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Number of elements in a factor ring
The 2019 Stack Overflow Developer Survey Results Are InFactor into a product of irreducible polynomialsNumber of elements in Quotient ring over $mathbb Z_5[x]$ and $mathbb Z_11[x]$Find element in factor ringHow many elements does this ring have?Find the number of elements in the factor ring $R/A$ and describe the cosets.Construct a field of 25 elements.Factor ring and prime elementsHow many elements does the factor ring $mathbbZ_5[x]/(x^2-2)$ have? Find the multiplicative inverses of $x, x+1$ and $3x+2$ in this ring.Constructing a non-commutative ring with $5^9$ elementsFactor Rings over Finite Fields
$begingroup$
I am presented with $f(x) = 2x^3 + 3x^2 + 1$ $in mathbbZ_5[x]$ and need to explain why $F = fracmathbbZ_5[x]f(x)$ is a field and also find how many elements are in F.
So far I have shown that $f(x)$ is an irreducible polynomial and I also know that,
If $f(x)$ is an irreducible polynomial in $mathbbZ_5[x]$, then the factor ring $fracmathbbZ_5[x]f(x)$ is also a field.
Basically I am not sure how to properly find the factor ring F and also determine how many elements are in it.
Thanks in advance
abstract-algebra
$endgroup$
add a comment |
$begingroup$
I am presented with $f(x) = 2x^3 + 3x^2 + 1$ $in mathbbZ_5[x]$ and need to explain why $F = fracmathbbZ_5[x]f(x)$ is a field and also find how many elements are in F.
So far I have shown that $f(x)$ is an irreducible polynomial and I also know that,
If $f(x)$ is an irreducible polynomial in $mathbbZ_5[x]$, then the factor ring $fracmathbbZ_5[x]f(x)$ is also a field.
Basically I am not sure how to properly find the factor ring F and also determine how many elements are in it.
Thanks in advance
abstract-algebra
$endgroup$
add a comment |
$begingroup$
I am presented with $f(x) = 2x^3 + 3x^2 + 1$ $in mathbbZ_5[x]$ and need to explain why $F = fracmathbbZ_5[x]f(x)$ is a field and also find how many elements are in F.
So far I have shown that $f(x)$ is an irreducible polynomial and I also know that,
If $f(x)$ is an irreducible polynomial in $mathbbZ_5[x]$, then the factor ring $fracmathbbZ_5[x]f(x)$ is also a field.
Basically I am not sure how to properly find the factor ring F and also determine how many elements are in it.
Thanks in advance
abstract-algebra
$endgroup$
I am presented with $f(x) = 2x^3 + 3x^2 + 1$ $in mathbbZ_5[x]$ and need to explain why $F = fracmathbbZ_5[x]f(x)$ is a field and also find how many elements are in F.
So far I have shown that $f(x)$ is an irreducible polynomial and I also know that,
If $f(x)$ is an irreducible polynomial in $mathbbZ_5[x]$, then the factor ring $fracmathbbZ_5[x]f(x)$ is also a field.
Basically I am not sure how to properly find the factor ring F and also determine how many elements are in it.
Thanks in advance
abstract-algebra
abstract-algebra
asked Apr 8 at 10:29
MathsRookieMathsRookie
1237
1237
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Well, in general, if a field $L$ is an extension field of some field $K$, then $L$ is also a $K$-vector space.
If $f$ is an irreducible polynomial of degree $n$ over, say, $Bbb Z_p$, then
the quotient field $Bbb Z_p[x]/langle frangle$ has $p^n$ elements and is a vector space over $Bbb Z_p$ of dimension $n$.
$endgroup$
add a comment |
$begingroup$
Hint: Use the division algorithm to find canonical representatives for each residue class $g(x) + (f(x))$.
$endgroup$
add a comment |
$begingroup$
Outline:
You know $Bbb Z_5[x]/I$ where $I=langle f(x) rangle $ is a field using the mentioned result. Now, the task is to find $|Bbb Z_5[x]/I |$. Here $Bbb Z_5[x]/I$ can be considered as an entension of $Bbb Z_5$. Also $$dim_Bbb Z_5 (Bbb Z_5[x]/I)=3=deg f$$
Call $v_1,v_2,v_3$ the basis of $Bbb Z_5[x]/I$ over $Bbb Z_5$
Now arbitrarily pick $f(x)+I in Bbb Z_5[x]/I$. Then $$f(x)+I=a_1v_1+a_2v_2+a_3v_3;;;a_i in Bbb Z_5$$
There are $5 times 5 times 5$ choices to choose the coefficients, so $$textnumber of such $f(x)+I$ =5^3=|Bbb Z_5[x]/I|$$
$endgroup$
add a comment |
$begingroup$
Note that two polynomials $g(x)$ and $h(x)$, both of degree 2 or less, will represent different elements in $F$. This is due to the fact that their difference
$g(x)-h(x)$ being of degree less than 3, cannot be a multiple of the cubic polynomials $f(x)=2x^3+3x^2+1$, and so belong to different cosets.
