Quotient $mathbf{F}_3[X]/(X^5+1)$
$begingroup$
Factor $X^5+1inmathbf{F}_3[X]$ into irreducibles. What does the quotient $mathbf{F}_3[X]/(X^5+1)$ look like?
Since $-1$ is a zero, we divide $X^5+1$ by $X+1$ using long division, to obtain $X^5+1=(X+1)(X^4-X^3+X^2-X+1)$. Now we claim that $X^4-X^3+X^2-X+1$ is irreducible. It has no zeros in $mathbf{F}_3$, and we can check that it can not be written as product of quadratic factors.
I am hesitating about the following part:
$(X^5+1)=(X+1,X^4-X^3+X^2-X+1)$.
By long division, we have $X^4-X^3+X^2-X+1=(X+1)(X^3-2X^2+3X-4)+5$, therefore $(X^5+1)=(X+1,5)$. We have $$frac{mathbf{F}_3[X]}{(X^5+1)}congfrac{mathbf{F}_3[X]}{(X+1,2)}congfrac{mathbf{Z}[X]/(3)}{(X+1,2)}cong frac{mathbf{Z}[X]/(X+1)}{(3,2)}congmathbf{Z}.$$
But $mathbf{Z}$ is not a field, and $frac{mathbf{F}_3[X]}{(X^5+1)}$ is I think. Could someone help me?
abstract-algebra polynomials ring-theory finite-fields irreducible-polynomials
edited Dec 14 '18 at 21:44
Batominovski
33.1k33293
asked Dec 14 '18 at 21:12
Heinz DoofenschmirtzHeinz Doofenschmirtz
615620
$endgroup$
add a comment |
$begingroup$
Factor $X^5+1inmathbf{F}_3[X]$ into irreducibles. What does the quotient $mathbf{F}_3[X]/(X^5+1)$ look like?
Since $-1$ is a zero, we divide $X^5+1$ by $X+1$ using long division, to obtain $X^5+1=(X+1)(X^4-X^3+X^2-X+1)$. Now we claim that $X^4-X^3+X^2-X+1$ is irreducible. It has no zeros in $mathbf{F}_3$, and we can check that it can not be written as product of quadratic factors.
I am hesitating about the following part:
$(X^5+1)=(X+1,X^4-X^3+X^2-X+1)$.
By long division, we have $X^4-X^3+X^2-X+1=(X+1)(X^3-2X^2+3X-4)+5$, therefore $(X^5+1)=(X+1,5)$. We have $$frac{mathbf{F}_3[X]}{(X^5+1)}congfrac{mathbf{F}_3[X]}{(X+1,2)}congfrac{mathbf{Z}[X]/(3)}{(X+1,2)}cong frac{mathbf{Z}[X]/(X+1)}{(3,2)}congmathbf{Z}.$$
But $mathbf{Z}$ is not a field, and $frac{mathbf{F}_3[X]}{(X^5+1)}$ is I think. Could someone help me?
abstract-algebra polynomials ring-theory finite-fields irreducible-polynomials
edited Dec 14 '18 at 21:44
Batominovski
33.1k33293
asked Dec 14 '18 at 21:12
Heinz DoofenschmirtzHeinz Doofenschmirtz
615620
$endgroup$
2
$begingroup$
Note $,5 = -1 in (X^5+1),Rightarrow,X^5+1$ is a unit (invertible) so that can't be true. The problem is the first equality after your "hesitating". Why do you believe that?
$endgroup$
– Bill Dubuque
Dec 14 '18 at 21:26
add a comment |
$begingroup$
Factor $X^5+1inmathbf{F}_3[X]$ into irreducibles. What does the quotient $mathbf{F}_3[X]/(X^5+1)$ look like?
Since $-1$ is a zero, we divide $X^5+1$ by $X+1$ using long division, to obtain $X^5+1=(X+1)(X^4-X^3+X^2-X+1)$. Now we claim that $X^4-X^3+X^2-X+1$ is irreducible. It has no zeros in $mathbf{F}_3$, and we can check that it can not be written as product of quadratic factors.
I am hesitating about the following part:
$(X^5+1)=(X+1,X^4-X^3+X^2-X+1)$.
By long division, we have $X^4-X^3+X^2-X+1=(X+1)(X^3-2X^2+3X-4)+5$, therefore $(X^5+1)=(X+1,5)$. We have $$frac{mathbf{F}_3[X]}{(X^5+1)}congfrac{mathbf{F}_3[X]}{(X+1,2)}congfrac{mathbf{Z}[X]/(3)}{(X+1,2)}cong frac{mathbf{Z}[X]/(X+1)}{(3,2)}congmathbf{Z}.$$
But $mathbf{Z}$ is not a field, and $frac{mathbf{F}_3[X]}{(X^5+1)}$ is I think. Could someone help me?
abstract-algebra polynomials ring-theory finite-fields irreducible-polynomials
edited Dec 14 '18 at 21:44
Batominovski
33.1k33293
asked Dec 14 '18 at 21:12
Heinz DoofenschmirtzHeinz Doofenschmirtz
615620
$endgroup$
Factor $X^5+1inmathbf{F}_3[X]$ into irreducibles. What does the quotient $mathbf{F}_3[X]/(X^5+1)$ look like?
