Mixing
Does there exist an element of order 4 in a finite ring with characteristic 2?
$begingroup$
It is obvious that every nonzero element in finite field $mathbb{F}_{2^n}$ has an odd order. So I was wondering if the following conjecture was true or false.
Notations: Let $(R,+,times)$ be a finite ring with an identity. The characteristic of ring $R$ is denoted by $text{char}(R)$. The multiplicative order of an invertible element $e$ is denoted by $text{ord}(e)$. The set of all invertible elements is denoted by $R^times$.
If $text{char}(R)=2$, then $text{ord}(e)equiv1pmod{2}$ for any $ein R^times$.
For a weaker conjecture, we have
If $text{char}(R)=2$, then $text{ord}(e)notequiv 0pmod{4}$ for any $ein R^times$.
For $R=mathbb{F}_{2^n}$, the conjecture is true. However I think characteristic and multiplicative order are two independent matters. Because I know few rings $R$ with $text{char}(R)=2$. Could anyone prove or disprove any of the two conjectures?
PS: the problem is araised from Here, in which I need to find out all the elements of order $4$.
=======================
Thanks to user26857 and Jyrki Lahtonen. The two conjectures are false.
For any $p$ and $d$, consider the ring $R=mathbb{F}_p[x]/(x^d-1)$ and the coset $e=x in R$. Then the multiplicative order of $e$ is $d$.
- Letting $p=2$ and $d=4$, it disproves the first conjectrure.
- Letting $p=2$ and $d=2$, it disproves the second conjectrure.
abstract-algebra ring-theory finite-groups
asked Jan 1 at 7:45
zongxiang yizongxiang yi
33019
$endgroup$
add a comment |
$begingroup$
It is obvious that every nonzero element in finite field $mathbb{F}_{2^n}$ has an odd order. So I was wondering if the following conjecture was true or false.
Notations: Let $(R,+,times)$ be a finite ring with an identity. The characteristic of ring $R$ is denoted by $text{char}(R)$. The multiplicative order of an invertible element $e$ is denoted by $text{ord}(e)$. The set of all invertible elements is denoted by $R^times$.
If $text{char}(R)=2$, then $text{ord}(e)equiv1pmod{2}$ for any $ein R^times$.
For a weaker conjecture, we have
If $text{char}(R)=2$, then $text{ord}(e)notequiv 0pmod{4}$ for any $ein R^times$.
For $R=mathbb{F}_{2^n}$, the conjecture is true. However I think characteristic and multiplicative order are two independent matters. Because I know few rings $R$ with $text{char}(R)=2$. Could anyone prove or disprove any of the two conjectures?
PS: the problem is araised from Here, in which I need to find out all the elements of order $4$.
=======================
Thanks to user26857 and Jyrki Lahtonen. The two conjectures are false.
For any $p$ and $d$, consider the ring $R=mathbb{F}_p[x]/(x^d-1)$ and the coset $e=x in R$. Then the multiplicative order of $e$ is $d$.
- Letting $p=2$ and $d=4$, it disproves the first conjectrure.
- Letting $p=2$ and $d=2$, it disproves the second conjectrure.
abstract-algebra ring-theory finite-groups
asked Jan 1 at 7:45
zongxiang yizongxiang yi
33019
$endgroup$
10
$begingroup$
What about $R=mathbb F_2[X]/(X^4-1)$ and $e$ the residue class of $X$? (If this does not work then maybe I misunderstood your question.)
$endgroup$
– user26857
Jan 1 at 8:50
5
$begingroup$
user26857's example can be rewritten to say the same about the coset of $1+x$ in $Bbb{F}_2[x]/langle x^4rangle$ (this example is an isomorphic image of theirs).
$endgroup$
– Jyrki Lahtonen
Jan 1 at 9:08
1
$begingroup$
@user26857 What a clever construction. I got it. Thanks. The construction may be generalized to obtain elements of any order in a ring with any characteristic.
$endgroup$
– zongxiang yi
Jan 2 at 3:30
add a comment |
$begingroup$
It is obvious that every nonzero element in finite field $mathbb{F}_{2^n}$ has an odd order. So I was wondering if the following conjecture was true or false.
Notations: Let $(R,+,times)$ be a finite ring with an identity. The characteristic of ring $R$ is denoted by $text{char}(R)$. The multiplicative order of an invertible element $e$ is denoted by $text{ord}(e)$. The set of all invertible elements is denoted by $R^times$.
If $text{char}(R)=2$, then $text{ord}(e)equiv1pmod{2}$ for any $ein R^times$.
For a weaker conjecture, we have
If $text{char}(R)=2$, then $text{ord}(e)notequiv 0pmod{4}$ for any $ein R^times$.
For $R=mathbb{F}_{2^n}$, the conjecture is true. However I think characteristic and multiplicative order are two independent matters. Because I know few rings $R$ with $text{char}(R)=2$. Could anyone prove or disprove any of the two conjectures?
PS: the problem is araised from Here, in which I need to find out all the elements of order $4$.
=======================
Thanks to user26857 and Jyrki Lahtonen. The two conjectures are false.
For any $p$ and $d$, consider the ring $R=mathbb{F}_p[x]/(x^d-1)$ and the coset $e=x in R$. Then the multiplicative order of $e$ is $d$.
- Letting $p=2$ and $d=4$, it disproves the first conjectrure.
- Letting $p=2$ and $d=2$, it disproves the second conjectrure.
abstract-algebra ring-theory finite-groups
asked Jan 1 at 7:45
zongxiang yizongxiang yi
33019
$endgroup$
It is obvious that every nonzero element in finite field $mathbb{F}_{2^n}$ has an odd order. So I was wondering if the following conjecture was true or false.
