Mixing
What is $(x+1)^4$ in $GF(3)[x]^*_{x^2+1}$
$begingroup$
Consider the field $GF(3)[x]^*_{x^2+1}$. My textbook says that $(x+1)^4 = 2$ in this field, but i don't understand how mulitplication works in this field
With $GF(3)[x]^*_{x^2+1} = { a(x) in GF(3)[x]_{x^2+1} | gcd(a(x),x^2+1= 1 }$
abstract-algebra
asked Jan 16 at 10:41
eraugeraug
12
$endgroup$
add a comment |
$begingroup$
Consider the field $GF(3)[x]^*_{x^2+1}$. My textbook says that $(x+1)^4 = 2$ in this field, but i don't understand how mulitplication works in this field
With $GF(3)[x]^*_{x^2+1} = { a(x) in GF(3)[x]_{x^2+1} | gcd(a(x),x^2+1= 1 }$
abstract-algebra
asked Jan 16 at 10:41
eraugeraug
12
$endgroup$
$begingroup$
What does the $^*$ mean?
$endgroup$
– Arthur
Jan 16 at 10:45
$begingroup$
The notation is, IMHO, non-standard, I would write this ring as $GF(3)[x]/langle x^2+1rangle$ (it's a field actually). But anyway, you are expected to give the remainder of a polynomial division by $x^2+1$. Remember that in the quotient ring $x^2+1$ and all its polynomial multiples become equated with zero. So $$(x+1)^2=x^2+2x+1=2x+1cdot(x^2+1)equiv 2xpmod{x^2+1}.$$ Now you should be able to do your exercise starting with $(x+1)^4=((x+1)^2)^2equiv(2x)^2=cdots$. Remember that you are to also reduce coefficient (but not exponents) modulo three, because in $GF(3)$ we have $3=0$.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 10:47
$begingroup$
Also, Welcome to MSE! Please also spend a while studying our guide for new askers. Some users are known to be rather unforgiving about shortcomings in questions that look like homework assignments. If you can use the hints/pointers in my first comment as means for either improving the question and/or answering it yourself, I will be that much happier.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 10:51
$begingroup$
@Arthur i added the meaning of this notation
$endgroup$
– eraug
Jan 16 at 11:00
$begingroup$
That's not a field. Without the $^*$ it's a field, but with it you don't have a $0$, so it's just a group under multiplication.
$endgroup$
– Arthur
Jan 16 at 11:10
add a comment |
$begingroup$
Consider the field $GF(3)[x]^*_{x^2+1}$. My textbook says that $(x+1)^4 = 2$ in this field, but i don't understand how mulitplication works in this field
With $GF(3)[x]^*_{x^2+1} = { a(x) in GF(3)[x]_{x^2+1} | gcd(a(x),x^2+1= 1 }$
abstract-algebra
asked Jan 16 at 10:41
eraugeraug
12
$endgroup$
Consider the field $GF(3)[x]^*_{x^2+1}$. My textbook says that $(x+1)^4 = 2$ in this field, but i don't understand how mulitplication works in this field
With $GF(3)[x]^*_{x^2+1} = { a(x) in GF(3)[x]_{x^2+1} | gcd(a(x),x^2+1= 1 }$
abstract-algebra
abstract-algebra
asked Jan 16 at 10:41
eraugeraug
12
asked Jan 16 at 10:41
eraugeraug
12
edited Jan 16 at 10:59
eraug
asked Jan 16 at 10:41
eraugeraug
12
asked Jan 16 at 10:41
eraugeraug
12
asked Jan 16 at 10:41
eraugeraug
12
12
$begingroup$
What does the $^*$ mean?
$endgroup$
– Arthur
Jan 16 at 10:45
$begingroup$
The notation is, IMHO, non-standard, I would write this ring as $GF(3)[x]/langle x^2+1rangle$ (it's a field actually). But anyway, you are expected to give the remainder of a polynomial division by $x^2+1$. Remember that in the quotient ring $x^2+1$ and all its polynomial multiples become equated with zero. So $$(x+1)^2=x^2+2x+1=2x+1cdot(x^2+1)equiv 2xpmod{x^2+1}.$$ Now you should be able to do your exercise starting with $(x+1)^4=((x+1)^2)^2equiv(2x)^2=cdots$. Remember that you are to also reduce coefficient (but not exponents) modulo three, because in $GF(3)$ we have $3=0$.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 10:47
$begingroup$
Also, Welcome to MSE! Please also spend a while studying our guide for new askers. Some users are known to be rather unforgiving about shortcomings in questions that look like homework assignments. If you can use the hints/pointers in my first comment as means for either improving the question and/or answering it yourself, I will be that much happier.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 10:51
$begingroup$
@Arthur i added the meaning of this notation
$endgroup$
– eraug
Jan 16 at 11:00
$begingroup$
That's not a field. Without the $^*$ it's a field, but with it you don't have a $0$, so it's just a group under multiplication.
