements has a multiplicative group with $p^k-1$ elements, so it has elements of order $5$ if and only if $5$ divides $p^k-1$. This is not the case for $p=7$ and $k=1,2,3$, but it is the case for $p=7,k=4$.
Therefore, the smallest field containing $mathbbF_7$ and roots of $x^4+x^3+x^2+x+1$ has $7^4$ elements. If that polynomial were reducible, there would be a smaller splitting field with such roots, but there isn’t.
answered Aug 6 at 19:43
Alon Amit
10.3k3765
Why would there be a smaller splitting field (than that with $7^4$ elements) with the roots of $x^4+x^3+x^2+x+1$ if that polynomial were reducible?
– user437309
Aug 6 at 20:00
Take a factor of that polynomial and consider the splitting field of that factor.
– Alon Amit
Aug 6 at 20:01
1
I have lost count of the questions I have settled here using this argument. Welcome to the club :-)
– Jyrki Lahtonen
Aug 6 at 21:12
@JyrkiLahtonen proud to be a member!
– Alon Amit
Aug 6 at 22:08
1
@user437309 yes, that’s right.
– Alon Amit
Aug 7 at 19:45
 |Â
show 10 more comments
up vote
11
down vote
accepted
up vote
11
down vote
accepted
Any element in any field satisfying $x^4+x^3+x^2+x+1=0$ also satisfies $x^5=1$. In other words, it has order $5$ in the multiplicative group of that field.
The field with $p^k$ elements has a multiplicative group with $p^k-1$ elements, so it has elements of order $5$ if and only if $5$ divides $p^k-1$. This is not the case for $p=7$ and $k=1,2,3$, but it is the case for $p=7,k=4$.
Therefore, the smallest field containing $mathbbF_7$ and roots of $x^4+x^3+x^2+x+1$ has $7^4$ elements. If that polynomial were reducible, there would be a smaller splitting field with such roots, but there isn’t.
answered Aug 6 at 19:43
Alon Amit
10.3k3765
Any element in any field satisfying $x^4+x^3+x^2+x+1=0$ also satisfies $x^5=1$. In other words, it has order $5$ in the multiplicative group of that field.
The field with $p^k$ elements has a multiplicative group with $p^k-1$ elements, so it has elements of order $5$ if and only if $5$ divides $p^k-1$. This is not the case for $p=7$ and $k=1,2,3$, but it is the case for $p=7,k=4$.
Therefore, the smallest field containing $mathbbF_7$ and roots of $x^4+x^3+x^2+x+1$ has $7^4$ elements. If that polynomial were reducible, there would be a smaller splitting field with such roots, but there isn’t.
answered Aug 6 at 19:43
Alon Amit
10.3k3765
answered Aug 6 at 19:43
Alon Amit
10.3k3765
answered Aug 6 at 19:43
Alon Amit
10.3k3765
answered Aug 6 at 19:43
Alon Amit
10.3k3765
10.3k3765
Why would there be a smaller splitting field (than that with $7^4$ elements) with the roots of $x^4+x^3+x^2+x+1$ if that polynomial were reducible?
– user437309
Aug 6 at 20:00
Take a factor of that polynomial and consider the splitting field of that factor.
– Alon Amit
Aug 6 at 20:01
1
I have lost count of the questions I have settled here using this argument. Welcome to the club :-)
– Jyrki Lahtonen
Aug 6 at 21:12
@JyrkiLahtonen proud to be a member!
– Alon Amit
Aug 6 at 22:08
1
@user437309 yes, that’s right.
– Alon Amit
Aug 7 at 19:45
 |Â
show 10 more comments
Why would there be a smaller splitting field (than that with $7^4$ elements) with the roots of $x^4+x^3+x^2+x+1$ if that polynomial were reducible?
– user437309
Aug 6 at 20:00
Take a factor of that polynomial and consider the splitting field of that factor.
– Alon Amit
Aug 6 at 20:01
1
I have lost count of the questions I have settled here using this argument. Welcome to the club :-)
– Jyrki Lahtonen
Aug 6 at 21:12
@JyrkiLahtonen proud to be a member!
– Alon Amit
Aug 6 at 22:08
1
@user437309 yes, that’s right.
– Alon Amit
Aug 7 at 19:45
Why would there be a smaller splitting field (than that with $7^4$ elements) with the roots of $x^4+x^3+x^2+x+1$ if that polynomial were reducible?
– user437309
Aug 6 at 20:00
Why would there be a smaller splitting field (than that with $7^4$ elements) with the roots of $x^4+x^3+x^2+x+1$ if that polynomial were reducible?
– user437309
Aug 6 at 20:00
Take a factor of that polynomial and consider the splitting field of that factor.
– Alon Amit
Aug 6 at 20:01
Take a factor of that polynomial and consider the splitting field of that factor.
