Find number of polynomials in $P_n(F)$ where $F$ is a finite field with cardinality $m$. [duplicate]
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Number of elements in a finite field extension for finite fields
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Firstly, I should mention that the above question hits my mind when I am solving a problem in Abstract Algebra which is- Find the number of elements in $fracBbbZ_3[x]<x^3+2x+1>$.
I have solved the problem in a following manner.
I know that $K[x]$ be a polynomial ring over field $K$. Let, $f(x)$ be a non-zero polynomial in $K[x]$ with deg$(f)=n>0$. Then the quotient ring $fracK[x]<f(x)>$ contains polynomials of form $g(x)+<f(x)>$ where $g(x)$ is a polynomial in $K[x]$ with degree at most $n-1$.
So, cardinality of $fracBbbZ_3[x]<x^3+2x+1>$ is same as of $P_2[BbbZ_3]$, set of all polynomials with degree less or equal to 2 and having coefficients from $BbbZ_3$. Again $P_2[BbbZ_3]$ has same cardinality as of the set of all 3-tuples(since the polynomials should have degree at most 2) with entries from $BbbZ_3$.
I have listed all 3-tuples with entries from $BbbZ_3$- $(0,0,0), (1,0,0), (2,0,0), (0,1,0),(0,2,0), (0,0,1), (0,0,2), (1,1,0), (0,1,1), (1,0,1), (2,2,0), (0,2,2), (2,0,2), (1,2,0), (0,1,2), (1,0,2), (2,1,0), (0,2,1), (2,0,1), (1,1,2), (1,2,1), (2,1,1), (2,2,1), (2,1,2), (1,2,2), (1,1,1), (2,2,2).$
So, number of elements in $fracBbbZ_3[x]<x^3+2x+1>$ is $27$.
But still the method is too laborious if the degree of polynomials increases of cardinality of field increases. So, I need to establish a easy way out or a formula.
Now, I try to find a general formula to find number of polynomials in $P_n(F)$ (set of all polynomials with degree at most n and having coefficients from F) where F is a finite field with cardinality $m$.($n,min BbbN$)
To do that I just need to find cardinality of the set $(a_0, a_1,...,a_n):a_iin Fquadforall i=0, 1,...,n$.
How can I find that? Can anybody help me do this?
Thanks for your help advance.
abstract-algebra polynomials ring-theory field-theory finite-fields
asked Aug 15 at 9:41
Biswarup Saha
2318
marked as duplicate by Jyrki Lahtonen abstract-algebra
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Aug 15 at 9:54
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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up vote
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This question already has an answer here:
Number of elements in a finite field extension for finite fields
1 answer
Firstly, I should mention that the above question hits my mind when I am solving a problem in Abstract Algebra which is- Find the number of elements in $fracBbbZ_3[x]<x^3+2x+1>$.
I have solved the problem in a following manner.
I know that $K[x]$ be a polynomial ring over field $K$. Let, $f(x)$ be a non-zero polynomial in $K[x]$ with deg$(f)=n>0$. Then the quotient ring $fracK[x]<f(x)>$ contains polynomials of form $g(x)+<f(x)>$ where $g(x)$ is a polynomial in $K[x]$ with degree at most $n-1$.
So, cardinality of $fracBbbZ_3[x]<x^3+2x+1>$ is same as of $P_2[BbbZ_3]$, set of all polynomials with degree less or equal to 2 and having coefficients from $BbbZ_3$. Again $P_2[BbbZ_3]$ has same cardinality as of the set of all 3-tuples(since the polynomials should have degree at most 2) with entries from $BbbZ_3$.
I have listed all 3-tuples with entries from $BbbZ_3$- $(0,0,0), (1,0,0), (2,0,0), (0,1,0),(0,2,0), (0,0,1), (0,0,2), (1,1,0), (0,1,1), (1,0,1), (2,2,0), (0,2,2), (2,0,2), (1,2,0), (0,1,2), (1,0,2), (2,1,0), (0,2,1), (2,0,1), (1,1,2), (1,2,1), (2,1,1), (2,2,1), (2,1,2), (1,2,2), (1,1,1), (2,2,2).$
So, number of elements in $fracBbbZ_3[x]<x^3+2x+1>$ is $27$.
But still the method is too laborious if the degree of polynomials increases of cardinality of field increases. So, I need to establish a easy way out or a formula.
