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How to write a polynomial in mod p
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Consider the polynomial $ x^3 - 3 x^2 +2 x -1$
How can this polynomial be written in mod 3 ?
What confuses me is that I thought we don't change any power . If we have to also consider the powers then will the leading coefficient be equal to zero then.
elementary-number-theory polynomials modular-arithmetic
asked Aug 9 at 22:37
Rosa1
686
add a comment |Â
up vote
3
down vote
favorite
Consider the polynomial $ x^3 - 3 x^2 +2 x -1$
How can this polynomial be written in mod 3 ?
What confuses me is that I thought we don't change any power . If we have to also consider the powers then will the leading coefficient be equal to zero then.
elementary-number-theory polynomials modular-arithmetic
asked Aug 9 at 22:37
Rosa1
686
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Consider the polynomial $ x^3 - 3 x^2 +2 x -1$
How can this polynomial be written in mod 3 ?
What confuses me is that I thought we don't change any power . If we have to also consider the powers then will the leading coefficient be equal to zero then.
elementary-number-theory polynomials modular-arithmetic
asked Aug 9 at 22:37
Rosa1
686
Consider the polynomial $ x^3 - 3 x^2 +2 x -1$
How can this polynomial be written in mod 3 ?
What confuses me is that I thought we don't change any power . If we have to also consider the powers then will the leading coefficient be equal to zero then.
elementary-number-theory polynomials modular-arithmetic
asked Aug 9 at 22:37
Rosa1
686
edited Aug 9 at 23:13
asked Aug 9 at 22:37
Rosa1
686
asked Aug 9 at 22:37
Rosa1
686
asked Aug 9 at 22:37
Rosa1
686
686
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
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up vote
2
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accepted
If $x$ is an integer, then
beginalign*
&x^3-3x^2+2x-1\[4pt]
equiv;&x^3-0x^2+2x-1;(textmod;3)\[4pt]
equiv;&x^3+2x-1;(textmod;3)\[4pt]
equiv;&x+2x-1;(textmod;3)
qquadtext[since $x^3equiv x;(textmod;3)$]\[4pt]
equiv;&3x-1;(textmod;3)\[4pt]
equiv;&0x-1;(textmod;3)\[4pt]
equiv;&-1;(textmod;3)\[4pt]
equiv;&2;(textmod;3)\[4pt]
endalign*
If $x$ is just an indeterminate, then you can't replace $x^3$ by $x$, but you can replace coefficients by their residues, mod $3$, hence, with that interpretation, you can get
beginalign*
&x^3-3x^2+2x-1
qquadqquadqquadqquadqquad;;;;;;,
\[4pt]
equiv;&x^3-0x^2+2x-1;(textmod;3)\[4pt]
equiv;&x^3+2x-1;(textmod;3)\[4pt]
equiv;&x^3+2x+2;(textmod;3)\[4pt]
endalign*
answered Aug 9 at 22:52
quasi
33.9k22461
actually in my note it is right up to your 4th line before "since" but I could not understand how we get x from $ x^3 $ although we should not change the powers
– Rosa1
Aug 9 at 23:11
If $x$ is intended as an indeterminate, then you can't change powers. If $x$ is just an unknown integer, then $x^3equiv x;(textmod;3)$ by Fermat's little Theorem, so with that interpretation, $x^3$ can be replaced by $x$. Thus, it depends on the intended interpretation of $x$. If you really meant polynomial, then $x$ should be regarded as an indeterminate, so in that case, you can't replace $x^3$ by $x$. In my answer, I showed both interpretations.
– quasi
Aug 9 at 23:21
add a comment |Â
up vote
4
down vote
What confuses me is that I thought we don't change any power . If we have to also consider the powers then will the leading coefficient be equal to zero then.
Your first thought is correct: we don't change any power. Even though $3$ is in the exponent, we don't take the exponents mod 3, but only the coefficients.
