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Find all zeros of $5x^3 + 4x^2 + 11x + 9 in mathbb{Z}_{12}[x]$
I want to find all of the zeros of the polynomial $5x^3 + 4x^2 + 11x + 9 in mathbb{Z}_{12}[x]$. I know that one simple way would be to check every value in the set $x={0,1,...,11}$, and when I did this I found that there were no zeros.
However, is there a more elegant way to show this? I was thinking that I could try to show the polynomial is irreducible, and then use that to help show there are no roots, but couldn't see a good way to show irreducibility in $mathbb{Z}_{12}[x]$
Thanks for any help you might have!
abstract-algebra polynomials ring-theory
asked Nov 19 '18 at 1:52
rocketPowered
224
add a comment |
I want to find all of the zeros of the polynomial $5x^3 + 4x^2 + 11x + 9 in mathbb{Z}_{12}[x]$. I know that one simple way would be to check every value in the set $x={0,1,...,11}$, and when I did this I found that there were no zeros.
However, is there a more elegant way to show this? I was thinking that I could try to show the polynomial is irreducible, and then use that to help show there are no roots, but couldn't see a good way to show irreducibility in $mathbb{Z}_{12}[x]$
Thanks for any help you might have!
abstract-algebra polynomials ring-theory
asked Nov 19 '18 at 1:52
rocketPowered
224
It is not over a field so, factor theorem does not hold
– Sorfosh
Nov 19 '18 at 2:14
Looks like you want $mathbb{Z}/12mathbb{Z}$, rather than $mathbb{Z}_{12}=varprojlim(mathbb{Z}/12^nmathbb{Z})$.
– user10354138
Nov 19 '18 at 2:23
add a comment |
I want to find all of the zeros of the polynomial $5x^3 + 4x^2 + 11x + 9 in mathbb{Z}_{12}[x]$. I know that one simple way would be to check every value in the set $x={0,1,...,11}$, and when I did this I found that there were no zeros.
However, is there a more elegant way to show this? I was thinking that I could try to show the polynomial is irreducible, and then use that to help show there are no roots, but couldn't see a good way to show irreducibility in $mathbb{Z}_{12}[x]$
Thanks for any help you might have!
abstract-algebra polynomials ring-theory
asked Nov 19 '18 at 1:52
rocketPowered
224
I want to find all of the zeros of the polynomial $5x^3 + 4x^2 + 11x + 9 in mathbb{Z}_{12}[x]$. I know that one simple way would be to check every value in the set $x={0,1,...,11}$, and when I did this I found that there were no zeros.
However, is there a more elegant way to show this? I was thinking that I could try to show the polynomial is irreducible, and then use that to help show there are no roots, but couldn't see a good way to show irreducibility in $mathbb{Z}_{12}[x]$
Thanks for any help you might have!
abstract-algebra polynomials ring-theory
abstract-algebra polynomials ring-theory
asked Nov 19 '18 at 1:52
rocketPowered
224
asked Nov 19 '18 at 1:52
rocketPowered
224
asked Nov 19 '18 at 1:52
rocketPowered
224
asked Nov 19 '18 at 1:52
rocketPowered
224
asked Nov 19 '18 at 1:52
rocketPowered
224
224
It is not over a field so, factor theorem does not hold
– Sorfosh
Nov 19 '18 at 2:14
Looks like you want $mathbb{Z}/12mathbb{Z}$, rather than $mathbb{Z}_{12}=varprojlim(mathbb{Z}/12^nmathbb{Z})$.
– user10354138
Nov 19 '18 at 2:23
add a comment |
It is not over a field so, factor theorem does not hold
– Sorfosh
Nov 19 '18 at 2:14
Looks like you want $mathbb{Z}/12mathbb{Z}$, rather than $mathbb{Z}_{12}=varprojlim(mathbb{Z}/12^nmathbb{Z})$.
– user10354138
Nov 19 '18 at 2:23
It is not over a field so, factor theorem does not hold
– Sorfosh
Nov 19 '18 at 2:14
It is not over a field so, factor theorem does not hold
– Sorfosh
Nov 19 '18 at 2:14
Looks like you want $mathbb{Z}/12mathbb{Z}$, rather than $mathbb{Z}_{12}=varprojlim(mathbb{Z}/12^nmathbb{Z})$.
– user10354138
Nov 19 '18 at 2:23
Looks like you want $mathbb{Z}/12mathbb{Z}$, rather than $mathbb{Z}_{12}=varprojlim(mathbb{Z}/12^nmathbb{Z})$.
– user10354138
Nov 19 '18 at 2:23
add a comment |
2 Answers
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If $x$ is even then the left side is odd from the odd constant term. If $x$ is odd then the left side is odd by having an odd number of odd terms. The left side will never be a multiple of $2$, let alone a multiple of $12$.
answered Nov 19 '18 at 2:23
Oscar Lanzi
12.1k12036
add a comment |
There are various procedures. Here's one using Euler's Totient function. x can also only be a solution in $Z_{mn}[x]$ , gcd(m,n)=1, iff x is a solution in $Z_m[x]$ and $Z_n[x]$.