– Alon Amit
Aug 6 at 20:01
1
1
I have lost count of the questions I have settled here using this argument. Welcome to the club :-)
– Jyrki Lahtonen
Aug 6 at 21:12
I have lost count of the questions I have settled here using this argument. Welcome to the club :-)
– Jyrki Lahtonen
Aug 6 at 21:12
@JyrkiLahtonen proud to be a member!
– Alon Amit
Aug 6 at 22:08
@JyrkiLahtonen proud to be a member!
– Alon Amit
Aug 6 at 22:08
1
1
@user437309 yes, that’s right.
– Alon Amit
Aug 7 at 19:45
@user437309 yes, that’s right.
– Alon Amit
Aug 7 at 19:45
 |Â
show 10 more comments
up vote
3
down vote
If you want to continue your approach in case you cannot come up with some clever argument like that in Alon Amit's answer, then you can do it as follows. First, note that $c=1-a$ and $d=b^-1$. Thus, the equation $ac+b+d=1$ becomes
$$a(1-a)+b+b^-1=1,.tag*$$
The equation $bc+ad=1$ is now
$$b(1-a)+ab^-1=1,.$$
Thus,
$$a^-1big(1-b(1-a)big)=b^-1=1-b-a(1-a),.$$
That is,
$$1-b(1-a)=a-ab-a^2(1-a),.$$
Consequently,
$$(2a-1)b=-1+a-a^2+a^3=(a-1)(a^2+1),.tag#$$
Thus, $anotin0,1,4$ for the equation above to be possible (noting that $aneq 0$ and $bneq 0$). You are left with four choices of $a$, namely, $ain2,3,5,6$.
If $a=2$, then, using (#), we get $b=3^-1cdot 5=5cdot 5=4$, so $b^-1=2$, but then $$a(1-a)+b+b^-1=2cdot(-1)+4+2=4neq 1,.$$
If $a=3$, then we have by (#) that $b=5^-1cdot20=-3=4$, whence
$$a(1-a)+b+b^-1=3cdot(-2)+4+2=0neq 1,.$$
If $a=5$, then $b=3^-1cdot (-1)=-5=2$, so $b^-1=4$, but then
$$a(1-a)+b+b^-1=5cdot(-4)+2+4=0neq 1,.$$
Finally, if $a=6$, then $b=4^-1cdot3=2cdot 3=-1$, whence $b^-1=-1$, and
$$a(1-a)+b+b^-1=6cdot(-5)+(-1)+(-1)=-4neq 1,.$$
Thus, (*) cannot be satisfied.
answered Aug 6 at 20:01
Batominovski
23.4k22779
add a comment |Â
up vote
3
down vote
If you want to continue your approach in case you cannot come up with some clever argument like that in Alon Amit's answer, then you can do it as follows. First, note that $c=1-a$ and $d=b^-1$. Thus, the equation $ac+b+d=1$ becomes
$$a(1-a)+b+b^-1=1,.tag*$$
The equation $bc+ad=1$ is now
$$b(1-a)+ab^-1=1,.$$
Thus,
$$a^-1big(1-b(1-a)big)=b^-1=1-b-a(1-a),.$$
That is,
$$1-b(1-a)=a-ab-a^2(1-a),.$$
Consequently,
$$(2a-1)b=-1+a-a^2+a^3=(a-1)(a^2+1),.tag#$$
Thus, $anotin0,1,4$ for the equation above to be possible (noting that $aneq 0$ and $bneq 0$). You are left with four choices of $a$, namely, $ain2,3,5,6$.
If $a=2$, then, using (#), we get $b=3^-1cdot 5=5cdot 5=4$, so $b^-1=2$, but then $$a(1-a)+b+b^-1=2cdot(-1)+4+2=4neq 1,.$$
If $a=3$, then we have by (#) that $b=5^-1cdot20=-3=4$, whence
$$a(1-a)+b+b^-1=3cdot(-2)+4+2=0neq 1,.$$
If $a=5$, then $b=3^-1cdot (-1)=-5=2$, so $b^-1=4$, but then
$$a(1-a)+b+b^-1=5cdot(-4)+2+4=0neq 1,.$$
Finally, if $a=6$, then $b=4^-1cdot3=2cdot 3=-1$, whence $b^-1=-1$, and
$$a(1-a)+b+b^-1=6cdot(-5)+(-1)+(-1)=-4neq 1,.$$
Thus, (*) cannot be satisfied.
answered Aug 6 at 20:01
Batominovski
23.4k22779
add a comment |Â
up vote
3
down vote
up vote
3
down vote
If you want to continue your approach in case you cannot come up with some clever argument like that in Alon Amit's answer, then you can do it as follows. First, note that $c=1-a$ and $d=b^-1$. Thus, the equation $ac+b+d=1$ becomes
$$a(1-a)+b+b^-1=1,.tag*$$
The equation $bc+ad=1$ is now
$$b(1-a)+ab^-1=1,.$$
Thus,
$$a^-1big(1-b(1-a)big)=b^-1=1-b-a(1-a),.$$
That is,
$$1-b(1-a)=a-ab-a^2(1-a),.$$
Consequently,
$$(2a-1)b=-1+a-a^2+a^3=(a-1)(a^2+1),.tag#$$
Thus, $anotin0,1,4$ for the equation above to be possible (noting that $aneq 0$ and $bneq 0$). You are left with four choices of $a$, namely, $ain2,3,5,6$.
