Let $mathbb{F}=mathbb{F}_3$ Find an irreducible polynomial of degree 2 and construct a field of 9 elements as...
$begingroup$
Find an irreducible polynomial p of degree 2, and use it construct a field of 9
elements as a quotient. Describe the cosets in the quotient explicitly, and use them to construct
the addition and multiplication tables for the field obtained by this quotient.
I'm not going to ask anyone to construct the tables. I'm more just unsure of how to go about solving this. I have an irreducible polynomial $[2]x^2+x+[1]$,
I know the elements should be remainders when dividing by other polynomials of degree less then 2. So I should have to use long division by this polynomial I believe I get remainders ${0,1,x,x+1,x+2,2x,2x+1,2x+2,2}$
but I'm not sure exactly how I go about this from here. I've done this for 4 elements but it seemed having so few elements made it obvious how to make it work.
linear-algebra quotient-spaces
asked Dec 9 '18 at 21:59
AColoredReptileAColoredReptile
18918
$endgroup$
add a comment |
$begingroup$
Find an irreducible polynomial p of degree 2, and use it construct a field of 9
elements as a quotient. Describe the cosets in the quotient explicitly, and use them to construct
the addition and multiplication tables for the field obtained by this quotient.
I'm not going to ask anyone to construct the tables. I'm more just unsure of how to go about solving this. I have an irreducible polynomial $[2]x^2+x+[1]$,
I know the elements should be remainders when dividing by other polynomials of degree less then 2. So I should have to use long division by this polynomial I believe I get remainders ${0,1,x,x+1,x+2,2x,2x+1,2x+2,2}$
but I'm not sure exactly how I go about this from here. I've done this for 4 elements but it seemed having so few elements made it obvious how to make it work.
linear-algebra quotient-spaces
asked Dec 9 '18 at 21:59
AColoredReptileAColoredReptile
18918
$endgroup$
$begingroup$
As remainders, you obtain all polynomials of degree $le 1$. with coefficients in $mathbf F_3$. This makes $9$ polynomials.
$endgroup$
– Bernard
Dec 9 '18 at 22:03
$begingroup$
You're making your life difficult: why not taking $x^2-x-1$?
$endgroup$
– egreg
Dec 9 '18 at 22:17
$begingroup$
@egreg this was the first polynomial I could think of. Is there someway to get an irreducible polynomial without checking it?
$endgroup$
– AColoredReptile
Dec 9 '18 at 22:19
1
$begingroup$
@AcoloredReptile: I think @egreg’s point is that it’s the same polynomial up to a constant! If you divide through by $[2]$, which is the same thing as multiplying through by $[2]$, you get $x^2 +[2]x+[2] = x^2 +[-1]x + [-1] = x^2-x-1$. You’ll need to check the polynomial, but you should always try to use a monic one, because it makes calculations easier.
$endgroup$
– Arturo Magidin
Dec 9 '18 at 22:21
add a comment |
$begingroup$
Find an irreducible polynomial p of degree 2, and use it construct a field of 9
elements as a quotient. Describe the cosets in the quotient explicitly, and use them to construct
the addition and multiplication tables for the field obtained by this quotient.
I'm not going to ask anyone to construct the tables. I'm more just unsure of how to go about solving this. I have an irreducible polynomial $[2]x^2+x+[1]$,
I know the elements should be remainders when dividing by other polynomials of degree less then 2. So I should have to use long division by this polynomial I believe I get remainders ${0,1,x,x+1,x+2,2x,2x+1,2x+2,2}$
but I'm not sure exactly how I go about this from here. I've done this for 4 elements but it seemed having so few elements made it obvious how to make it work.
linear-algebra quotient-spaces
asked Dec 9 '18 at 21:59
AColoredReptileAColoredReptile
18918
$endgroup$
Find an irreducible polynomial p of degree 2, and use it construct a field of 9
elements as a quotient. Describe the cosets in the quotient explicitly, and use them to construct
the addition and multiplication tables for the field obtained by this quotient.