Since $-1$ is a zero, we divide $X^5+1$ by $X+1$ using long division, to obtain $X^5+1=(X+1)(X^4-X^3+X^2-X+1)$. Now we claim that $X^4-X^3+X^2-X+1$ is irreducible. It has no zeros in $mathbf{F}_3$, and we can check that it can not be written as product of quadratic factors.
I am hesitating about the following part:
$(X^5+1)=(X+1,X^4-X^3+X^2-X+1)$.
By long division, we have $X^4-X^3+X^2-X+1=(X+1)(X^3-2X^2+3X-4)+5$, therefore $(X^5+1)=(X+1,5)$. We have $$frac{mathbf{F}_3[X]}{(X^5+1)}congfrac{mathbf{F}_3[X]}{(X+1,2)}congfrac{mathbf{Z}[X]/(3)}{(X+1,2)}cong frac{mathbf{Z}[X]/(X+1)}{(3,2)}congmathbf{Z}.$$
But $mathbf{Z}$ is not a field, and $frac{mathbf{F}_3[X]}{(X^5+1)}$ is I think. Could someone help me?
abstract-algebra polynomials ring-theory finite-fields irreducible-polynomials
abstract-algebra polynomials ring-theory finite-fields irreducible-polynomials
edited Dec 14 '18 at 21:44
Batominovski
33.1k33293
asked Dec 14 '18 at 21:12
Heinz DoofenschmirtzHeinz Doofenschmirtz
615620
edited Dec 14 '18 at 21:44
Batominovski
33.1k33293
asked Dec 14 '18 at 21:12
Heinz DoofenschmirtzHeinz Doofenschmirtz
615620
edited Dec 14 '18 at 21:44
Batominovski
33.1k33293
edited Dec 14 '18 at 21:44
Batominovski
33.1k33293
edited Dec 14 '18 at 21:44
Batominovski
33.1k33293
33.1k33293
asked Dec 14 '18 at 21:12
Heinz DoofenschmirtzHeinz Doofenschmirtz
615620
asked Dec 14 '18 at 21:12
Heinz DoofenschmirtzHeinz Doofenschmirtz
615620
asked Dec 14 '18 at 21:12
Heinz DoofenschmirtzHeinz Doofenschmirtz
615620
615620
2
$begingroup$
Note $,5 = -1 in (X^5+1),Rightarrow,X^5+1$ is a unit (invertible) so that can't be true. The problem is the first equality after your "hesitating". Why do you believe that?
$endgroup$
– Bill Dubuque
Dec 14 '18 at 21:26
add a comment |
2
$begingroup$
Note $,5 = -1 in (X^5+1),Rightarrow,X^5+1$ is a unit (invertible) so that can't be true. The problem is the first equality after your "hesitating". Why do you believe that?
$endgroup$
– Bill Dubuque
Dec 14 '18 at 21:26
2
2
$begingroup$
Note $,5 = -1 in (X^5+1),Rightarrow,X^5+1$ is a unit (invertible) so that can't be true. The problem is the first equality after your "hesitating". Why do you believe that?
$endgroup$
– Bill Dubuque
Dec 14 '18 at 21:26
$begingroup$
Note $,5 = -1 in (X^5+1),Rightarrow,X^5+1$ is a unit (invertible) so that can't be true. The problem is the first equality after your "hesitating". Why do you believe that?
$endgroup$
– Bill Dubuque
Dec 14 '18 at 21:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think the mistake you made is that the ideal $(X^5 + 1) neq (X+1, X^4 - X^3 + X^2 - X + 1)$. Similarly $(X^5 + 1) neq (X + 1,5)$. Long division only gives us "$subset$" in both cases, but not the other direction.
What you want to do is factor $X^5 + 1 = (X + 1) cdot (X^4 - X^3 + X^2 - X + 1)$. Note the ideal generated by $X^5 + 1$ is also the product of the ideal generated by the two factors, but not the sum.
If you want to compute the quotient $mathbb{F}[X] / (X^5 + 1)$, you can now apply the Chinese remainder theorem, that tells us
$$mathbb{F}[X] / (X^5 + 1) cong mathbb{F}[X] / (X+1) times mathbb{F}[X]/(X^4 - X^3+X^2-X+1).$$
Also note that this is not a field, because $X^5+1$ is not irreducible. This is because $X+1$ and $X^4 - X^3 + X^2 - X + 1$ are not in the ideal, i.e. they are not $0$ in the quotient, but if multiplied they give $0$, so they are zero divisors of the quotient.
answered Dec 14 '18 at 22:09
red_trumpetred_trumpet
991319
$endgroup$
add a comment |
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$begingroup$
I think the mistake you made is that the ideal $(X^5 + 1) neq (X+1, X^4 - X^3 + X^2 - X + 1)$. Similarly $(X^5 + 1) neq (X + 1,5)$. Long division only gives us "$subset$" in both cases, but not the other direction.
