What are the Characteristics of the Ring $mathbf F_p[X]/g$ (mod $g$, $g$ is non-irreducible)?

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$f(x)$ is a primitive polynomial of degree $m$ over a field $mathbf F_p^m$. As such, $mathbf F_p^m[x]/f(x)$ forms a ring, with addition and multiplication ($mkern-4mubmod f(x)$), and furthermore is a field since $f(x)$ is irreducible. Call it $F$

Now, replace $f(x)$ with $g(x)$, a non-primitive, non-irreducible polynomial of degree $m$ over $mathbf F_p^m$ and where $g(0) = 1$. Call it $G$. I am having trouble finding information on line or referred to about the characteristics of such a $g$ or the algebraic structure of $G$ (a ring?). This is probably because I don't know the right terminology and keywords regarding these to search by. Fields and Rings are new to me.

Some questions I have are (assume $m > 2$):

1. There are a finite number of elements in $G$ (because it is modulo $g$), but it doesn't form a field. What is it called? A "finite-ring"?

EDIT removed prior questions about cycle.

2. For any non-zero $alpha in F$, $exists n > 0$ such that $alpha^n = 1$. This is not true for all $alpha in G$. Can we easily determine the ones that do have such an $n$?


3. For any non-zero $alpha in F$, the powers of $alpha$ form a closed group under multiplication. Again, not true $G$. In $G exists alpha$ whose powers never equal 1, so they cannot form a group (no multiplicative identity). And when $g(x)$ is allowed to have factors with multiplicity > 1, $exists alpha in G$ such that $alpha^k = 0, k ge |alpha|$. However, there $exists alpha in G$ that do have powers that form cyclic multiplicative groups. Can we easily determine which ones they are?

Can you point me to a link or a paper (or even just the proper terms to use for searching) that discusses more about this non-irreducible case?

polynomials finite-fields

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edited Aug 6 at 18:59

asked Aug 1 at 21:32

Les

1214

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Why should there be a finite number of elements in the ring?
  – Bernard
  Aug 1 at 21:37
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What do you mean by a "cycle"?
  – Robert Lewis
  Aug 1 at 22:02
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  But $mathbf F[x]/(g)$ isn't finite, unless $mathbf F$ itself is: $mathbf R[x]/(x^2+1)simeq mathbf C$ is the simplest counter-example.
  – Bernard
  Aug 1 at 23:06
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  By the Chinese remainder theorem, it is an artinian $F$-algebra. If $g$ has no multiple irreducible factors (.i.e; if the algebra is a reduced ring), it is a product of field extensions of $F$. For the rest, I can't tell: I don't know what you call a ‘cycle’ here.
  – Bernard
  Aug 1 at 23:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Bernard has put his finger on the solution to the problem. When $g$ has an irreducible factor with multiplicity $>1$, there will be nilpotents in the resulting ring. Try $mathbf F[x]/((x-a)^m)$, for instance, the simplest case of this.
  – Lubin
  Aug 2 at 0:28
  

  
  
  
  

 | 
show 3 more comments

up vote
4
down vote

favorite

$f(x)$ is a primitive polynomial of degree $m$ over a field $mathbf F_p^m$. As such, $mathbf F_p^m[x]/f(x)$ forms a ring, with addition and multiplication ($mkern-4mubmod f(x)$), and furthermore is a field since $f(x)$ is irreducible. Call it $F$

Now, replace $f(x)$ with $g(x)$, a non-primitive, non-irreducible polynomial of degree $m$ over $mathbf F_p^m$ and where $g(0) = 1$. Call it $G$. I am having trouble finding information on line or referred to about the characteristics of such a $g$ or the algebraic structure of $G$ (a ring?). This is probably because I don't know the right terminology and keywords regarding these to search by. Fields and Rings are new to me.

Some questions I have are (assume $m > 2$):

1. There are a finite number of elements in $G$ (because it is modulo $g$), but it doesn't form a field. What is it called? A "finite-ring"?

EDIT removed prior questions about cycle.

2. For any non-zero $alpha in F$, $exists n > 0$ such that $alpha^n = 1$. This is not true for all $alpha in G$. Can we easily determine the ones that do have such an $n$?


3. For any non-zero $alpha in F$, the powers of $alpha$ form a closed group under multiplication. Again, not true $G$. In $G exists alpha$ whose powers never equal 1, so they cannot form a group (no multiplicative identity). And when $g(x)$ is allowed to have factors with multiplicity > 1, $exists alpha in G$ such that $alpha^k = 0, k ge |alpha|$. However, there $exists alpha in G$ that do have powers that form cyclic multiplicative groups. Can we easily determine which ones they are?