If $a=2$, then, using (#), we get $b=3^-1cdot 5=5cdot 5=4$, so $b^-1=2$, but then $$a(1-a)+b+b^-1=2cdot(-1)+4+2=4neq 1,.$$
If $a=3$, then we have by (#) that $b=5^-1cdot20=-3=4$, whence
$$a(1-a)+b+b^-1=3cdot(-2)+4+2=0neq 1,.$$
If $a=5$, then $b=3^-1cdot (-1)=-5=2$, so $b^-1=4$, but then
$$a(1-a)+b+b^-1=5cdot(-4)+2+4=0neq 1,.$$
Finally, if $a=6$, then $b=4^-1cdot3=2cdot 3=-1$, whence $b^-1=-1$, and
$$a(1-a)+b+b^-1=6cdot(-5)+(-1)+(-1)=-4neq 1,.$$
Thus, (*) cannot be satisfied.
answered Aug 6 at 20:01
Batominovski
23.4k22779
If you want to continue your approach in case you cannot come up with some clever argument like that in Alon Amit's answer, then you can do it as follows. First, note that $c=1-a$ and $d=b^-1$. Thus, the equation $ac+b+d=1$ becomes
$$a(1-a)+b+b^-1=1,.tag*$$
The equation $bc+ad=1$ is now
$$b(1-a)+ab^-1=1,.$$
Thus,
$$a^-1big(1-b(1-a)big)=b^-1=1-b-a(1-a),.$$
That is,
$$1-b(1-a)=a-ab-a^2(1-a),.$$
Consequently,
$$(2a-1)b=-1+a-a^2+a^3=(a-1)(a^2+1),.tag#$$
Thus, $anotin0,1,4$ for the equation above to be possible (noting that $aneq 0$ and $bneq 0$). You are left with four choices of $a$, namely, $ain2,3,5,6$.
If $a=2$, then, using (#), we get $b=3^-1cdot 5=5cdot 5=4$, so $b^-1=2$, but then $$a(1-a)+b+b^-1=2cdot(-1)+4+2=4neq 1,.$$
If $a=3$, then we have by (#) that $b=5^-1cdot20=-3=4$, whence
$$a(1-a)+b+b^-1=3cdot(-2)+4+2=0neq 1,.$$
If $a=5$, then $b=3^-1cdot (-1)=-5=2$, so $b^-1=4$, but then
$$a(1-a)+b+b^-1=5cdot(-4)+2+4=0neq 1,.$$
Finally, if $a=6$, then $b=4^-1cdot3=2cdot 3=-1$, whence $b^-1=-1$, and
$$a(1-a)+b+b^-1=6cdot(-5)+(-1)+(-1)=-4neq 1,.$$
Thus, (*) cannot be satisfied.
answered Aug 6 at 20:01
Batominovski
23.4k22779
answered Aug 6 at 20:01
Batominovski
23.4k22779
answered Aug 6 at 20:01
Batominovski
23.4k22779
answered Aug 6 at 20:01
Batominovski
23.4k22779
23.4k22779
add a comment |Â
add a comment |Â
up vote
1
down vote
Yet an other solution. It is based on the fact that
$$ f = x^4+x^3+x^2+x+1inBbb F_7[x] $$
is reciprocal, and tries to use the idea in the OP. (There should be no splitting in two polynomials of degree two.)
Assume there is a factorization
$$
f=(x^2+ax+b)(x^2+cx+d) ,qquad a,b,c,dinBbb F_7 .$$
Let $une 0$ be a root of the first factor in some extension (of degree two). Then $1/u$ is also a root. We distinguish two cases.
First case: $1/u$ is also a root of the first factor. Then $b=1$, then $d=1$, $c+a=1$, the product
$$(x^2+ax+1)(x^2+(1-a)x+1)$$
is reciprocal,
and we have to look for a match of the coefficient in $x^2$, for a few values of $a$ (modulo the action $aleftrightarrow (1-a)$). There is no such $a$ with $1+a(1-a)+1=1$.
Second case: $1/u$ is not a root of the first factor. Let $u$, $v$ be the two roots of $x^2+ax+b$. Then $1/u$, $1/v$ are the roots of the second factor, which is the reciprocal polynomial, made monic, so we expect an equality of reciprocal polynomials
$$
(x^2+ax+b)(1+ax+bx^2) = b(x^4+x^3+x^2+x+1) .
$$
The two equations obtained, $ab+a=b$ and $1+a^2+b^2=b$, have no solution in $Bbb F_7$. (The substitution of $a=b/(b+1)$ in the second equation leads to... $(b^4+b^3+b^2+b+1)/(b+1)^2=0$.)