By one of Euler's theorems $a^{phi(n)}=1 (mod n)$, where $phi(n)$ is the Euler Totient function, i.e. the number of integers less than n that are relatively prime to n.
So take $5x^3+4x^2+11x+9∈Z_{12}[x]$.
$phi(4)=2$.
$phi(3)=2$.
We have a solution if we can solve the polynomial over $Z_{4}[x]$ and $Z_3[x]$.
In mod 4, the polynomial becomes :
$$5x^3+4x^2+11x+9$$
$$(4+1)x^{(phi(4)+1)}+(4+0)x^2+(3*4-1)x+(2*4+1)$$
$$x-x+1=0( mod 4)$$
Where coefficients have been replaced with their lowest congruences mod 4 and the exponent has been reduced by Euler's theorem. We see we have no solution in mod 4.
Let's try with 3. $phi(3)=2$
$$5x^3+4x^2+11x+9$$
$$(3+2)x^{phi(3)+1}+(3+1)x^{phi(3)}+(3*4-1)x+3*3$$
$$2x+1-x=0 ( mod 3)$$
$$x=-1 ( mod 3)$$
So we have a solution in mod 3. Since there is no solution in mod 4, we have no solution in mod 12.
However, suppose we did have a solution in mod 4, that our solutions were:
$x=p ( mod 4)$
$x=q ( mod 3 )$
Then you could use the Chinese Remainder theorem to find solutions for $Z_{12}[x]$
answered Nov 21 '18 at 2:27
TurlocTheRed
838311
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
If $x$ is even then the left side is odd from the odd constant term. If $x$ is odd then the left side is odd by having an odd number of odd terms. The left side will never be a multiple of $2$, let alone a multiple of $12$.
answered Nov 19 '18 at 2:23
Oscar Lanzi
12.1k12036
add a comment |
If $x$ is even then the left side is odd from the odd constant term. If $x$ is odd then the left side is odd by having an odd number of odd terms. The left side will never be a multiple of $2$, let alone a multiple of $12$.
answered Nov 19 '18 at 2:23
Oscar Lanzi
12.1k12036
add a comment |
If $x$ is even then the left side is odd from the odd constant term. If $x$ is odd then the left side is odd by having an odd number of odd terms. The left side will never be a multiple of $2$, let alone a multiple of $12$.
answered Nov 19 '18 at 2:23
Oscar Lanzi
12.1k12036
If $x$ is even then the left side is odd from the odd constant term. If $x$ is odd then the left side is odd by having an odd number of odd terms. The left side will never be a multiple of $2$, let alone a multiple of $12$.
answered Nov 19 '18 at 2:23
Oscar Lanzi
12.1k12036
answered Nov 19 '18 at 2:23
Oscar Lanzi
12.1k12036
answered Nov 19 '18 at 2:23
Oscar Lanzi
12.1k12036
answered Nov 19 '18 at 2:23
Oscar Lanzi
12.1k12036
12.1k12036
add a comment |
add a comment |
There are various procedures. Here's one using Euler's Totient function. x can also only be a solution in $Z_{mn}[x]$ , gcd(m,n)=1, iff x is a solution in $Z_m[x]$ and $Z_n[x]$.
By one of Euler's theorems $a^{phi(n)}=1 (mod n)$, where $phi(n)$ is the Euler Totient function, i.e. the number of integers less than n that are relatively prime to n.
So take $5x^3+4x^2+11x+9∈Z_{12}[x]$.
$phi(4)=2$.
$phi(3)=2$.
We have a solution if we can solve the polynomial over $Z_{4}[x]$ and $Z_3[x]$.
In mod 4, the polynomial becomes :
$$5x^3+4x^2+11x+9$$
$$(4+1)x^{(phi(4)+1)}+(4+0)x^2+(3*4-1)x+(2*4+1)$$
$$x-x+1=0( mod 4)$$
Where coefficients have been replaced with their lowest congruences mod 4 and the exponent has been reduced by Euler's theorem. We see we have no solution in mod 4.
Let's try with 3. $phi(3)=2$
$$5x^3+4x^2+11x+9$$
$$(3+2)x^{phi(3)+1}+(3+1)x^{phi(3)}+(3*4-1)x+3*3$$
$$2x+1-x=0 ( mod 3)$$
$$x=-1 ( mod 3)$$
So we have a solution in mod 3. Since there is no solution in mod 4, we have no solution in mod 12.