I'm not going to ask anyone to construct the tables. I'm more just unsure of how to go about solving this. I have an irreducible polynomial $[2]x^2+x+[1]$,
I know the elements should be remainders when dividing by other polynomials of degree less then 2. So I should have to use long division by this polynomial I believe I get remainders ${0,1,x,x+1,x+2,2x,2x+1,2x+2,2}$
but I'm not sure exactly how I go about this from here. I've done this for 4 elements but it seemed having so few elements made it obvious how to make it work.
linear-algebra quotient-spaces
linear-algebra quotient-spaces
asked Dec 9 '18 at 21:59
AColoredReptileAColoredReptile
18918
asked Dec 9 '18 at 21:59
AColoredReptileAColoredReptile
18918
edited Dec 9 '18 at 22:17
AColoredReptile
asked Dec 9 '18 at 21:59
AColoredReptileAColoredReptile
18918
asked Dec 9 '18 at 21:59
AColoredReptileAColoredReptile
18918
asked Dec 9 '18 at 21:59
AColoredReptileAColoredReptile
18918
18918
$begingroup$
As remainders, you obtain all polynomials of degree $le 1$. with coefficients in $mathbf F_3$. This makes $9$ polynomials.
$endgroup$
– Bernard
Dec 9 '18 at 22:03
$begingroup$
You're making your life difficult: why not taking $x^2-x-1$?
$endgroup$
– egreg
Dec 9 '18 at 22:17
$begingroup$
@egreg this was the first polynomial I could think of. Is there someway to get an irreducible polynomial without checking it?
$endgroup$
– AColoredReptile
Dec 9 '18 at 22:19
1
$begingroup$
@AcoloredReptile: I think @egreg’s point is that it’s the same polynomial up to a constant! If you divide through by $[2]$, which is the same thing as multiplying through by $[2]$, you get $x^2 +[2]x+[2] = x^2 +[-1]x + [-1] = x^2-x-1$. You’ll need to check the polynomial, but you should always try to use a monic one, because it makes calculations easier.
$endgroup$
– Arturo Magidin
Dec 9 '18 at 22:21
add a comment |
$begingroup$
As remainders, you obtain all polynomials of degree $le 1$. with coefficients in $mathbf F_3$. This makes $9$ polynomials.
$endgroup$
– Bernard
Dec 9 '18 at 22:03
$begingroup$
You're making your life difficult: why not taking $x^2-x-1$?
$endgroup$
– egreg
Dec 9 '18 at 22:17
$begingroup$
@egreg this was the first polynomial I could think of. Is there someway to get an irreducible polynomial without checking it?
$endgroup$
– AColoredReptile
Dec 9 '18 at 22:19
1
$begingroup$
@AcoloredReptile: I think @egreg’s point is that it’s the same polynomial up to a constant! If you divide through by $[2]$, which is the same thing as multiplying through by $[2]$, you get $x^2 +[2]x+[2] = x^2 +[-1]x + [-1] = x^2-x-1$. You’ll need to check the polynomial, but you should always try to use a monic one, because it makes calculations easier.
$endgroup$
– Arturo Magidin
Dec 9 '18 at 22:21
$begingroup$
As remainders, you obtain all polynomials of degree $le 1$. with coefficients in $mathbf F_3$. This makes $9$ polynomials.
$endgroup$
– Bernard
Dec 9 '18 at 22:03
$begingroup$
As remainders, you obtain all polynomials of degree $le 1$. with coefficients in $mathbf F_3$. This makes $9$ polynomials.
$endgroup$
– Bernard
Dec 9 '18 at 22:03
$begingroup$
You're making your life difficult: why not taking $x^2-x-1$?
$endgroup$
– egreg
Dec 9 '18 at 22:17
$begingroup$
You're making your life difficult: why not taking $x^2-x-1$?
$endgroup$
– egreg
Dec 9 '18 at 22:17
$begingroup$
@egreg this was the first polynomial I could think of. Is there someway to get an irreducible polynomial without checking it?
$endgroup$
– AColoredReptile
Dec 9 '18 at 22:19
$begingroup$
@egreg this was the first polynomial I could think of. Is there someway to get an irreducible polynomial without checking it?
$endgroup$
– AColoredReptile
Dec 9 '18 at 22:19
1
1
$begingroup$
@AcoloredReptile: I think @egreg’s point is that it’s the same polynomial up to a constant! If you divide through by $[2]$, which is the same thing as multiplying through by $[2]$, you get $x^2 +[2]x+[2] = x^2 +[-1]x + [-1] = x^2-x-1$. You’ll need to check the polynomial, but you should always try to use a monic one, because it makes calculations easier.