Now, I try to find a general formula to find number of polynomials in $P_n(F)$ (set of all polynomials with degree at most n and having coefficients from F) where F is a finite field with cardinality $m$.($n,min BbbN$)
To do that I just need to find cardinality of the set $(a_0, a_1,...,a_n):a_iin Fquadforall i=0, 1,...,n$.
How can I find that? Can anybody help me do this?
Thanks for your help advance.
abstract-algebra polynomials ring-theory field-theory finite-fields
asked Aug 15 at 9:41
Biswarup Saha
2318
marked as duplicate by Jyrki Lahtonen abstract-algebra
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Aug 15 at 9:54
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
@JyrkiLahtonen, I am also thinking that, order matters, even $n$ may be less than $m$.
– Biswarup Saha
Aug 15 at 9:57
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up vote
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This question already has an answer here:
Number of elements in a finite field extension for finite fields
1 answer
Firstly, I should mention that the above question hits my mind when I am solving a problem in Abstract Algebra which is- Find the number of elements in $fracBbbZ_3[x]<x^3+2x+1>$.
I have solved the problem in a following manner.
I know that $K[x]$ be a polynomial ring over field $K$. Let, $f(x)$ be a non-zero polynomial in $K[x]$ with deg$(f)=n>0$. Then the quotient ring $fracK[x]<f(x)>$ contains polynomials of form $g(x)+<f(x)>$ where $g(x)$ is a polynomial in $K[x]$ with degree at most $n-1$.
So, cardinality of $fracBbbZ_3[x]<x^3+2x+1>$ is same as of $P_2[BbbZ_3]$, set of all polynomials with degree less or equal to 2 and having coefficients from $BbbZ_3$. Again $P_2[BbbZ_3]$ has same cardinality as of the set of all 3-tuples(since the polynomials should have degree at most 2) with entries from $BbbZ_3$.
I have listed all 3-tuples with entries from $BbbZ_3$- $(0,0,0), (1,0,0), (2,0,0), (0,1,0),(0,2,0), (0,0,1), (0,0,2), (1,1,0), (0,1,1), (1,0,1), (2,2,0), (0,2,2), (2,0,2), (1,2,0), (0,1,2), (1,0,2), (2,1,0), (0,2,1), (2,0,1), (1,1,2), (1,2,1), (2,1,1), (2,2,1), (2,1,2), (1,2,2), (1,1,1), (2,2,2).$
So, number of elements in $fracBbbZ_3[x]<x^3+2x+1>$ is $27$.
But still the method is too laborious if the degree of polynomials increases of cardinality of field increases. So, I need to establish a easy way out or a formula.
Now, I try to find a general formula to find number of polynomials in $P_n(F)$ (set of all polynomials with degree at most n and having coefficients from F) where F is a finite field with cardinality $m$.($n,min BbbN$)
To do that I just need to find cardinality of the set $(a_0, a_1,...,a_n):a_iin Fquadforall i=0, 1,...,n$.
How can I find that? Can anybody help me do this?
Thanks for your help advance.
abstract-algebra polynomials ring-theory field-theory finite-fields
asked Aug 15 at 9:41
Biswarup Saha
2318
This question already has an answer here:
Number of elements in a finite field extension for finite fields
1 answer
Firstly, I should mention that the above question hits my mind when I am solving a problem in Abstract Algebra which is- Find the number of elements in $fracBbbZ_3[x]<x^3+2x+1>$.
I have solved the problem in a following manner.
I know that $K[x]$ be a polynomial ring over field $K$. Let, $f(x)$ be a non-zero polynomial in $K[x]$ with deg$(f)=n>0$. Then the quotient ring $fracK[x]<f(x)>$ contains polynomials of form $g(x)+<f(x)>$ where $g(x)$ is a polynomial in $K[x]$ with degree at most $n-1$.
So, cardinality of $fracBbbZ_3[x]<x^3+2x+1>$ is same as of $P_2[BbbZ_3]$, set of all polynomials with degree less or equal to 2 and having coefficients from $BbbZ_3$. Again $P_2[BbbZ_3]$ has same cardinality as of the set of all 3-tuples(since the polynomials should have degree at most 2) with entries from $BbbZ_3$.
I have listed all 3-tuples with entries from $BbbZ_3$- $(0,0,0), (1,0,0), (2,0,0), (0,1,0),(0,2,0), (0,0,1), (0,0,2), (1,1,0), (0,1,1), (1,0,1), (2,2,0), (0,2,2), (2,0,2), (1,2,0), (0,1,2), (1,0,2), (2,1,0), (0,2,1), (2,0,1), (1,1,2), (1,2,1), (2,1,1), (2,2,1), (2,1,2), (1,2,2), (1,1,1), (2,2,2).$
So, number of elements in $fracBbbZ_3[x]<x^3+2x+1>$ is $27$.