Notations: Let $(R,+,times)$ be a finite ring with an identity. The characteristic of ring $R$ is denoted by $text{char}(R)$. The multiplicative order of an invertible element $e$ is denoted by $text{ord}(e)$. The set of all invertible elements is denoted by $R^times$.
If $text{char}(R)=2$, then $text{ord}(e)equiv1pmod{2}$ for any $ein R^times$.
For a weaker conjecture, we have
If $text{char}(R)=2$, then $text{ord}(e)notequiv 0pmod{4}$ for any $ein R^times$.
For $R=mathbb{F}_{2^n}$, the conjecture is true. However I think characteristic and multiplicative order are two independent matters. Because I know few rings $R$ with $text{char}(R)=2$. Could anyone prove or disprove any of the two conjectures?
PS: the problem is araised from Here, in which I need to find out all the elements of order $4$.
=======================
Thanks to user26857 and Jyrki Lahtonen. The two conjectures are false.
For any $p$ and $d$, consider the ring $R=mathbb{F}_p[x]/(x^d-1)$ and the coset $e=x in R$. Then the multiplicative order of $e$ is $d$.
- Letting $p=2$ and $d=4$, it disproves the first conjectrure.
- Letting $p=2$ and $d=2$, it disproves the second conjectrure.
abstract-algebra ring-theory finite-groups
abstract-algebra ring-theory finite-groups
asked Jan 1 at 7:45
zongxiang yizongxiang yi
33019
asked Jan 1 at 7:45
zongxiang yizongxiang yi
33019
edited Jan 2 at 3:39
zongxiang yi
asked Jan 1 at 7:45
zongxiang yizongxiang yi
33019
asked Jan 1 at 7:45
zongxiang yizongxiang yi
33019
asked Jan 1 at 7:45
zongxiang yizongxiang yi
33019
33019
10
$begingroup$
What about $R=mathbb F_2[X]/(X^4-1)$ and $e$ the residue class of $X$? (If this does not work then maybe I misunderstood your question.)
$endgroup$
– user26857
Jan 1 at 8:50
5
$begingroup$
user26857's example can be rewritten to say the same about the coset of $1+x$ in $Bbb{F}_2[x]/langle x^4rangle$ (this example is an isomorphic image of theirs).
$endgroup$
– Jyrki Lahtonen
Jan 1 at 9:08
1
$begingroup$
@user26857 What a clever construction. I got it. Thanks. The construction may be generalized to obtain elements of any order in a ring with any characteristic.
$endgroup$
– zongxiang yi
Jan 2 at 3:30
add a comment |
10
$begingroup$
What about $R=mathbb F_2[X]/(X^4-1)$ and $e$ the residue class of $X$? (If this does not work then maybe I misunderstood your question.)
$endgroup$
– user26857
Jan 1 at 8:50
5
$begingroup$
user26857's example can be rewritten to say the same about the coset of $1+x$ in $Bbb{F}_2[x]/langle x^4rangle$ (this example is an isomorphic image of theirs).
$endgroup$
– Jyrki Lahtonen
Jan 1 at 9:08
1
$begingroup$
@user26857 What a clever construction. I got it. Thanks. The construction may be generalized to obtain elements of any order in a ring with any characteristic.
$endgroup$
– zongxiang yi
Jan 2 at 3:30
10
10
$begingroup$
What about $R=mathbb F_2[X]/(X^4-1)$ and $e$ the residue class of $X$? (If this does not work then maybe I misunderstood your question.)
$endgroup$
– user26857
Jan 1 at 8:50
$begingroup$
What about $R=mathbb F_2[X]/(X^4-1)$ and $e$ the residue class of $X$? (If this does not work then maybe I misunderstood your question.)
$endgroup$
– user26857
Jan 1 at 8:50
5
5
$begingroup$
user26857's example can be rewritten to say the same about the coset of $1+x$ in $Bbb{F}_2[x]/langle x^4rangle$ (this example is an isomorphic image of theirs).
$endgroup$
– Jyrki Lahtonen
Jan 1 at 9:08
$begingroup$
user26857's example can be rewritten to say the same about the coset of $1+x$ in $Bbb{F}_2[x]/langle x^4rangle$ (this example is an isomorphic image of theirs).
$endgroup$
– Jyrki Lahtonen
Jan 1 at 9:08
1
1
$begingroup$
@user26857 What a clever construction. I got it. Thanks. The construction may be generalized to obtain elements of any order in a ring with any characteristic.
$endgroup$
– zongxiang yi
Jan 2 at 3:30
$begingroup$
@user26857 What a clever construction. I got it. Thanks. The construction may be generalized to obtain elements of any order in a ring with any characteristic.
$endgroup$
– zongxiang yi
Jan 2 at 3:30
add a comment |
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$begingroup$
What about $R=mathbb F_2[X]/(X^4-1)$ and $e$ the residue class of $X$? (If this does not work then maybe I misunderstood your question.)
$endgroup$
– user26857
Jan 1 at 8:50
5
$begingroup$
user26857's example can be rewritten to say the same about the coset of $1+x$ in $Bbb{F}_2[x]/langle x^4rangle$ (this example is an isomorphic image of theirs).
$endgroup$
– Jyrki Lahtonen
Jan 1 at 9:08
1
$begingroup$
@user26857 What a clever construction. I got it. Thanks. The construction may be generalized to obtain elements of any order in a ring with any characteristic.
$endgroup$
– zongxiang yi
Jan 2 at 3:30