$endgroup$
– Arthur
Jan 16 at 11:10
add a comment |
$begingroup$
What does the $^*$ mean?
$endgroup$
– Arthur
Jan 16 at 10:45
$begingroup$
The notation is, IMHO, non-standard, I would write this ring as $GF(3)[x]/langle x^2+1rangle$ (it's a field actually). But anyway, you are expected to give the remainder of a polynomial division by $x^2+1$. Remember that in the quotient ring $x^2+1$ and all its polynomial multiples become equated with zero. So $$(x+1)^2=x^2+2x+1=2x+1cdot(x^2+1)equiv 2xpmod{x^2+1}.$$ Now you should be able to do your exercise starting with $(x+1)^4=((x+1)^2)^2equiv(2x)^2=cdots$. Remember that you are to also reduce coefficient (but not exponents) modulo three, because in $GF(3)$ we have $3=0$.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 10:47
$begingroup$
Also, Welcome to MSE! Please also spend a while studying our guide for new askers. Some users are known to be rather unforgiving about shortcomings in questions that look like homework assignments. If you can use the hints/pointers in my first comment as means for either improving the question and/or answering it yourself, I will be that much happier.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 10:51
$begingroup$
@Arthur i added the meaning of this notation
$endgroup$
– eraug
Jan 16 at 11:00
$begingroup$
That's not a field. Without the $^*$ it's a field, but with it you don't have a $0$, so it's just a group under multiplication.
$endgroup$
– Arthur
Jan 16 at 11:10
$begingroup$
What does the $^*$ mean?
$endgroup$
– Arthur
Jan 16 at 10:45
$begingroup$
What does the $^*$ mean?
$endgroup$
– Arthur
Jan 16 at 10:45
$begingroup$
The notation is, IMHO, non-standard, I would write this ring as $GF(3)[x]/langle x^2+1rangle$ (it's a field actually). But anyway, you are expected to give the remainder of a polynomial division by $x^2+1$. Remember that in the quotient ring $x^2+1$ and all its polynomial multiples become equated with zero. So $$(x+1)^2=x^2+2x+1=2x+1cdot(x^2+1)equiv 2xpmod{x^2+1}.$$ Now you should be able to do your exercise starting with $(x+1)^4=((x+1)^2)^2equiv(2x)^2=cdots$. Remember that you are to also reduce coefficient (but not exponents) modulo three, because in $GF(3)$ we have $3=0$.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 10:47
$begingroup$
The notation is, IMHO, non-standard, I would write this ring as $GF(3)[x]/langle x^2+1rangle$ (it's a field actually). But anyway, you are expected to give the remainder of a polynomial division by $x^2+1$. Remember that in the quotient ring $x^2+1$ and all its polynomial multiples become equated with zero. So $$(x+1)^2=x^2+2x+1=2x+1cdot(x^2+1)equiv 2xpmod{x^2+1}.$$ Now you should be able to do your exercise starting with $(x+1)^4=((x+1)^2)^2equiv(2x)^2=cdots$. Remember that you are to also reduce coefficient (but not exponents) modulo three, because in $GF(3)$ we have $3=0$.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 10:47
$begingroup$
Also, Welcome to MSE! Please also spend a while studying our guide for new askers. Some users are known to be rather unforgiving about shortcomings in questions that look like homework assignments. If you can use the hints/pointers in my first comment as means for either improving the question and/or answering it yourself, I will be that much happier.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 10:51
$begingroup$
Also, Welcome to MSE! Please also spend a while studying our guide for new askers. Some users are known to be rather unforgiving about shortcomings in questions that look like homework assignments. If you can use the hints/pointers in my first comment as means for either improving the question and/or answering it yourself, I will be that much happier.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 10:51
$begingroup$
@Arthur i added the meaning of this notation
$endgroup$
– eraug
Jan 16 at 11:00
$begingroup$
@Arthur i added the meaning of this notation
$endgroup$
– eraug
Jan 16 at 11:00
$begingroup$
That's not a field. Without the $^*$ it's a field, but with it you don't have a $0$, so it's just a group under multiplication.
$endgroup$
– Arthur
Jan 16 at 11:10
$begingroup$
That's not a field. Without the $^*$ it's a field, but with it you don't have a $0$, so it's just a group under multiplication.