Note: This is arguably shorter then taking all possible polynomials $x^2+ax+b$, $bne 0$, associating $c=1-a$, $d=1/b$, and checking if the other conditions among $a,b,c,d$ are satisfied.
Note: Computer algebra programs show that we have an irreducible polynomial in our hands. For instance using sage:
sage: R.<x> = PolynomialRing(GF(7))
sage: f = x^4 + x^3 + x^2 + x + 1
sage: f.is_irreducible()
True
One can use then computer power also as follows to conclude. (With a slightly bigger effort than for the typing, one can also conclude humanly.) Assume $f$ splits in two (or more factors). One can check there is no root in $Bbb F_7$. Then there is a root in $Bbb F_7^2=Bbb F_49$, so the polynomials $f$ and $x^49-x$ have a common divisor. The gcd is but... we type
sage: gcd( x^49 - x, f )
1
sage: gcd( x^(7^4) - x, f )
x^4 + x^3 + x^2 + x + 1
For this, one can also compute easily $(x^49-x,f)$ as a human by noting that $f$ divides $x^5-1$, so $(x^49-x,f)=(x^49-x^4+x^4-x,f)=(x^4-x,f)=(x^3-1,f)=(x^2+x+1,f)=(x^2+x+1,x+1)=(1,x+1)=1$.
answered Aug 6 at 21:32
dan_fulea
4,2171211
add a comment |Â
up vote
1
down vote
Yet an other solution. It is based on the fact that
$$ f = x^4+x^3+x^2+x+1inBbb F_7[x] $$
is reciprocal, and tries to use the idea in the OP. (There should be no splitting in two polynomials of degree two.)
Assume there is a factorization
$$
f=(x^2+ax+b)(x^2+cx+d) ,qquad a,b,c,dinBbb F_7 .$$
Let $une 0$ be a root of the first factor in some extension (of degree two). Then $1/u$ is also a root. We distinguish two cases.
First case: $1/u$ is also a root of the first factor. Then $b=1$, then $d=1$, $c+a=1$, the product
$$(x^2+ax+1)(x^2+(1-a)x+1)$$
is reciprocal,
and we have to look for a match of the coefficient in $x^2$, for a few values of $a$ (modulo the action $aleftrightarrow (1-a)$). There is no such $a$ with $1+a(1-a)+1=1$.
Second case: $1/u$ is not a root of the first factor. Let $u$, $v$ be the two roots of $x^2+ax+b$. Then $1/u$, $1/v$ are the roots of the second factor, which is the reciprocal polynomial, made monic, so we expect an equality of reciprocal polynomials
$$
(x^2+ax+b)(1+ax+bx^2) = b(x^4+x^3+x^2+x+1) .
$$
The two equations obtained, $ab+a=b$ and $1+a^2+b^2=b$, have no solution in $Bbb F_7$. (The substitution of $a=b/(b+1)$ in the second equation leads to... $(b^4+b^3+b^2+b+1)/(b+1)^2=0$.)
Note: This is arguably shorter then taking all possible polynomials $x^2+ax+b$, $bne 0$, associating $c=1-a$, $d=1/b$, and checking if the other conditions among $a,b,c,d$ are satisfied.
Note: Computer algebra programs show that we have an irreducible polynomial in our hands. For instance using sage:
sage: R.<x> = PolynomialRing(GF(7))
sage: f = x^4 + x^3 + x^2 + x + 1
sage: f.is_irreducible()
True
One can use then computer power also as follows to conclude. (With a slightly bigger effort than for the typing, one can also conclude humanly.) Assume $f$ splits in two (or more factors). One can check there is no root in $Bbb F_7$. Then there is a root in $Bbb F_7^2=Bbb F_49$, so the polynomials $f$ and $x^49-x$ have a common divisor. The gcd is but... we type
sage: gcd( x^49 - x, f )
1
sage: gcd( x^(7^4) - x, f )
x^4 + x^3 + x^2 + x + 1
For this, one can also compute easily $(x^49-x,f)$ as a human by noting that $f$ divides $x^5-1$, so $(x^49-x,f)=(x^49-x^4+x^4-x,f)=(x^4-x,f)=(x^3-1,f)=(x^2+x+1,f)=(x^2+x+1,x+1)=(1,x+1)=1$.
answered Aug 6 at 21:32
dan_fulea
4,2171211
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Yet an other solution. It is based on the fact that
$$ f = x^4+x^3+x^2+x+1inBbb F_7[x] $$
is reciprocal, and tries to use the idea in the OP. (There should be no splitting in two polynomials of degree two.)
Assume there is a factorization
$$
f=(x^2+ax+b)(x^2+cx+d) ,qquad a,b,c,dinBbb F_7 .$$
Let $une 0$ be a root of the first factor in some extension (of degree two). Then $1/u$ is also a root. We distinguish two cases.