$endgroup$
– Arturo Magidin
Dec 9 '18 at 22:21
$begingroup$
@AcoloredReptile: I think @egreg’s point is that it’s the same polynomial up to a constant! If you divide through by $[2]$, which is the same thing as multiplying through by $[2]$, you get $x^2 +[2]x+[2] = x^2 +[-1]x + [-1] = x^2-x-1$. You’ll need to check the polynomial, but you should always try to use a monic one, because it makes calculations easier.
$endgroup$
– Arturo Magidin
Dec 9 '18 at 22:21
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The polynomial $p(x)=2x^2+x+1$ is irreducible because it has no roots: indeed $p(0)=1$, $p(1)=1$ and $p(2)=2$. It would be more complicated to find, say, an irreducible polynomial of degree four or more, because the “no root” criterion doesn't apply.
The remainders are ${0,1,2,x,x+1,x+2,2x,2x+1,2x+2}$. How do you multiply two of them? Actually, it's better to think of them as the elements
$$
0,1,2,xi,xi+1,xi+2,2xi,2xi+1,2xi+2
$$
with $2xi^2+xi+1=0$, that is, $xi^2=xi+1$. Then
$$
(xi+2)(2xi+1)=2xi^2+xi+xi+2=2(xi+1)+2xi+2=xi+1
$$
(recall that $2+2=1$ in this field). Similarly for the other pairs.
With your polynomial, the remainders are the same, and the same is the relation.
answered Dec 9 '18 at 22:24
egregegreg
181k1485203
$endgroup$
$begingroup$
Ok. And are the cosets the sets ${[v]+2x^2+x+1}$? where $[v]in mathbb{F_3}/(2x^2+x+1)$ So for instance I have a coset of $mathbb{F}_3/(2x^2+x+1)$, which is $(2x+1)+(2x^2+x+1)$?
$endgroup$
– AColoredReptile
Dec 9 '18 at 22:42
$begingroup$
@AColoredReptile You are confusing things; if $vinmathbb{F}_3$, then you have the coset $[v]=v+(2x^2+x+1)$. In my notation, $xi=[x]=x+(2x^2+x+1)$.
$endgroup$
– egreg
Dec 9 '18 at 22:57
$begingroup$
But if I want to write out all the cosets aren't that just my remainders plus the irreducible polynomial?
$endgroup$
– AColoredReptile
Dec 9 '18 at 23:53
$begingroup$
@AColoredReptile No, the cosets are the remainders plus the ideal generated by the irreducible polynomial.
$endgroup$
– egreg
Dec 9 '18 at 23:56
$begingroup$
I dont know what an ideal is. The definition of coset I know is ${v} + W={ v+w: win W}$.
$endgroup$
– AColoredReptile
Dec 10 '18 at 0:05
add a comment |
$begingroup$
You have several remainders with degree 2. How do you get a remainder of degree 2 when what you're dividing by has degree 2?
The short answer is that you don't. You get every single possible first degree (or lower) polynomial as remainders: any such polynomial appears as a remainder, for instance when trying to divide that polynomial itself by $2x+x+1$, and any remainder has at most degree 1 because second degree or higher terms may be divided (possibly with some remainder, but that's the point). There are nine of those polynomials, so those are the elements of your field.
answered Dec 9 '18 at 22:06
ArthurArthur
114k7115197
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add a comment |
$begingroup$
$F_3[x]/(2x^2+x+1)cong F_3(alpha)$ is the vector space over $F_3$ with basis ${1,alpha}$, where $2alpha^2+alpha +1=0$.
Two polynomials in $F_3[x]$ are equivalent if they differ by an element of $(2x^2+x+1)$ (which is to say by a multiple of $2x^2+x+1$, since $F_3[x]$ is commutative).
Now to complete the exercise you have only to write out the multiplication and addition tables for the $9$ elements.
answered Dec 9 '18 at 22:18
Chris CusterChris Custer
12.7k3825
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The polynomial $p(x)=2x^2+x+1$ is irreducible because it has no roots: indeed $p(0)=1$, $p(1)=1$ and $p(2)=2$. It would be more complicated to find, say, an irreducible polynomial of degree four or more, because the “no root” criterion doesn't apply.