Can you point me to a link or a paper (or even just the proper terms to use for searching) that discusses more about this non-irreducible case?

polynomials finite-fields

share|cite|improve this question

edited Aug 6 at 18:59

asked Aug 1 at 21:32

Les

1214

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Why should there be a finite number of elements in the ring?
  – Bernard
  Aug 1 at 21:37
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What do you mean by a "cycle"?
  – Robert Lewis
  Aug 1 at 22:02
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  But $mathbf F[x]/(g)$ isn't finite, unless $mathbf F$ itself is: $mathbf R[x]/(x^2+1)simeq mathbf C$ is the simplest counter-example.
  – Bernard
  Aug 1 at 23:06
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  By the Chinese remainder theorem, it is an artinian $F$-algebra. If $g$ has no multiple irreducible factors (.i.e; if the algebra is a reduced ring), it is a product of field extensions of $F$. For the rest, I can't tell: I don't know what you call a ‘cycle’ here.
  – Bernard
  Aug 1 at 23:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Bernard has put his finger on the solution to the problem. When $g$ has an irreducible factor with multiplicity $>1$, there will be nilpotents in the resulting ring. Try $mathbf F[x]/((x-a)^m)$, for instance, the simplest case of this.
  – Lubin
  Aug 2 at 0:28
  

  
  
  
  

 | 
show 3 more comments

up vote
4
down vote

favorite

up vote
4
down vote

favorite

$f(x)$ is a primitive polynomial of degree $m$ over a field $mathbf F_p^m$. As such, $mathbf F_p^m[x]/f(x)$ forms a ring, with addition and multiplication ($mkern-4mubmod f(x)$), and furthermore is a field since $f(x)$ is irreducible. Call it $F$

Now, replace $f(x)$ with $g(x)$, a non-primitive, non-irreducible polynomial of degree $m$ over $mathbf F_p^m$ and where $g(0) = 1$. Call it $G$. I am having trouble finding information on line or referred to about the characteristics of such a $g$ or the algebraic structure of $G$ (a ring?). This is probably because I don't know the right terminology and keywords regarding these to search by. Fields and Rings are new to me.

Some questions I have are (assume $m > 2$):

1. There are a finite number of elements in $G$ (because it is modulo $g$), but it doesn't form a field. What is it called? A "finite-ring"?

EDIT removed prior questions about cycle.

2. For any non-zero $alpha in F$, $exists n > 0$ such that $alpha^n = 1$. This is not true for all $alpha in G$. Can we easily determine the ones that do have such an $n$?


3. For any non-zero $alpha in F$, the powers of $alpha$ form a closed group under multiplication. Again, not true $G$. In $G exists alpha$ whose powers never equal 1, so they cannot form a group (no multiplicative identity). And when $g(x)$ is allowed to have factors with multiplicity > 1, $exists alpha in G$ such that $alpha^k = 0, k ge |alpha|$. However, there $exists alpha in G$ that do have powers that form cyclic multiplicative groups. Can we easily determine which ones they are?

Can you point me to a link or a paper (or even just the proper terms to use for searching) that discusses more about this non-irreducible case?

polynomials finite-fields

share|cite|improve this question

edited Aug 6 at 18:59

asked Aug 1 at 21:32

Les

1214

$f(x)$ is a primitive polynomial of degree $m$ over a field $mathbf F_p^m$. As such, $mathbf F_p^m[x]/f(x)$ forms a ring, with addition and multiplication ($mkern-4mubmod f(x)$), and furthermore is a field since $f(x)$ is irreducible. Call it $F$

Now, replace $f(x)$ with $g(x)$, a non-primitive, non-irreducible polynomial of degree $m$ over $mathbf F_p^m$ and where $g(0) = 1$. Call it $G$. I am having trouble finding information on line or referred to about the characteristics of such a $g$ or the algebraic structure of $G$ (a ring?). This is probably because I don't know the right terminology and keywords regarding these to search by. Fields and Rings are new to me.

Some questions I have are (assume $m > 2$):

1. There are a finite number of elements in $G$ (because it is modulo $g$), but it doesn't form a field. What is it called? A "finite-ring"?

EDIT removed prior questions about cycle.

2. For any non-zero $alpha in F$, $exists n > 0$ such that $alpha^n = 1$. This is not true for all $alpha in G$. Can we easily determine the ones that do have such an $n$?


3. For any non-zero $alpha in F$, the powers of $alpha$ form a closed group under multiplication. Again, not true $G$. In $G exists alpha$ whose powers never equal 1, so they cannot form a group (no multiplicative identity). And when $g(x)$ is allowed to have factors with multiplicity > 1, $exists alpha in G$ such that $alpha^k = 0, k ge |alpha|$. However, there $exists alpha in G$ that do have powers that form cyclic multiplicative groups. Can we easily determine which ones they are?