First case: $1/u$ is also a root of the first factor. Then $b=1$, then $d=1$, $c+a=1$, the product
$$(x^2+ax+1)(x^2+(1-a)x+1)$$
is reciprocal,
and we have to look for a match of the coefficient in $x^2$, for a few values of $a$ (modulo the action $aleftrightarrow (1-a)$). There is no such $a$ with $1+a(1-a)+1=1$.
Second case: $1/u$ is not a root of the first factor. Let $u$, $v$ be the two roots of $x^2+ax+b$. Then $1/u$, $1/v$ are the roots of the second factor, which is the reciprocal polynomial, made monic, so we expect an equality of reciprocal polynomials
$$
(x^2+ax+b)(1+ax+bx^2) = b(x^4+x^3+x^2+x+1) .
$$
The two equations obtained, $ab+a=b$ and $1+a^2+b^2=b$, have no solution in $Bbb F_7$. (The substitution of $a=b/(b+1)$ in the second equation leads to... $(b^4+b^3+b^2+b+1)/(b+1)^2=0$.)
Note: This is arguably shorter then taking all possible polynomials $x^2+ax+b$, $bne 0$, associating $c=1-a$, $d=1/b$, and checking if the other conditions among $a,b,c,d$ are satisfied.
Note: Computer algebra programs show that we have an irreducible polynomial in our hands. For instance using sage:
sage: R.<x> = PolynomialRing(GF(7))
sage: f = x^4 + x^3 + x^2 + x + 1
sage: f.is_irreducible()
True
One can use then computer power also as follows to conclude. (With a slightly bigger effort than for the typing, one can also conclude humanly.) Assume $f$ splits in two (or more factors). One can check there is no root in $Bbb F_7$. Then there is a root in $Bbb F_7^2=Bbb F_49$, so the polynomials $f$ and $x^49-x$ have a common divisor. The gcd is but... we type
sage: gcd( x^49 - x, f )
1
sage: gcd( x^(7^4) - x, f )
x^4 + x^3 + x^2 + x + 1
For this, one can also compute easily $(x^49-x,f)$ as a human by noting that $f$ divides $x^5-1$, so $(x^49-x,f)=(x^49-x^4+x^4-x,f)=(x^4-x,f)=(x^3-1,f)=(x^2+x+1,f)=(x^2+x+1,x+1)=(1,x+1)=1$.
answered Aug 6 at 21:32
dan_fulea
4,2171211
Yet an other solution. It is based on the fact that
$$ f = x^4+x^3+x^2+x+1inBbb F_7[x] $$
is reciprocal, and tries to use the idea in the OP. (There should be no splitting in two polynomials of degree two.)
Assume there is a factorization
$$
f=(x^2+ax+b)(x^2+cx+d) ,qquad a,b,c,dinBbb F_7 .$$
Let $une 0$ be a root of the first factor in some extension (of degree two). Then $1/u$ is also a root. We distinguish two cases.
First case: $1/u$ is also a root of the first factor. Then $b=1$, then $d=1$, $c+a=1$, the product
$$(x^2+ax+1)(x^2+(1-a)x+1)$$
is reciprocal,
and we have to look for a match of the coefficient in $x^2$, for a few values of $a$ (modulo the action $aleftrightarrow (1-a)$). There is no such $a$ with $1+a(1-a)+1=1$.
Second case: $1/u$ is not a root of the first factor. Let $u$, $v$ be the two roots of $x^2+ax+b$. Then $1/u$, $1/v$ are the roots of the second factor, which is the reciprocal polynomial, made monic, so we expect an equality of reciprocal polynomials
$$
(x^2+ax+b)(1+ax+bx^2) = b(x^4+x^3+x^2+x+1) .
$$
The two equations obtained, $ab+a=b$ and $1+a^2+b^2=b$, have no solution in $Bbb F_7$. (The substitution of $a=b/(b+1)$ in the second equation leads to... $(b^4+b^3+b^2+b+1)/(b+1)^2=0$.)
Note: This is arguably shorter then taking all possible polynomials $x^2+ax+b$, $bne 0$, associating $c=1-a$, $d=1/b$, and checking if the other conditions among $a,b,c,d$ are satisfied.
Note: Computer algebra programs show that we have an irreducible polynomial in our hands. For instance using sage:
sage: R.<x> = PolynomialRing(GF(7))
sage: f = x^4 + x^3 + x^2 + x + 1
sage: f.is_irreducible()
True
One can use then computer power also as follows to conclude. (With a slightly bigger effort than for the typing, one can also conclude humanly.) Assume $f$ splits in two (or more factors). One can check there is no root in $Bbb F_7$. Then there is a root in $Bbb F_7^2=Bbb F_49$, so the polynomials $f$ and $x^49-x$ have a common divisor. The gcd is but... we type
sage: gcd( x^49 - x, f )
1
sage: gcd( x^(7^4) - x, f )
x^4 + x^3 + x^2 + x + 1
For this, one can also compute easily $(x^49-x,f)$ as a human by noting that $f$ divides $x^5-1$, so $(x^49-x,f)=(x^49-x^4+x^4-x,f)=(x^4-x,f)=(x^3-1,f)=(x^2+x+1,f)=(x^2+x+1,x+1)=(1,x+1)=1$.
answered Aug 6 at 21:32
dan_fulea
4,2171211
answered Aug 6 at 21:32
dan_fulea
4,2171211
answered Aug 6 at 21:32
dan_fulea
4,2171211
answered Aug 6 at 21:32
dan_fulea
4,2171211
4,2171211
add a comment |Â
add a comment |Â
up vote
0
down vote
HINT.-If $x^4+x^3+x^2+x+1$ were reducible then we have either a linear factor with a cubic one or two quadratic factors.