The remainders are ${0,1,2,x,x+1,x+2,2x,2x+1,2x+2}$. How do you multiply two of them? Actually, it's better to think of them as the elements
$$
0,1,2,xi,xi+1,xi+2,2xi,2xi+1,2xi+2
$$
with $2xi^2+xi+1=0$, that is, $xi^2=xi+1$. Then
$$
(xi+2)(2xi+1)=2xi^2+xi+xi+2=2(xi+1)+2xi+2=xi+1
$$
(recall that $2+2=1$ in this field). Similarly for the other pairs.
With your polynomial, the remainders are the same, and the same is the relation.
answered Dec 9 '18 at 22:24
egregegreg
181k1485203
$endgroup$
$begingroup$
Ok. And are the cosets the sets ${[v]+2x^2+x+1}$? where $[v]in mathbb{F_3}/(2x^2+x+1)$ So for instance I have a coset of $mathbb{F}_3/(2x^2+x+1)$, which is $(2x+1)+(2x^2+x+1)$?
$endgroup$
– AColoredReptile
Dec 9 '18 at 22:42
$begingroup$
@AColoredReptile You are confusing things; if $vinmathbb{F}_3$, then you have the coset $[v]=v+(2x^2+x+1)$. In my notation, $xi=[x]=x+(2x^2+x+1)$.
$endgroup$
– egreg
Dec 9 '18 at 22:57
$begingroup$
But if I want to write out all the cosets aren't that just my remainders plus the irreducible polynomial?
$endgroup$
– AColoredReptile
Dec 9 '18 at 23:53
$begingroup$
@AColoredReptile No, the cosets are the remainders plus the ideal generated by the irreducible polynomial.
$endgroup$
– egreg
Dec 9 '18 at 23:56
$begingroup$
I dont know what an ideal is. The definition of coset I know is ${v} + W={ v+w: win W}$.
$endgroup$
– AColoredReptile
Dec 10 '18 at 0:05
add a comment |
$begingroup$
The polynomial $p(x)=2x^2+x+1$ is irreducible because it has no roots: indeed $p(0)=1$, $p(1)=1$ and $p(2)=2$. It would be more complicated to find, say, an irreducible polynomial of degree four or more, because the “no root” criterion doesn't apply.
The remainders are ${0,1,2,x,x+1,x+2,2x,2x+1,2x+2}$. How do you multiply two of them? Actually, it's better to think of them as the elements
$$
0,1,2,xi,xi+1,xi+2,2xi,2xi+1,2xi+2
$$
with $2xi^2+xi+1=0$, that is, $xi^2=xi+1$. Then
$$
(xi+2)(2xi+1)=2xi^2+xi+xi+2=2(xi+1)+2xi+2=xi+1
$$
(recall that $2+2=1$ in this field). Similarly for the other pairs.
With your polynomial, the remainders are the same, and the same is the relation.
answered Dec 9 '18 at 22:24
egregegreg
181k1485203
$endgroup$
$begingroup$
Ok. And are the cosets the sets ${[v]+2x^2+x+1}$? where $[v]in mathbb{F_3}/(2x^2+x+1)$ So for instance I have a coset of $mathbb{F}_3/(2x^2+x+1)$, which is $(2x+1)+(2x^2+x+1)$?
$endgroup$
– AColoredReptile
Dec 9 '18 at 22:42
$begingroup$
@AColoredReptile You are confusing things; if $vinmathbb{F}_3$, then you have the coset $[v]=v+(2x^2+x+1)$. In my notation, $xi=[x]=x+(2x^2+x+1)$.
$endgroup$
– egreg
Dec 9 '18 at 22:57
$begingroup$
But if I want to write out all the cosets aren't that just my remainders plus the irreducible polynomial?
$endgroup$
– AColoredReptile
Dec 9 '18 at 23:53
$begingroup$
@AColoredReptile No, the cosets are the remainders plus the ideal generated by the irreducible polynomial.
$endgroup$
– egreg
Dec 9 '18 at 23:56
$begingroup$
I dont know what an ideal is. The definition of coset I know is ${v} + W={ v+w: win W}$.
$endgroup$
– AColoredReptile
Dec 10 '18 at 0:05
add a comment |
$begingroup$
The polynomial $p(x)=2x^2+x+1$ is irreducible because it has no roots: indeed $p(0)=1$, $p(1)=1$ and $p(2)=2$. It would be more complicated to find, say, an irreducible polynomial of degree four or more, because the “no root” criterion doesn't apply.