But still the method is too laborious if the degree of polynomials increases of cardinality of field increases. So, I need to establish a easy way out or a formula.
Now, I try to find a general formula to find number of polynomials in $P_n(F)$ (set of all polynomials with degree at most n and having coefficients from F) where F is a finite field with cardinality $m$.($n,min BbbN$)
To do that I just need to find cardinality of the set $(a_0, a_1,...,a_n):a_iin Fquadforall i=0, 1,...,n$.
How can I find that? Can anybody help me do this?
Thanks for your help advance.
This question already has an answer here:
Number of elements in a finite field extension for finite fields
1 answer
abstract-algebra polynomials ring-theory field-theory finite-fields
asked Aug 15 at 9:41
Biswarup Saha
2318
asked Aug 15 at 9:41
Biswarup Saha
2318
asked Aug 15 at 9:41
Biswarup Saha
2318
asked Aug 15 at 9:41
Biswarup Saha
2318
2318
marked as duplicate by Jyrki Lahtonen abstract-algebra
Users with the  abstract-algebra badge can single-handedly close abstract-algebra questions as duplicates and reopen them as needed.
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Aug 15 at 9:54
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Jyrki Lahtonen abstract-algebra
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Aug 15 at 9:54
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
@JyrkiLahtonen, I am also thinking that, order matters, even $n$ may be less than $m$.
– Biswarup Saha
Aug 15 at 9:57
add a comment |Â
@JyrkiLahtonen, I am also thinking that, order matters, even $n$ may be less than $m$.
– Biswarup Saha
Aug 15 at 9:57
@JyrkiLahtonen, I am also thinking that, order matters, even $n$ may be less than $m$.
– Biswarup Saha
Aug 15 at 9:57
@JyrkiLahtonen, I am also thinking that, order matters, even $n$ may be less than $m$.
– Biswarup Saha
Aug 15 at 9:57
add a comment |Â
1 Answer
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I take it that $P_n(F)$ in your case means polynomials with degree less than or equal to $n$.
In that case, every coefficient can be any of the elements in $F$ so you have $|F|$ choices in each coefficient.
Thus the answer would be $|F|^n+1$. (And in general if $F$ finite $|F|=p^k$ for some prime number $p$.)
answered Aug 15 at 9:54
daruma
905612
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1 Answer
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1 Answer
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active
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active
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up vote
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down vote
I take it that $P_n(F)$ in your case means polynomials with degree less than or equal to $n$.
In that case, every coefficient can be any of the elements in $F$ so you have $|F|$ choices in each coefficient.
Thus the answer would be $|F|^n+1$. (And in general if $F$ finite $|F|=p^k$ for some prime number $p$.)
answered Aug 15 at 9:54
daruma
905612
add a comment |Â
up vote
1
down vote
I take it that $P_n(F)$ in your case means polynomials with degree less than or equal to $n$.
In that case, every coefficient can be any of the elements in $F$ so you have $|F|$ choices in each coefficient.
Thus the answer would be $|F|^n+1$. (And in general if $F$ finite $|F|=p^k$ for some prime number $p$.)
answered Aug 15 at 9:54
daruma
905612
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I take it that $P_n(F)$ in your case means polynomials with degree less than or equal to $n$.
In that case, every coefficient can be any of the elements in $F$ so you have $|F|$ choices in each coefficient.
Thus the answer would be $|F|^n+1$. (And in general if $F$ finite $|F|=p^k$ for some prime number $p$.)
answered Aug 15 at 9:54
daruma
905612
I take it that $P_n(F)$ in your case means polynomials with degree less than or equal to $n$.
In that case, every coefficient can be any of the elements in $F$ so you have $|F|$ choices in each coefficient.
Thus the answer would be $|F|^n+1$. (And in general if $F$ finite $|F|=p^k$ for some prime number $p$.)
answered Aug 15 at 9:54
daruma
905612
answered Aug 15 at 9:54
daruma
905612
answered Aug 15 at 9:54
daruma
905612
answered Aug 15 at 9:54
daruma
905612
905612
add a comment |Â
add a comment |Â
@JyrkiLahtonen, I am also thinking that, order matters, even $n$ may be less than $m$.
– Biswarup Saha
Aug 15 at 9:57