So, taking a polynomial mod 3 just means taking each coefficient mod 3 (and leaving the exponents). For example, $6x^3 + 4x^2 - 2 mod 3$ would be $0x^3 + x^2 + 1$, which is the same as $x^2 + 1$ since we can drop $0$s. This is because $6 mod 3$ is $0$, $4 mod 3$ is $1$, and $-2 mod 3$ is $1$.
answered Aug 9 at 22:46
6005
34.9k750124
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If $x$ is an integer, then
beginalign*
&x^3-3x^2+2x-1\[4pt]
equiv;&x^3-0x^2+2x-1;(textmod;3)\[4pt]
equiv;&x^3+2x-1;(textmod;3)\[4pt]
equiv;&x+2x-1;(textmod;3)
qquadtext[since $x^3equiv x;(textmod;3)$]\[4pt]
equiv;&3x-1;(textmod;3)\[4pt]
equiv;&0x-1;(textmod;3)\[4pt]
equiv;&-1;(textmod;3)\[4pt]
equiv;&2;(textmod;3)\[4pt]
endalign*
If $x$ is just an indeterminate, then you can't replace $x^3$ by $x$, but you can replace coefficients by their residues, mod $3$, hence, with that interpretation, you can get
beginalign*
&x^3-3x^2+2x-1
qquadqquadqquadqquadqquad;;;;;;,
\[4pt]
equiv;&x^3-0x^2+2x-1;(textmod;3)\[4pt]
equiv;&x^3+2x-1;(textmod;3)\[4pt]
equiv;&x^3+2x+2;(textmod;3)\[4pt]
endalign*
answered Aug 9 at 22:52
quasi
33.9k22461
actually in my note it is right up to your 4th line before "since" but I could not understand how we get x from $ x^3 $ although we should not change the powers
– Rosa1
Aug 9 at 23:11
If $x$ is intended as an indeterminate, then you can't change powers. If $x$ is just an unknown integer, then $x^3equiv x;(textmod;3)$ by Fermat's little Theorem, so with that interpretation, $x^3$ can be replaced by $x$. Thus, it depends on the intended interpretation of $x$. If you really meant polynomial, then $x$ should be regarded as an indeterminate, so in that case, you can't replace $x^3$ by $x$. In my answer, I showed both interpretations.
– quasi
Aug 9 at 23:21
add a comment |Â
up vote
2
down vote
accepted
If $x$ is an integer, then
beginalign*
&x^3-3x^2+2x-1\[4pt]
equiv;&x^3-0x^2+2x-1;(textmod;3)\[4pt]
equiv;&x^3+2x-1;(textmod;3)\[4pt]
equiv;&x+2x-1;(textmod;3)
qquadtext[since $x^3equiv x;(textmod;3)$]\[4pt]
equiv;&3x-1;(textmod;3)\[4pt]
equiv;&0x-1;(textmod;3)\[4pt]
equiv;&-1;(textmod;3)\[4pt]
equiv;&2;(textmod;3)\[4pt]
endalign*
If $x$ is just an indeterminate, then you can't replace $x^3$ by $x$, but you can replace coefficients by their residues, mod $3$, hence, with that interpretation, you can get
beginalign*
&x^3-3x^2+2x-1
qquadqquadqquadqquadqquad;;;;;;,
\[4pt]
equiv;&x^3-0x^2+2x-1;(textmod;3)\[4pt]
equiv;&x^3+2x-1;(textmod;3)\[4pt]
equiv;&x^3+2x+2;(textmod;3)\[4pt]
endalign*
answered Aug 9 at 22:52
quasi
33.9k22461
actually in my note it is right up to your 4th line before "since" but I could not understand how we get x from $ x^3 $ although we should not change the powers
– Rosa1
Aug 9 at 23:11
If $x$ is intended as an indeterminate, then you can't change powers. If $x$ is just an unknown integer, then $x^3equiv x;(textmod;3)$ by Fermat's little Theorem, so with that interpretation, $x^3$ can be replaced by $x$. Thus, it depends on the intended interpretation of $x$. If you really meant polynomial, then $x$ should be regarded as an indeterminate, so in that case, you can't replace $x^3$ by $x$. In my answer, I showed both interpretations.