The remainders are ${0,1,2,x,x+1,x+2,2x,2x+1,2x+2}$. How do you multiply two of them? Actually, it's better to think of them as the elements
$$
0,1,2,xi,xi+1,xi+2,2xi,2xi+1,2xi+2
$$
with $2xi^2+xi+1=0$, that is, $xi^2=xi+1$. Then
$$
(xi+2)(2xi+1)=2xi^2+xi+xi+2=2(xi+1)+2xi+2=xi+1
$$
(recall that $2+2=1$ in this field). Similarly for the other pairs.
With your polynomial, the remainders are the same, and the same is the relation.
answered Dec 9 '18 at 22:24
egregegreg
181k1485203
$endgroup$
The polynomial $p(x)=2x^2+x+1$ is irreducible because it has no roots: indeed $p(0)=1$, $p(1)=1$ and $p(2)=2$. It would be more complicated to find, say, an irreducible polynomial of degree four or more, because the “no root” criterion doesn't apply.
The remainders are ${0,1,2,x,x+1,x+2,2x,2x+1,2x+2}$. How do you multiply two of them? Actually, it's better to think of them as the elements
$$
0,1,2,xi,xi+1,xi+2,2xi,2xi+1,2xi+2
$$
with $2xi^2+xi+1=0$, that is, $xi^2=xi+1$. Then
$$
(xi+2)(2xi+1)=2xi^2+xi+xi+2=2(xi+1)+2xi+2=xi+1
$$
(recall that $2+2=1$ in this field). Similarly for the other pairs.
With your polynomial, the remainders are the same, and the same is the relation.
answered Dec 9 '18 at 22:24
egregegreg
181k1485203
answered Dec 9 '18 at 22:24
egregegreg
181k1485203
answered Dec 9 '18 at 22:24
egregegreg
181k1485203
answered Dec 9 '18 at 22:24
egregegreg
181k1485203
181k1485203
$begingroup$
Ok. And are the cosets the sets ${[v]+2x^2+x+1}$? where $[v]in mathbb{F_3}/(2x^2+x+1)$ So for instance I have a coset of $mathbb{F}_3/(2x^2+x+1)$, which is $(2x+1)+(2x^2+x+1)$?
$endgroup$
– AColoredReptile
Dec 9 '18 at 22:42
$begingroup$
@AColoredReptile You are confusing things; if $vinmathbb{F}_3$, then you have the coset $[v]=v+(2x^2+x+1)$. In my notation, $xi=[x]=x+(2x^2+x+1)$.
$endgroup$
– egreg
Dec 9 '18 at 22:57
$begingroup$
But if I want to write out all the cosets aren't that just my remainders plus the irreducible polynomial?
$endgroup$
– AColoredReptile
Dec 9 '18 at 23:53
$begingroup$
@AColoredReptile No, the cosets are the remainders plus the ideal generated by the irreducible polynomial.
$endgroup$
– egreg
Dec 9 '18 at 23:56
$begingroup$
I dont know what an ideal is. The definition of coset I know is ${v} + W={ v+w: win W}$.
$endgroup$
– AColoredReptile
Dec 10 '18 at 0:05
add a comment |
$begingroup$
Ok. And are the cosets the sets ${[v]+2x^2+x+1}$? where $[v]in mathbb{F_3}/(2x^2+x+1)$ So for instance I have a coset of $mathbb{F}_3/(2x^2+x+1)$, which is $(2x+1)+(2x^2+x+1)$?
$endgroup$
– AColoredReptile
Dec 9 '18 at 22:42
$begingroup$
@AColoredReptile You are confusing things; if $vinmathbb{F}_3$, then you have the coset $[v]=v+(2x^2+x+1)$. In my notation, $xi=[x]=x+(2x^2+x+1)$.
$endgroup$
– egreg
Dec 9 '18 at 22:57
$begingroup$
But if I want to write out all the cosets aren't that just my remainders plus the irreducible polynomial?
$endgroup$
– AColoredReptile
Dec 9 '18 at 23:53
$begingroup$
@AColoredReptile No, the cosets are the remainders plus the ideal generated by the irreducible polynomial.
$endgroup$
– egreg
Dec 9 '18 at 23:56
$begingroup$
I dont know what an ideal is. The definition of coset I know is ${v} + W={ v+w: win W}$.