Now a polynomial $h(x)$ of degree 3 or more can be divided by $f(x)$ to get quotient $q(x)$ and remainder $r(x)$
such that $h(x)=q(x)f(x) + r(x)$
Clearly $r(x)$ and $h(x)$ will represent the same coset. This shows that the field $F$ in your question simply consists of polynomials of degree less than $3$. This is easy to count as each coefficient has only a finite number of choices.
$endgroup$
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Well, in general, if a field $L$ is an extension field of some field $K$, then $L$ is also a $K$-vector space.
If $f$ is an irreducible polynomial of degree $n$ over, say, $Bbb Z_p$, then
the quotient field $Bbb Z_p[x]/langle frangle$ has $p^n$ elements and is a vector space over $Bbb Z_p$ of dimension $n$.
$endgroup$
add a comment |
$begingroup$
Well, in general, if a field $L$ is an extension field of some field $K$, then $L$ is also a $K$-vector space.
If $f$ is an irreducible polynomial of degree $n$ over, say, $Bbb Z_p$, then
the quotient field $Bbb Z_p[x]/langle frangle$ has $p^n$ elements and is a vector space over $Bbb Z_p$ of dimension $n$.
$endgroup$
add a comment |
$begingroup$
Well, in general, if a field $L$ is an extension field of some field $K$, then $L$ is also a $K$-vector space.
If $f$ is an irreducible polynomial of degree $n$ over, say, $Bbb Z_p$, then
the quotient field $Bbb Z_p[x]/langle frangle$ has $p^n$ elements and is a vector space over $Bbb Z_p$ of dimension $n$.
$endgroup$
Well, in general, if a field $L$ is an extension field of some field $K$, then $L$ is also a $K$-vector space.
If $f$ is an irreducible polynomial of degree $n$ over, say, $Bbb Z_p$, then
the quotient field $Bbb Z_p[x]/langle frangle$ has $p^n$ elements and is a vector space over $Bbb Z_p$ of dimension $n$.
edited Apr 8 at 14:14
answered Apr 8 at 10:35
WuestenfuxWuestenfux
5,5131513
5,5131513
add a comment |
add a comment |
$begingroup$
Hint: Use the division algorithm to find canonical representatives for each residue class $g(x) + (f(x))$.
$endgroup$
add a comment |
$begingroup$
Hint: Use the division algorithm to find canonical representatives for each residue class $g(x) + (f(x))$.
$endgroup$
add a comment |
$begingroup$
Hint: Use the division algorithm to find canonical representatives for each residue class $g(x) + (f(x))$.
$endgroup$
Hint: Use the division algorithm to find canonical representatives for each residue class $g(x) + (f(x))$.
answered Apr 8 at 10:34
lhflhf
168k11172404
168k11172404
add a comment |
add a comment |
$begingroup$
Outline:
You know $Bbb Z_5[x]/I$ where $I=langle f(x) rangle $ is a field using the mentioned result. Now, the task is to find $|Bbb Z_5[x]/I |$. Here $Bbb Z_5[x]/I$ can be considered as an entension of $Bbb Z_5$. Also $$dim_Bbb Z_5 (Bbb Z_5[x]/I)=3=deg f$$
Call $v_1,v_2,v_3$ the basis of $Bbb Z_5[x]/I$ over $Bbb Z_5$
Now arbitrarily pick $f(x)+I in Bbb Z_5[x]/I$. Then $$f(x)+I=a_1v_1+a_2v_2+a_3v_3;;;a_i in Bbb Z_5$$
There are $5 times 5 times 5$ choices to choose the coefficients, so $$textnumber of such $f(x)+I$ =5^3=|Bbb Z_5[x]/I|$$
$endgroup$
add a comment |
$begingroup$
Outline:
You know $Bbb Z_5[x]/I$ where $I=langle f(x) rangle $ is a field using the mentioned result. Now, the task is to find $|Bbb Z_5[x]/I |$. Here $Bbb Z_5[x]/I$ can be considered as an entension of $Bbb Z_5$. Also $$dim_Bbb Z_5 (Bbb Z_5[x]/I)=3=deg f$$
Call $v_1,v_2,v_3$ the basis of $Bbb Z_5[x]/I$ over $Bbb Z_5$
Now arbitrarily pick $f(x)+I in Bbb Z_5[x]/I$. Then $$f(x)+I=a_1v_1+a_2v_2+a_3v_3;;;a_i in Bbb Z_5$$
There are $5 times 5 times 5$ choices to choose the coefficients, so $$textnumber of such $f(x)+I$ =5^3=|Bbb Z_5[x]/I|$$
$endgroup$
add a comment |
$begingroup$
Outline:
You know $Bbb Z_5[x]/I$ where $I=langle f(x) rangle $ is a field using the mentioned result. Now, the task is to find $|Bbb Z_5[x]/I |$. Here $Bbb Z_5[x]/I$ can be considered as an entension of $Bbb Z_5$. Also $$dim_Bbb Z_5 (Bbb Z_5[x]/I)=3=deg f$$