$endgroup$
– Arthur
Jan 16 at 11:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Assuming you mean what I would call $Bbb F_3[x]/(x^2+1)$, i.e. take a polynomial ring over the field of three elements, and divide out by the ideal generated by $x^2+1$, then this is a field with nine elements:
$$begin{matrix}
0&1&2\x&x+1&x+2\2x&2x+1&2x+2end{matrix}
$$
and multiplication is simple: First you multiply two elements as you would any polynomial. Then you take any $x^2$ you see, and you swap it out with $2$ (note: $x^2+1 = 0$ means $x^2=2$ in $Bbb F_3$, so $x^2$ is, in the quotient ring, literally the same as $2$).
For instance, we have
$$
(2x+1)cdot (x+2) = 2x^2 + 2x + 2\
= 2cdot 2 + 2x + 2 = 2x
$$
In your example, using the binomial theorem (in $Bbb F_3$ we have $binom41 = 1$ and $binom42 = 0$), we get
$$
(x + 1)^4 = x^4 + x^3 + x + 1\
= (x^2)^2 + x^2cdot x + x + 1\
= 2^2 + 2x + x + 1\
= 2
$$
answered Jan 16 at 10:52
ArthurArthur
116k7116198
$endgroup$
$begingroup$
As an aside, I have no idea why teachers and books aren't using this approach to introduce complex numbers (i.e. highlight $i^2 = -1$ rather than $i = sqrt{-1}$).
$endgroup$
– Arthur
Jan 16 at 10:55
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
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votes
$begingroup$
Assuming you mean what I would call $Bbb F_3[x]/(x^2+1)$, i.e. take a polynomial ring over the field of three elements, and divide out by the ideal generated by $x^2+1$, then this is a field with nine elements:
$$begin{matrix}
0&1&2\x&x+1&x+2\2x&2x+1&2x+2end{matrix}
$$
and multiplication is simple: First you multiply two elements as you would any polynomial. Then you take any $x^2$ you see, and you swap it out with $2$ (note: $x^2+1 = 0$ means $x^2=2$ in $Bbb F_3$, so $x^2$ is, in the quotient ring, literally the same as $2$).
For instance, we have
$$
(2x+1)cdot (x+2) = 2x^2 + 2x + 2\
= 2cdot 2 + 2x + 2 = 2x
$$
In your example, using the binomial theorem (in $Bbb F_3$ we have $binom41 = 1$ and $binom42 = 0$), we get
$$
(x + 1)^4 = x^4 + x^3 + x + 1\
= (x^2)^2 + x^2cdot x + x + 1\
= 2^2 + 2x + x + 1\
= 2
$$
answered Jan 16 at 10:52
ArthurArthur
116k7116198
$endgroup$
$begingroup$
As an aside, I have no idea why teachers and books aren't using this approach to introduce complex numbers (i.e. highlight $i^2 = -1$ rather than $i = sqrt{-1}$).
$endgroup$
– Arthur
Jan 16 at 10:55
add a comment |
$begingroup$
Assuming you mean what I would call $Bbb F_3[x]/(x^2+1)$, i.e. take a polynomial ring over the field of three elements, and divide out by the ideal generated by $x^2+1$, then this is a field with nine elements:
$$begin{matrix}
0&1&2\x&x+1&x+2\2x&2x+1&2x+2end{matrix}
$$
and multiplication is simple: First you multiply two elements as you would any polynomial. Then you take any $x^2$ you see, and you swap it out with $2$ (note: $x^2+1 = 0$ means $x^2=2$ in $Bbb F_3$, so $x^2$ is, in the quotient ring, literally the same as $2$).
For instance, we have
$$
(2x+1)cdot (x+2) = 2x^2 + 2x + 2\
= 2cdot 2 + 2x + 2 = 2x
$$
In your example, using the binomial theorem (in $Bbb F_3$ we have $binom41 = 1$ and $binom42 = 0$), we get
$$
(x + 1)^4 = x^4 + x^3 + x + 1\
= (x^2)^2 + x^2cdot x + x + 1\
= 2^2 + 2x + x + 1\
= 2
$$
answered Jan 16 at 10:52
ArthurArthur
116k7116198
$endgroup$
$begingroup$
As an aside, I have no idea why teachers and books aren't using this approach to introduce complex numbers (i.e. highlight $i^2 = -1$ rather than $i = sqrt{-1}$).
$endgroup$
– Arthur
Jan 16 at 10:55
add a comment |
$begingroup$
Assuming you mean what I would call $Bbb F_3[x]/(x^2+1)$, i.e. take a polynomial ring over the field of three elements, and divide out by the ideal generated by $x^2+1$, then this is a field with nine elements:
$$begin{matrix}
0&1&2\x&x+1&x+2\2x&2x+1&2x+2end{matrix}
$$
and multiplication is simple: First you multiply two elements as you would any polynomial. Then you take any $x^2$ you see, and you swap it out with $2$ (note: $x^2+1 = 0$ means $x^2=2$ in $Bbb F_3$, so $x^2$ is, in the quotient ring, literally the same as $2$).