$endgroup$
– AColoredReptile
Dec 10 '18 at 0:05
$begingroup$
Ok. And are the cosets the sets ${[v]+2x^2+x+1}$? where $[v]in mathbb{F_3}/(2x^2+x+1)$ So for instance I have a coset of $mathbb{F}_3/(2x^2+x+1)$, which is $(2x+1)+(2x^2+x+1)$?
$endgroup$
– AColoredReptile
Dec 9 '18 at 22:42
$begingroup$
Ok. And are the cosets the sets ${[v]+2x^2+x+1}$? where $[v]in mathbb{F_3}/(2x^2+x+1)$ So for instance I have a coset of $mathbb{F}_3/(2x^2+x+1)$, which is $(2x+1)+(2x^2+x+1)$?
$endgroup$
– AColoredReptile
Dec 9 '18 at 22:42
$begingroup$
@AColoredReptile You are confusing things; if $vinmathbb{F}_3$, then you have the coset $[v]=v+(2x^2+x+1)$. In my notation, $xi=[x]=x+(2x^2+x+1)$.
$endgroup$
– egreg
Dec 9 '18 at 22:57
$begingroup$
@AColoredReptile You are confusing things; if $vinmathbb{F}_3$, then you have the coset $[v]=v+(2x^2+x+1)$. In my notation, $xi=[x]=x+(2x^2+x+1)$.
$endgroup$
– egreg
Dec 9 '18 at 22:57
$begingroup$
But if I want to write out all the cosets aren't that just my remainders plus the irreducible polynomial?
$endgroup$
– AColoredReptile
Dec 9 '18 at 23:53
$begingroup$
But if I want to write out all the cosets aren't that just my remainders plus the irreducible polynomial?
$endgroup$
– AColoredReptile
Dec 9 '18 at 23:53
$begingroup$
@AColoredReptile No, the cosets are the remainders plus the ideal generated by the irreducible polynomial.
$endgroup$
– egreg
Dec 9 '18 at 23:56
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@AColoredReptile No, the cosets are the remainders plus the ideal generated by the irreducible polynomial.
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– egreg
Dec 9 '18 at 23:56
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I dont know what an ideal is. The definition of coset I know is ${v} + W={ v+w: win W}$.
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– AColoredReptile
Dec 10 '18 at 0:05
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I dont know what an ideal is. The definition of coset I know is ${v} + W={ v+w: win W}$.
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– AColoredReptile
Dec 10 '18 at 0:05
add a comment |
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You have several remainders with degree 2. How do you get a remainder of degree 2 when what you're dividing by has degree 2?
The short answer is that you don't. You get every single possible first degree (or lower) polynomial as remainders: any such polynomial appears as a remainder, for instance when trying to divide that polynomial itself by $2x+x+1$, and any remainder has at most degree 1 because second degree or higher terms may be divided (possibly with some remainder, but that's the point). There are nine of those polynomials, so those are the elements of your field.
answered Dec 9 '18 at 22:06
ArthurArthur
114k7115197
$endgroup$
add a comment |
$begingroup$
You have several remainders with degree 2. How do you get a remainder of degree 2 when what you're dividing by has degree 2?
The short answer is that you don't. You get every single possible first degree (or lower) polynomial as remainders: any such polynomial appears as a remainder, for instance when trying to divide that polynomial itself by $2x+x+1$, and any remainder has at most degree 1 because second degree or higher terms may be divided (possibly with some remainder, but that's the point). There are nine of those polynomials, so those are the elements of your field.
answered Dec 9 '18 at 22:06
ArthurArthur
114k7115197
$endgroup$
add a comment |
$begingroup$
You have several remainders with degree 2. How do you get a remainder of degree 2 when what you're dividing by has degree 2?
The short answer is that you don't. You get every single possible first degree (or lower) polynomial as remainders: any such polynomial appears as a remainder, for instance when trying to divide that polynomial itself by $2x+x+1$, and any remainder has at most degree 1 because second degree or higher terms may be divided (possibly with some remainder, but that's the point). There are nine of those polynomials, so those are the elements of your field.
answered Dec 9 '18 at 22:06
ArthurArthur
114k7115197
$endgroup$
You have several remainders with degree 2. How do you get a remainder of degree 2 when what you're dividing by has degree 2?