Call $v_1,v_2,v_3$ the basis of $Bbb Z_5[x]/I$ over $Bbb Z_5$
Now arbitrarily pick $f(x)+I in Bbb Z_5[x]/I$. Then $$f(x)+I=a_1v_1+a_2v_2+a_3v_3;;;a_i in Bbb Z_5$$
There are $5 times 5 times 5$ choices to choose the coefficients, so $$textnumber of such $f(x)+I$ =5^3=|Bbb Z_5[x]/I|$$
$endgroup$
Outline:
You know $Bbb Z_5[x]/I$ where $I=langle f(x) rangle $ is a field using the mentioned result. Now, the task is to find $|Bbb Z_5[x]/I |$. Here $Bbb Z_5[x]/I$ can be considered as an entension of $Bbb Z_5$. Also $$dim_Bbb Z_5 (Bbb Z_5[x]/I)=3=deg f$$
Call $v_1,v_2,v_3$ the basis of $Bbb Z_5[x]/I$ over $Bbb Z_5$
Now arbitrarily pick $f(x)+I in Bbb Z_5[x]/I$. Then $$f(x)+I=a_1v_1+a_2v_2+a_3v_3;;;a_i in Bbb Z_5$$
There are $5 times 5 times 5$ choices to choose the coefficients, so $$textnumber of such $f(x)+I$ =5^3=|Bbb Z_5[x]/I|$$
answered Apr 8 at 10:39
Chinnapparaj RChinnapparaj R
6,34521029
6,34521029
add a comment |
add a comment |
$begingroup$
Note that two polynomials $g(x)$ and $h(x)$, both of degree 2 or less, will represent different elements in $F$. This is due to the fact that their difference
$g(x)-h(x)$ being of degree less than 3, cannot be a multiple of the cubic polynomials $f(x)=2x^3+3x^2+1$, and so belong to different cosets.
Now a polynomial $h(x)$ of degree 3 or more can be divided by $f(x)$ to get quotient $q(x)$ and remainder $r(x)$
such that $h(x)=q(x)f(x) + r(x)$
Clearly $r(x)$ and $h(x)$ will represent the same coset. This shows that the field $F$ in your question simply consists of polynomials of degree less than $3$. This is easy to count as each coefficient has only a finite number of choices.
$endgroup$
add a comment |
$begingroup$
Note that two polynomials $g(x)$ and $h(x)$, both of degree 2 or less, will represent different elements in $F$. This is due to the fact that their difference
$g(x)-h(x)$ being of degree less than 3, cannot be a multiple of the cubic polynomials $f(x)=2x^3+3x^2+1$, and so belong to different cosets.
Now a polynomial $h(x)$ of degree 3 or more can be divided by $f(x)$ to get quotient $q(x)$ and remainder $r(x)$
such that $h(x)=q(x)f(x) + r(x)$
Clearly $r(x)$ and $h(x)$ will represent the same coset. This shows that the field $F$ in your question simply consists of polynomials of degree less than $3$. This is easy to count as each coefficient has only a finite number of choices.
$endgroup$
add a comment |
$begingroup$
Note that two polynomials $g(x)$ and $h(x)$, both of degree 2 or less, will represent different elements in $F$. This is due to the fact that their difference
$g(x)-h(x)$ being of degree less than 3, cannot be a multiple of the cubic polynomials $f(x)=2x^3+3x^2+1$, and so belong to different cosets.
Now a polynomial $h(x)$ of degree 3 or more can be divided by $f(x)$ to get quotient $q(x)$ and remainder $r(x)$
such that $h(x)=q(x)f(x) + r(x)$
Clearly $r(x)$ and $h(x)$ will represent the same coset. This shows that the field $F$ in your question simply consists of polynomials of degree less than $3$. This is easy to count as each coefficient has only a finite number of choices.
$endgroup$
Note that two polynomials $g(x)$ and $h(x)$, both of degree 2 or less, will represent different elements in $F$. This is due to the fact that their difference
$g(x)-h(x)$ being of degree less than 3, cannot be a multiple of the cubic polynomials $f(x)=2x^3+3x^2+1$, and so belong to different cosets.
Now a polynomial $h(x)$ of degree 3 or more can be divided by $f(x)$ to get quotient $q(x)$ and remainder $r(x)$
such that $h(x)=q(x)f(x) + r(x)$
Clearly $r(x)$ and $h(x)$ will represent the same coset. This shows that the field $F$ in your question simply consists of polynomials of degree less than $3$. This is easy to count as each coefficient has only a finite number of choices.
answered Apr 8 at 10:47
P VanchinathanP Vanchinathan
15.6k12136
15.6k12136
add a comment |
add a comment |
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