We have $$x^4+x^3+x^2+x+1=-dfrac 14(-2x^2+(sqrt5-1)x-2)(2x^2+(sqrt5+1)x+2)$$
Since $mathbb F_7^2=1,2,4,0$ we see that $5$ is not a square modulo $7$. It follows $x^4+x^3+x^2+x+1$ is not a factor of two quadratics modulo $7$.
Besides that a linear factor is not possible is checked easily.
answered Aug 6 at 21:33
Piquito
17.4k31234
Explain the reason of your downvote, please. If there exists a factorization in two quadratics, then it would have two distinct factorizations in two quadratics over an extension of $mathbb F_7$. This is not possible.
– Piquito
Aug 6 at 22:38
I don't know who downvoted this answer, I just upvoted it.
– user437309
Aug 6 at 23:05
@user437309: I did not say, dear friend, that it was you who put the downvote. It should be mandatory in S.E. that a downvote is always with justification exposed. That the person that put it refuses to respond shows its (mathematically speaking) ilk.
– Piquito
Aug 7 at 0:07
Hmm. Over $BbbC$ the roots of the polynomial $x^4+x^3+x^2+x+1$ are $zeta^k$ with $zeta=e^2pi i/5$, $k=1,2,3,4$. Your argument shows that the quadratic factors with real coefficients, $(x-zeta)(x-zeta^4)$ and $(x-zeta^2)(x-zeta^3)$, don't exist in $BbbF_7$ because they involve $sqrt5$. But how do you rule out a factor like $(x-zeta)(x-zeta^2)$ showing up? True, Galois theory shows that $BbbQ(sqrt5)$ is the only quadratic subfield of $BbbQ(zeta)$. That does lead to an argument, but the connection to Galois theory of finite fields runs a bit deep.
– Jyrki Lahtonen
Aug 7 at 5:34
(cont'd) In if you really want to use Galois theory here, a simpler way is to use the Frobenius automorphism $F$ of all extensions of $BbbF_7$: If $zeta$ is a root of some polynomial with coefficients in $BbbF_7$ then so is $F(zeta)=zeta^7=zeta^2$, and therefore also $F(zeta^2)=zeta^14=zeta^4$, and therefore also $F(zeta^4)=zeta^3$. That's four zeros, so any polynomial with $zeta$ as a root has degree four.
– Jyrki Lahtonen
Aug 7 at 5:37
 |Â
show 5 more comments
up vote
0
down vote
HINT.-If $x^4+x^3+x^2+x+1$ were reducible then we have either a linear factor with a cubic one or two quadratic factors.
We have $$x^4+x^3+x^2+x+1=-dfrac 14(-2x^2+(sqrt5-1)x-2)(2x^2+(sqrt5+1)x+2)$$
Since $mathbb F_7^2=1,2,4,0$ we see that $5$ is not a square modulo $7$. It follows $x^4+x^3+x^2+x+1$ is not a factor of two quadratics modulo $7$.
Besides that a linear factor is not possible is checked easily.
answered Aug 6 at 21:33
Piquito
17.4k31234
Explain the reason of your downvote, please. If there exists a factorization in two quadratics, then it would have two distinct factorizations in two quadratics over an extension of $mathbb F_7$. This is not possible.
– Piquito
Aug 6 at 22:38
I don't know who downvoted this answer, I just upvoted it.
– user437309
Aug 6 at 23:05
@user437309: I did not say, dear friend, that it was you who put the downvote. It should be mandatory in S.E. that a downvote is always with justification exposed. That the person that put it refuses to respond shows its (mathematically speaking) ilk.
– Piquito
Aug 7 at 0:07
Hmm. Over $BbbC$ the roots of the polynomial $x^4+x^3+x^2+x+1$ are $zeta^k$ with $zeta=e^2pi i/5$, $k=1,2,3,4$. Your argument shows that the quadratic factors with real coefficients, $(x-zeta)(x-zeta^4)$ and $(x-zeta^2)(x-zeta^3)$, don't exist in $BbbF_7$ because they involve $sqrt5$. But how do you rule out a factor like $(x-zeta)(x-zeta^2)$ showing up? True, Galois theory shows that $BbbQ(sqrt5)$ is the only quadratic subfield of $BbbQ(zeta)$. That does lead to an argument, but the connection to Galois theory of finite fields runs a bit deep.