What you want to do is factor $X^5 + 1 = (X + 1) cdot (X^4 - X^3 + X^2 - X + 1)$. Note the ideal generated by $X^5 + 1$ is also the product of the ideal generated by the two factors, but not the sum.
If you want to compute the quotient $mathbb{F}[X] / (X^5 + 1)$, you can now apply the Chinese remainder theorem, that tells us
$$mathbb{F}[X] / (X^5 + 1) cong mathbb{F}[X] / (X+1) times mathbb{F}[X]/(X^4 - X^3+X^2-X+1).$$
Also note that this is not a field, because $X^5+1$ is not irreducible. This is because $X+1$ and $X^4 - X^3 + X^2 - X + 1$ are not in the ideal, i.e. they are not $0$ in the quotient, but if multiplied they give $0$, so they are zero divisors of the quotient.
answered Dec 14 '18 at 22:09
red_trumpetred_trumpet
991319
$endgroup$
add a comment |
$begingroup$
I think the mistake you made is that the ideal $(X^5 + 1) neq (X+1, X^4 - X^3 + X^2 - X + 1)$. Similarly $(X^5 + 1) neq (X + 1,5)$. Long division only gives us "$subset$" in both cases, but not the other direction.
What you want to do is factor $X^5 + 1 = (X + 1) cdot (X^4 - X^3 + X^2 - X + 1)$. Note the ideal generated by $X^5 + 1$ is also the product of the ideal generated by the two factors, but not the sum.
If you want to compute the quotient $mathbb{F}[X] / (X^5 + 1)$, you can now apply the Chinese remainder theorem, that tells us
$$mathbb{F}[X] / (X^5 + 1) cong mathbb{F}[X] / (X+1) times mathbb{F}[X]/(X^4 - X^3+X^2-X+1).$$
Also note that this is not a field, because $X^5+1$ is not irreducible. This is because $X+1$ and $X^4 - X^3 + X^2 - X + 1$ are not in the ideal, i.e. they are not $0$ in the quotient, but if multiplied they give $0$, so they are zero divisors of the quotient.
answered Dec 14 '18 at 22:09
red_trumpetred_trumpet
991319
$endgroup$
add a comment |
$begingroup$
I think the mistake you made is that the ideal $(X^5 + 1) neq (X+1, X^4 - X^3 + X^2 - X + 1)$. Similarly $(X^5 + 1) neq (X + 1,5)$. Long division only gives us "$subset$" in both cases, but not the other direction.
What you want to do is factor $X^5 + 1 = (X + 1) cdot (X^4 - X^3 + X^2 - X + 1)$. Note the ideal generated by $X^5 + 1$ is also the product of the ideal generated by the two factors, but not the sum.
If you want to compute the quotient $mathbb{F}[X] / (X^5 + 1)$, you can now apply the Chinese remainder theorem, that tells us
$$mathbb{F}[X] / (X^5 + 1) cong mathbb{F}[X] / (X+1) times mathbb{F}[X]/(X^4 - X^3+X^2-X+1).$$
Also note that this is not a field, because $X^5+1$ is not irreducible. This is because $X+1$ and $X^4 - X^3 + X^2 - X + 1$ are not in the ideal, i.e. they are not $0$ in the quotient, but if multiplied they give $0$, so they are zero divisors of the quotient.
answered Dec 14 '18 at 22:09
red_trumpetred_trumpet
991319
$endgroup$
I think the mistake you made is that the ideal $(X^5 + 1) neq (X+1, X^4 - X^3 + X^2 - X + 1)$. Similarly $(X^5 + 1) neq (X + 1,5)$. Long division only gives us "$subset$" in both cases, but not the other direction.
What you want to do is factor $X^5 + 1 = (X + 1) cdot (X^4 - X^3 + X^2 - X + 1)$. Note the ideal generated by $X^5 + 1$ is also the product of the ideal generated by the two factors, but not the sum.
If you want to compute the quotient $mathbb{F}[X] / (X^5 + 1)$, you can now apply the Chinese remainder theorem, that tells us
$$mathbb{F}[X] / (X^5 + 1) cong mathbb{F}[X] / (X+1) times mathbb{F}[X]/(X^4 - X^3+X^2-X+1).$$
Also note that this is not a field, because $X^5+1$ is not irreducible. This is because $X+1$ and $X^4 - X^3 + X^2 - X + 1$ are not in the ideal, i.e. they are not $0$ in the quotient, but if multiplied they give $0$, so they are zero divisors of the quotient.
answered Dec 14 '18 at 22:09
red_trumpetred_trumpet
991319
edited Dec 14 '18 at 22:17
answered Dec 14 '18 at 22:09
red_trumpetred_trumpet
991319
answered Dec 14 '18 at 22:09
red_trumpetred_trumpet
991319
answered Dec 14 '18 at 22:09
red_trumpetred_trumpet
991319
991319
add a comment |
add a comment |
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$begingroup$
Note $,5 = -1 in (X^5+1),Rightarrow,X^5+1$ is a unit (invertible) so that can't be true. The problem is the first equality after your "hesitating". Why do you believe that?
$endgroup$
– Bill Dubuque
Dec 14 '18 at 21:26