– quasi
Aug 9 at 23:21
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If $x$ is an integer, then
beginalign*
&x^3-3x^2+2x-1\[4pt]
equiv;&x^3-0x^2+2x-1;(textmod;3)\[4pt]
equiv;&x^3+2x-1;(textmod;3)\[4pt]
equiv;&x+2x-1;(textmod;3)
qquadtext[since $x^3equiv x;(textmod;3)$]\[4pt]
equiv;&3x-1;(textmod;3)\[4pt]
equiv;&0x-1;(textmod;3)\[4pt]
equiv;&-1;(textmod;3)\[4pt]
equiv;&2;(textmod;3)\[4pt]
endalign*
If $x$ is just an indeterminate, then you can't replace $x^3$ by $x$, but you can replace coefficients by their residues, mod $3$, hence, with that interpretation, you can get
beginalign*
&x^3-3x^2+2x-1
qquadqquadqquadqquadqquad;;;;;;,
\[4pt]
equiv;&x^3-0x^2+2x-1;(textmod;3)\[4pt]
equiv;&x^3+2x-1;(textmod;3)\[4pt]
equiv;&x^3+2x+2;(textmod;3)\[4pt]
endalign*
answered Aug 9 at 22:52
quasi
33.9k22461
If $x$ is an integer, then
beginalign*
&x^3-3x^2+2x-1\[4pt]
equiv;&x^3-0x^2+2x-1;(textmod;3)\[4pt]
equiv;&x^3+2x-1;(textmod;3)\[4pt]
equiv;&x+2x-1;(textmod;3)
qquadtext[since $x^3equiv x;(textmod;3)$]\[4pt]
equiv;&3x-1;(textmod;3)\[4pt]
equiv;&0x-1;(textmod;3)\[4pt]
equiv;&-1;(textmod;3)\[4pt]
equiv;&2;(textmod;3)\[4pt]
endalign*
If $x$ is just an indeterminate, then you can't replace $x^3$ by $x$, but you can replace coefficients by their residues, mod $3$, hence, with that interpretation, you can get
beginalign*
&x^3-3x^2+2x-1
qquadqquadqquadqquadqquad;;;;;;,
\[4pt]
equiv;&x^3-0x^2+2x-1;(textmod;3)\[4pt]
equiv;&x^3+2x-1;(textmod;3)\[4pt]
equiv;&x^3+2x+2;(textmod;3)\[4pt]
endalign*
answered Aug 9 at 22:52
quasi
33.9k22461
edited Aug 9 at 22:59
answered Aug 9 at 22:52
quasi
33.9k22461
answered Aug 9 at 22:52
quasi
33.9k22461
answered Aug 9 at 22:52
quasi
33.9k22461
33.9k22461
actually in my note it is right up to your 4th line before "since" but I could not understand how we get x from $ x^3 $ although we should not change the powers
– Rosa1
Aug 9 at 23:11
If $x$ is intended as an indeterminate, then you can't change powers. If $x$ is just an unknown integer, then $x^3equiv x;(textmod;3)$ by Fermat's little Theorem, so with that interpretation, $x^3$ can be replaced by $x$. Thus, it depends on the intended interpretation of $x$. If you really meant polynomial, then $x$ should be regarded as an indeterminate, so in that case, you can't replace $x^3$ by $x$. In my answer, I showed both interpretations.
– quasi
Aug 9 at 23:21
add a comment |Â
actually in my note it is right up to your 4th line before "since" but I could not understand how we get x from $ x^3 $ although we should not change the powers
– Rosa1
Aug 9 at 23:11
If $x$ is intended as an indeterminate, then you can't change powers. If $x$ is just an unknown integer, then $x^3equiv x;(textmod;3)$ by Fermat's little Theorem, so with that interpretation, $x^3$ can be replaced by $x$. Thus, it depends on the intended interpretation of $x$. If you really meant polynomial, then $x$ should be regarded as an indeterminate, so in that case, you can't replace $x^3$ by $x$. In my answer, I showed both interpretations.