– Jyrki Lahtonen
Aug 7 at 5:34
(cont'd) In if you really want to use Galois theory here, a simpler way is to use the Frobenius automorphism $F$ of all extensions of $BbbF_7$: If $zeta$ is a root of some polynomial with coefficients in $BbbF_7$ then so is $F(zeta)=zeta^7=zeta^2$, and therefore also $F(zeta^2)=zeta^14=zeta^4$, and therefore also $F(zeta^4)=zeta^3$. That's four zeros, so any polynomial with $zeta$ as a root has degree four.
– Jyrki Lahtonen
Aug 7 at 5:37
 |Â
show 5 more comments
up vote
0
down vote
up vote
0
down vote
HINT.-If $x^4+x^3+x^2+x+1$ were reducible then we have either a linear factor with a cubic one or two quadratic factors.
We have $$x^4+x^3+x^2+x+1=-dfrac 14(-2x^2+(sqrt5-1)x-2)(2x^2+(sqrt5+1)x+2)$$
Since $mathbb F_7^2=1,2,4,0$ we see that $5$ is not a square modulo $7$. It follows $x^4+x^3+x^2+x+1$ is not a factor of two quadratics modulo $7$.
Besides that a linear factor is not possible is checked easily.
answered Aug 6 at 21:33
Piquito
17.4k31234
HINT.-If $x^4+x^3+x^2+x+1$ were reducible then we have either a linear factor with a cubic one or two quadratic factors.
We have $$x^4+x^3+x^2+x+1=-dfrac 14(-2x^2+(sqrt5-1)x-2)(2x^2+(sqrt5+1)x+2)$$
Since $mathbb F_7^2=1,2,4,0$ we see that $5$ is not a square modulo $7$. It follows $x^4+x^3+x^2+x+1$ is not a factor of two quadratics modulo $7$.
Besides that a linear factor is not possible is checked easily.
answered Aug 6 at 21:33
Piquito
17.4k31234
edited Aug 7 at 1:16
answered Aug 6 at 21:33
Piquito
17.4k31234
answered Aug 6 at 21:33
Piquito
17.4k31234
answered Aug 6 at 21:33
Piquito
17.4k31234
17.4k31234
Explain the reason of your downvote, please. If there exists a factorization in two quadratics, then it would have two distinct factorizations in two quadratics over an extension of $mathbb F_7$. This is not possible.
– Piquito
Aug 6 at 22:38
I don't know who downvoted this answer, I just upvoted it.
– user437309
Aug 6 at 23:05
@user437309: I did not say, dear friend, that it was you who put the downvote. It should be mandatory in S.E. that a downvote is always with justification exposed. That the person that put it refuses to respond shows its (mathematically speaking) ilk.
– Piquito
Aug 7 at 0:07
Hmm. Over $BbbC$ the roots of the polynomial $x^4+x^3+x^2+x+1$ are $zeta^k$ with $zeta=e^2pi i/5$, $k=1,2,3,4$. Your argument shows that the quadratic factors with real coefficients, $(x-zeta)(x-zeta^4)$ and $(x-zeta^2)(x-zeta^3)$, don't exist in $BbbF_7$ because they involve $sqrt5$. But how do you rule out a factor like $(x-zeta)(x-zeta^2)$ showing up? True, Galois theory shows that $BbbQ(sqrt5)$ is the only quadratic subfield of $BbbQ(zeta)$. That does lead to an argument, but the connection to Galois theory of finite fields runs a bit deep.
– Jyrki Lahtonen
Aug 7 at 5:34
(cont'd) In if you really want to use Galois theory here, a simpler way is to use the Frobenius automorphism $F$ of all extensions of $BbbF_7$: If $zeta$ is a root of some polynomial with coefficients in $BbbF_7$ then so is $F(zeta)=zeta^7=zeta^2$, and therefore also $F(zeta^2)=zeta^14=zeta^4$, and therefore also $F(zeta^4)=zeta^3$. That's four zeros, so any polynomial with $zeta$ as a root has degree four.
– Jyrki Lahtonen
Aug 7 at 5:37
 |Â
show 5 more comments
Explain the reason of your downvote, please. If there exists a factorization in two quadratics, then it would have two distinct factorizations in two quadratics over an extension of $mathbb F_7$. This is not possible.
– Piquito
Aug 6 at 22:38
I don't know who downvoted this answer, I just upvoted it.
– user437309
Aug 6 at 23:05
@user437309: I did not say, dear friend, that it was you who put the downvote. It should be mandatory in S.E. that a downvote is always with justification exposed. That the person that put it refuses to respond shows its (mathematically speaking) ilk.
– Piquito
Aug 7 at 0:07
Hmm. Over $BbbC$ the roots of the polynomial $x^4+x^3+x^2+x+1$ are $zeta^k$ with $zeta=e^2pi i/5$, $k=1,2,3,4$. Your argument shows that the quadratic factors with real coefficients, $(x-zeta)(x-zeta^4)$ and $(x-zeta^2)(x-zeta^3)$, don't exist in $BbbF_7$ because they involve $sqrt5$. But how do you rule out a factor like $(x-zeta)(x-zeta^2)$ showing up? True, Galois theory shows that $BbbQ(sqrt5)$ is the only quadratic subfield of $BbbQ(zeta)$. That does lead to an argument, but the connection to Galois theory of finite fields runs a bit deep.