For instance, we have
$$
(2x+1)cdot (x+2) = 2x^2 + 2x + 2\
= 2cdot 2 + 2x + 2 = 2x
$$
In your example, using the binomial theorem (in $Bbb F_3$ we have $binom41 = 1$ and $binom42 = 0$), we get
$$
(x + 1)^4 = x^4 + x^3 + x + 1\
= (x^2)^2 + x^2cdot x + x + 1\
= 2^2 + 2x + x + 1\
= 2
$$
answered Jan 16 at 10:52
ArthurArthur
116k7116198
$endgroup$
Assuming you mean what I would call $Bbb F_3[x]/(x^2+1)$, i.e. take a polynomial ring over the field of three elements, and divide out by the ideal generated by $x^2+1$, then this is a field with nine elements:
$$begin{matrix}
0&1&2\x&x+1&x+2\2x&2x+1&2x+2end{matrix}
$$
and multiplication is simple: First you multiply two elements as you would any polynomial. Then you take any $x^2$ you see, and you swap it out with $2$ (note: $x^2+1 = 0$ means $x^2=2$ in $Bbb F_3$, so $x^2$ is, in the quotient ring, literally the same as $2$).
For instance, we have
$$
(2x+1)cdot (x+2) = 2x^2 + 2x + 2\
= 2cdot 2 + 2x + 2 = 2x
$$
In your example, using the binomial theorem (in $Bbb F_3$ we have $binom41 = 1$ and $binom42 = 0$), we get
$$
(x + 1)^4 = x^4 + x^3 + x + 1\
= (x^2)^2 + x^2cdot x + x + 1\
= 2^2 + 2x + x + 1\
= 2
$$
answered Jan 16 at 10:52
ArthurArthur
116k7116198
answered Jan 16 at 10:52
ArthurArthur
116k7116198
answered Jan 16 at 10:52
ArthurArthur
116k7116198
answered Jan 16 at 10:52
ArthurArthur
116k7116198
116k7116198
$begingroup$
As an aside, I have no idea why teachers and books aren't using this approach to introduce complex numbers (i.e. highlight $i^2 = -1$ rather than $i = sqrt{-1}$).
$endgroup$
– Arthur
Jan 16 at 10:55
add a comment |
$begingroup$
As an aside, I have no idea why teachers and books aren't using this approach to introduce complex numbers (i.e. highlight $i^2 = -1$ rather than $i = sqrt{-1}$).
$endgroup$
– Arthur
Jan 16 at 10:55
$begingroup$
As an aside, I have no idea why teachers and books aren't using this approach to introduce complex numbers (i.e. highlight $i^2 = -1$ rather than $i = sqrt{-1}$).
$endgroup$
– Arthur
Jan 16 at 10:55
$begingroup$
As an aside, I have no idea why teachers and books aren't using this approach to introduce complex numbers (i.e. highlight $i^2 = -1$ rather than $i = sqrt{-1}$).
$endgroup$
– Arthur
Jan 16 at 10:55
add a comment |
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$begingroup$
What does the $^*$ mean?
$endgroup$
– Arthur
Jan 16 at 10:45
$begingroup$
The notation is, IMHO, non-standard, I would write this ring as $GF(3)[x]/langle x^2+1rangle$ (it's a field actually). But anyway, you are expected to give the remainder of a polynomial division by $x^2+1$. Remember that in the quotient ring $x^2+1$ and all its polynomial multiples become equated with zero. So $$(x+1)^2=x^2+2x+1=2x+1cdot(x^2+1)equiv 2xpmod{x^2+1}.$$ Now you should be able to do your exercise starting with $(x+1)^4=((x+1)^2)^2equiv(2x)^2=cdots$. Remember that you are to also reduce coefficient (but not exponents) modulo three, because in $GF(3)$ we have $3=0$.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 10:47
$begingroup$
Also, Welcome to MSE! Please also spend a while studying our guide for new askers. Some users are known to be rather unforgiving about shortcomings in questions that look like homework assignments. If you can use the hints/pointers in my first comment as means for either improving the question and/or answering it yourself, I will be that much happier.
$endgroup$
– Jyrki Lahtonen
Jan 16 at 10:51
$begingroup$
@Arthur i added the meaning of this notation
$endgroup$
– eraug
Jan 16 at 11:00
$begingroup$
That's not a field. Without the $^*$ it's a field, but with it you don't have a $0$, so it's just a group under multiplication.
$endgroup$
– Arthur
Jan 16 at 11:10