– quasi
Aug 9 at 23:21
actually in my note it is right up to your 4th line before "since" but I could not understand how we get x from $ x^3 $ although we should not change the powers
– Rosa1
Aug 9 at 23:11
actually in my note it is right up to your 4th line before "since" but I could not understand how we get x from $ x^3 $ although we should not change the powers
– Rosa1
Aug 9 at 23:11
If $x$ is intended as an indeterminate, then you can't change powers. If $x$ is just an unknown integer, then $x^3equiv x;(textmod;3)$ by Fermat's little Theorem, so with that interpretation, $x^3$ can be replaced by $x$. Thus, it depends on the intended interpretation of $x$. If you really meant polynomial, then $x$ should be regarded as an indeterminate, so in that case, you can't replace $x^3$ by $x$. In my answer, I showed both interpretations.
– quasi
Aug 9 at 23:21
If $x$ is intended as an indeterminate, then you can't change powers. If $x$ is just an unknown integer, then $x^3equiv x;(textmod;3)$ by Fermat's little Theorem, so with that interpretation, $x^3$ can be replaced by $x$. Thus, it depends on the intended interpretation of $x$. If you really meant polynomial, then $x$ should be regarded as an indeterminate, so in that case, you can't replace $x^3$ by $x$. In my answer, I showed both interpretations.
– quasi
Aug 9 at 23:21
add a comment |Â
up vote
4
down vote
What confuses me is that I thought we don't change any power . If we have to also consider the powers then will the leading coefficient be equal to zero then.
Your first thought is correct: we don't change any power. Even though $3$ is in the exponent, we don't take the exponents mod 3, but only the coefficients.
So, taking a polynomial mod 3 just means taking each coefficient mod 3 (and leaving the exponents). For example, $6x^3 + 4x^2 - 2 mod 3$ would be $0x^3 + x^2 + 1$, which is the same as $x^2 + 1$ since we can drop $0$s. This is because $6 mod 3$ is $0$, $4 mod 3$ is $1$, and $-2 mod 3$ is $1$.
answered Aug 9 at 22:46
6005
34.9k750124
add a comment |Â
up vote
4
down vote
What confuses me is that I thought we don't change any power . If we have to also consider the powers then will the leading coefficient be equal to zero then.
Your first thought is correct: we don't change any power. Even though $3$ is in the exponent, we don't take the exponents mod 3, but only the coefficients.
So, taking a polynomial mod 3 just means taking each coefficient mod 3 (and leaving the exponents). For example, $6x^3 + 4x^2 - 2 mod 3$ would be $0x^3 + x^2 + 1$, which is the same as $x^2 + 1$ since we can drop $0$s. This is because $6 mod 3$ is $0$, $4 mod 3$ is $1$, and $-2 mod 3$ is $1$.
answered Aug 9 at 22:46
6005
34.9k750124
add a comment |Â
up vote
4
down vote
up vote
4
down vote
What confuses me is that I thought we don't change any power . If we have to also consider the powers then will the leading coefficient be equal to zero then.
Your first thought is correct: we don't change any power. Even though $3$ is in the exponent, we don't take the exponents mod 3, but only the coefficients.
So, taking a polynomial mod 3 just means taking each coefficient mod 3 (and leaving the exponents). For example, $6x^3 + 4x^2 - 2 mod 3$ would be $0x^3 + x^2 + 1$, which is the same as $x^2 + 1$ since we can drop $0$s. This is because $6 mod 3$ is $0$, $4 mod 3$ is $1$, and $-2 mod 3$ is $1$.
answered Aug 9 at 22:46
6005
34.9k750124
What confuses me is that I thought we don't change any power . If we have to also consider the powers then will the leading coefficient be equal to zero then.
Your first thought is correct: we don't change any power. Even though $3$ is in the exponent, we don't take the exponents mod 3, but only the coefficients.
So, taking a polynomial mod 3 just means taking each coefficient mod 3 (and leaving the exponents). For example, $6x^3 + 4x^2 - 2 mod 3$ would be $0x^3 + x^2 + 1$, which is the same as $x^2 + 1$ since we can drop $0$s. This is because $6 mod 3$ is $0$, $4 mod 3$ is $1$, and $-2 mod 3$ is $1$.
answered Aug 9 at 22:46
6005
34.9k750124
answered Aug 9 at 22:46
6005
34.9k750124
answered Aug 9 at 22:46
6005
34.9k750124
answered Aug 9 at 22:46
6005
34.9k750124
34.9k750124
add a comment |Â
add a comment |Â
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