– Jyrki Lahtonen
Aug 7 at 5:34
(cont'd) In if you really want to use Galois theory here, a simpler way is to use the Frobenius automorphism $F$ of all extensions of $BbbF_7$: If $zeta$ is a root of some polynomial with coefficients in $BbbF_7$ then so is $F(zeta)=zeta^7=zeta^2$, and therefore also $F(zeta^2)=zeta^14=zeta^4$, and therefore also $F(zeta^4)=zeta^3$. That's four zeros, so any polynomial with $zeta$ as a root has degree four.
– Jyrki Lahtonen
Aug 7 at 5:37
Explain the reason of your downvote, please. If there exists a factorization in two quadratics, then it would have two distinct factorizations in two quadratics over an extension of $mathbb F_7$. This is not possible.
– Piquito
Aug 6 at 22:38
Explain the reason of your downvote, please. If there exists a factorization in two quadratics, then it would have two distinct factorizations in two quadratics over an extension of $mathbb F_7$. This is not possible.
– Piquito
Aug 6 at 22:38
I don't know who downvoted this answer, I just upvoted it.
– user437309
Aug 6 at 23:05
I don't know who downvoted this answer, I just upvoted it.
– user437309
Aug 6 at 23:05
@user437309: I did not say, dear friend, that it was you who put the downvote. It should be mandatory in S.E. that a downvote is always with justification exposed. That the person that put it refuses to respond shows its (mathematically speaking) ilk.
– Piquito
Aug 7 at 0:07
@user437309: I did not say, dear friend, that it was you who put the downvote. It should be mandatory in S.E. that a downvote is always with justification exposed. That the person that put it refuses to respond shows its (mathematically speaking) ilk.
– Piquito
Aug 7 at 0:07
Hmm. Over $BbbC$ the roots of the polynomial $x^4+x^3+x^2+x+1$ are $zeta^k$ with $zeta=e^2pi i/5$, $k=1,2,3,4$. Your argument shows that the quadratic factors with real coefficients, $(x-zeta)(x-zeta^4)$ and $(x-zeta^2)(x-zeta^3)$, don't exist in $BbbF_7$ because they involve $sqrt5$. But how do you rule out a factor like $(x-zeta)(x-zeta^2)$ showing up? True, Galois theory shows that $BbbQ(sqrt5)$ is the only quadratic subfield of $BbbQ(zeta)$. That does lead to an argument, but the connection to Galois theory of finite fields runs a bit deep.
– Jyrki Lahtonen
Aug 7 at 5:34
Hmm. Over $BbbC$ the roots of the polynomial $x^4+x^3+x^2+x+1$ are $zeta^k$ with $zeta=e^2pi i/5$, $k=1,2,3,4$. Your argument shows that the quadratic factors with real coefficients, $(x-zeta)(x-zeta^4)$ and $(x-zeta^2)(x-zeta^3)$, don't exist in $BbbF_7$ because they involve $sqrt5$. But how do you rule out a factor like $(x-zeta)(x-zeta^2)$ showing up? True, Galois theory shows that $BbbQ(sqrt5)$ is the only quadratic subfield of $BbbQ(zeta)$. That does lead to an argument, but the connection to Galois theory of finite fields runs a bit deep.
– Jyrki Lahtonen
Aug 7 at 5:34
(cont'd) In if you really want to use Galois theory here, a simpler way is to use the Frobenius automorphism $F$ of all extensions of $BbbF_7$: If $zeta$ is a root of some polynomial with coefficients in $BbbF_7$ then so is $F(zeta)=zeta^7=zeta^2$, and therefore also $F(zeta^2)=zeta^14=zeta^4$, and therefore also $F(zeta^4)=zeta^3$. That's four zeros, so any polynomial with $zeta$ as a root has degree four.
– Jyrki Lahtonen
Aug 7 at 5:37
(cont'd) In if you really want to use Galois theory here, a simpler way is to use the Frobenius automorphism $F$ of all extensions of $BbbF_7$: If $zeta$ is a root of some polynomial with coefficients in $BbbF_7$ then so is $F(zeta)=zeta^7=zeta^2$, and therefore also $F(zeta^2)=zeta^14=zeta^4$, and therefore also $F(zeta^4)=zeta^3$. That's four zeros, so any polynomial with $zeta$ as a root has degree four.
– Jyrki Lahtonen
Aug 7 at 5:37
 |Â
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Worst case scenario, check the different choices for $b$ and $a$. $c$ and $d$ are dictated at that point. I'd like to see a slicker way.
– Randall
Aug 6 at 19:38
this question may help.
– lulu
Aug 6 at 19:43