Can you point me to a link or a paper (or even just the proper terms to use for searching) that discusses more about this non-irreducible case?

polynomials finite-fields

share|cite|improve this question

edited Aug 6 at 18:59

asked Aug 1 at 21:32

Les

1214

share|cite|improve this question

share|cite|improve this question

edited Aug 6 at 18:59

edited Aug 6 at 18:59

edited Aug 6 at 18:59

asked Aug 1 at 21:32

Les

1214

asked Aug 1 at 21:32

Les

1214

asked Aug 1 at 21:32

Les

1214

1214

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Why should there be a finite number of elements in the ring?
  – Bernard
  Aug 1 at 21:37
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What do you mean by a "cycle"?
  – Robert Lewis
  Aug 1 at 22:02
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  But $mathbf F[x]/(g)$ isn't finite, unless $mathbf F$ itself is: $mathbf R[x]/(x^2+1)simeq mathbf C$ is the simplest counter-example.
  – Bernard
  Aug 1 at 23:06
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  By the Chinese remainder theorem, it is an artinian $F$-algebra. If $g$ has no multiple irreducible factors (.i.e; if the algebra is a reduced ring), it is a product of field extensions of $F$. For the rest, I can't tell: I don't know what you call a ‘cycle’ here.
  – Bernard
  Aug 1 at 23:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Bernard has put his finger on the solution to the problem. When $g$ has an irreducible factor with multiplicity $>1$, there will be nilpotents in the resulting ring. Try $mathbf F[x]/((x-a)^m)$, for instance, the simplest case of this.
  – Lubin
  Aug 2 at 0:28
  

  
  
  
  

 | 
show 3 more comments

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Why should there be a finite number of elements in the ring?
  – Bernard
  Aug 1 at 21:37
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  What do you mean by a "cycle"?
  – Robert Lewis
  Aug 1 at 22:02
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  But $mathbf F[x]/(g)$ isn't finite, unless $mathbf F$ itself is: $mathbf R[x]/(x^2+1)simeq mathbf C$ is the simplest counter-example.
  – Bernard
  Aug 1 at 23:06
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  By the Chinese remainder theorem, it is an artinian $F$-algebra. If $g$ has no multiple irreducible factors (.i.e; if the algebra is a reduced ring), it is a product of field extensions of $F$. For the rest, I can't tell: I don't know what you call a ‘cycle’ here.
  – Bernard
  Aug 1 at 23:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Bernard has put his finger on the solution to the problem. When $g$ has an irreducible factor with multiplicity $>1$, there will be nilpotents in the resulting ring. Try $mathbf F[x]/((x-a)^m)$, for instance, the simplest case of this.
  – Lubin
  Aug 2 at 0:28
  

  
  
  
  

2

2

Why should there be a finite number of elements in the ring?
– Bernard
Aug 1 at 21:37

Why should there be a finite number of elements in the ring?
– Bernard
Aug 1 at 21:37

1

1

What do you mean by a "cycle"?
– Robert Lewis
Aug 1 at 22:02

What do you mean by a "cycle"?
– Robert Lewis
Aug 1 at 22:02

1

1

But $mathbf F[x]/(g)$ isn't finite, unless $mathbf F$ itself is: $mathbf R[x]/(x^2+1)simeq mathbf C$ is the simplest counter-example.
– Bernard
Aug 1 at 23:06

But $mathbf F[x]/(g)$ isn't finite, unless $mathbf F$ itself is: $mathbf R[x]/(x^2+1)simeq mathbf C$ is the simplest counter-example.
– Bernard
Aug 1 at 23:06

2

2

By the Chinese remainder theorem, it is an artinian $F$-algebra. If $g$ has no multiple irreducible factors (.i.e; if the algebra is a reduced ring), it is a product of field extensions of $F$. For the rest, I can't tell: I don't know what you call a ‘cycle’ here.
– Bernard
Aug 1 at 23:19

By the Chinese remainder theorem, it is an artinian $F$-algebra. If $g$ has no multiple irreducible factors (.i.e; if the algebra is a reduced ring), it is a product of field extensions of $F$. For the rest, I can't tell: I don't know what you call a ‘cycle’ here.
– Bernard
Aug 1 at 23:19

1

1

@Bernard has put his finger on the solution to the problem. When $g$ has an irreducible factor with multiplicity $>1$, there will be nilpotents in the resulting ring. Try $mathbf F[x]/((x-a)^m)$, for instance, the simplest case of this.
– Lubin
Aug 2 at 0:28

@Bernard has put his finger on the solution to the problem. When $g$ has an irreducible factor with multiplicity $>1$, there will be nilpotents in the resulting ring. Try $mathbf F[x]/((x-a)^m)$, for instance, the simplest case of this.
– Lubin
Aug 2 at 0:28

 | 
show 3 more comments

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