The short answer is that you don't. You get every single possible first degree (or lower) polynomial as remainders: any such polynomial appears as a remainder, for instance when trying to divide that polynomial itself by $2x+x+1$, and any remainder has at most degree 1 because second degree or higher terms may be divided (possibly with some remainder, but that's the point). There are nine of those polynomials, so those are the elements of your field.
answered Dec 9 '18 at 22:06
ArthurArthur
114k7115197
answered Dec 9 '18 at 22:06
ArthurArthur
114k7115197
answered Dec 9 '18 at 22:06
ArthurArthur
114k7115197
answered Dec 9 '18 at 22:06
ArthurArthur
114k7115197
114k7115197
add a comment |
add a comment |
$begingroup$
$F_3[x]/(2x^2+x+1)cong F_3(alpha)$ is the vector space over $F_3$ with basis ${1,alpha}$, where $2alpha^2+alpha +1=0$.
Two polynomials in $F_3[x]$ are equivalent if they differ by an element of $(2x^2+x+1)$ (which is to say by a multiple of $2x^2+x+1$, since $F_3[x]$ is commutative).
Now to complete the exercise you have only to write out the multiplication and addition tables for the $9$ elements.
answered Dec 9 '18 at 22:18
Chris CusterChris Custer
12.7k3825
$endgroup$
add a comment |
$begingroup$
$F_3[x]/(2x^2+x+1)cong F_3(alpha)$ is the vector space over $F_3$ with basis ${1,alpha}$, where $2alpha^2+alpha +1=0$.
Two polynomials in $F_3[x]$ are equivalent if they differ by an element of $(2x^2+x+1)$ (which is to say by a multiple of $2x^2+x+1$, since $F_3[x]$ is commutative).
Now to complete the exercise you have only to write out the multiplication and addition tables for the $9$ elements.
answered Dec 9 '18 at 22:18
Chris CusterChris Custer
12.7k3825
$endgroup$
add a comment |
$begingroup$
$F_3[x]/(2x^2+x+1)cong F_3(alpha)$ is the vector space over $F_3$ with basis ${1,alpha}$, where $2alpha^2+alpha +1=0$.
Two polynomials in $F_3[x]$ are equivalent if they differ by an element of $(2x^2+x+1)$ (which is to say by a multiple of $2x^2+x+1$, since $F_3[x]$ is commutative).
Now to complete the exercise you have only to write out the multiplication and addition tables for the $9$ elements.
answered Dec 9 '18 at 22:18
Chris CusterChris Custer
12.7k3825
$endgroup$
$F_3[x]/(2x^2+x+1)cong F_3(alpha)$ is the vector space over $F_3$ with basis ${1,alpha}$, where $2alpha^2+alpha +1=0$.
Two polynomials in $F_3[x]$ are equivalent if they differ by an element of $(2x^2+x+1)$ (which is to say by a multiple of $2x^2+x+1$, since $F_3[x]$ is commutative).
Now to complete the exercise you have only to write out the multiplication and addition tables for the $9$ elements.
answered Dec 9 '18 at 22:18
Chris CusterChris Custer
12.7k3825
edited Dec 9 '18 at 23:00
answered Dec 9 '18 at 22:18
Chris CusterChris Custer
12.7k3825
answered Dec 9 '18 at 22:18
Chris CusterChris Custer
12.7k3825
answered Dec 9 '18 at 22:18
Chris CusterChris Custer
12.7k3825
12.7k3825
add a comment |
add a comment |
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$begingroup$
As remainders, you obtain all polynomials of degree $le 1$. with coefficients in $mathbf F_3$. This makes $9$ polynomials.
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– Bernard
Dec 9 '18 at 22:03
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You're making your life difficult: why not taking $x^2-x-1$?
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– egreg
Dec 9 '18 at 22:17
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@egreg this was the first polynomial I could think of. Is there someway to get an irreducible polynomial without checking it?
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– AColoredReptile
Dec 9 '18 at 22:19
1
$begingroup$
@AcoloredReptile: I think @egreg’s point is that it’s the same polynomial up to a constant! If you divide through by $[2]$, which is the same thing as multiplying through by $[2]$, you get $x^2 +[2]x+[2] = x^2 +[-1]x + [-1] = x^2-x-1$. You’ll need to check the polynomial, but you should always try to use a monic one, because it makes calculations easier.
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– Arturo Magidin
Dec 9 '18 at 22:21