However, suppose we did have a solution in mod 4, that our solutions were:
$x=p ( mod 4)$
$x=q ( mod 3 )$
Then you could use the Chinese Remainder theorem to find solutions for $Z_{12}[x]$
answered Nov 21 '18 at 2:27
TurlocTheRed
838311
add a comment |
There are various procedures. Here's one using Euler's Totient function. x can also only be a solution in $Z_{mn}[x]$ , gcd(m,n)=1, iff x is a solution in $Z_m[x]$ and $Z_n[x]$.
By one of Euler's theorems $a^{phi(n)}=1 (mod n)$, where $phi(n)$ is the Euler Totient function, i.e. the number of integers less than n that are relatively prime to n.
So take $5x^3+4x^2+11x+9∈Z_{12}[x]$.
$phi(4)=2$.
$phi(3)=2$.
We have a solution if we can solve the polynomial over $Z_{4}[x]$ and $Z_3[x]$.
In mod 4, the polynomial becomes :
$$5x^3+4x^2+11x+9$$
$$(4+1)x^{(phi(4)+1)}+(4+0)x^2+(3*4-1)x+(2*4+1)$$
$$x-x+1=0( mod 4)$$
Where coefficients have been replaced with their lowest congruences mod 4 and the exponent has been reduced by Euler's theorem. We see we have no solution in mod 4.
Let's try with 3. $phi(3)=2$
$$5x^3+4x^2+11x+9$$
$$(3+2)x^{phi(3)+1}+(3+1)x^{phi(3)}+(3*4-1)x+3*3$$
$$2x+1-x=0 ( mod 3)$$
$$x=-1 ( mod 3)$$
So we have a solution in mod 3. Since there is no solution in mod 4, we have no solution in mod 12.
However, suppose we did have a solution in mod 4, that our solutions were:
$x=p ( mod 4)$
$x=q ( mod 3 )$
Then you could use the Chinese Remainder theorem to find solutions for $Z_{12}[x]$
answered Nov 21 '18 at 2:27
TurlocTheRed
838311
add a comment |
There are various procedures. Here's one using Euler's Totient function. x can also only be a solution in $Z_{mn}[x]$ , gcd(m,n)=1, iff x is a solution in $Z_m[x]$ and $Z_n[x]$.
By one of Euler's theorems $a^{phi(n)}=1 (mod n)$, where $phi(n)$ is the Euler Totient function, i.e. the number of integers less than n that are relatively prime to n.
So take $5x^3+4x^2+11x+9∈Z_{12}[x]$.
$phi(4)=2$.
$phi(3)=2$.
We have a solution if we can solve the polynomial over $Z_{4}[x]$ and $Z_3[x]$.
In mod 4, the polynomial becomes :
$$5x^3+4x^2+11x+9$$
$$(4+1)x^{(phi(4)+1)}+(4+0)x^2+(3*4-1)x+(2*4+1)$$
$$x-x+1=0( mod 4)$$
Where coefficients have been replaced with their lowest congruences mod 4 and the exponent has been reduced by Euler's theorem. We see we have no solution in mod 4.
Let's try with 3. $phi(3)=2$
$$5x^3+4x^2+11x+9$$
$$(3+2)x^{phi(3)+1}+(3+1)x^{phi(3)}+(3*4-1)x+3*3$$
$$2x+1-x=0 ( mod 3)$$
$$x=-1 ( mod 3)$$
So we have a solution in mod 3. Since there is no solution in mod 4, we have no solution in mod 12.
However, suppose we did have a solution in mod 4, that our solutions were:
$x=p ( mod 4)$
$x=q ( mod 3 )$
Then you could use the Chinese Remainder theorem to find solutions for $Z_{12}[x]$
answered Nov 21 '18 at 2:27
TurlocTheRed
838311
There are various procedures. Here's one using Euler's Totient function. x can also only be a solution in $Z_{mn}[x]$ , gcd(m,n)=1, iff x is a solution in $Z_m[x]$ and $Z_n[x]$.
By one of Euler's theorems $a^{phi(n)}=1 (mod n)$, where $phi(n)$ is the Euler Totient function, i.e. the number of integers less than n that are relatively prime to n.
So take $5x^3+4x^2+11x+9∈Z_{12}[x]$.
$phi(4)=2$.
$phi(3)=2$.
We have a solution if we can solve the polynomial over $Z_{4}[x]$ and $Z_3[x]$.
In mod 4, the polynomial becomes :
$$5x^3+4x^2+11x+9$$
$$(4+1)x^{(phi(4)+1)}+(4+0)x^2+(3*4-1)x+(2*4+1)$$
$$x-x+1=0( mod 4)$$
Where coefficients have been replaced with their lowest congruences mod 4 and the exponent has been reduced by Euler's theorem. We see we have no solution in mod 4.
Let's try with 3. $phi(3)=2$
$$5x^3+4x^2+11x+9$$
$$(3+2)x^{phi(3)+1}+(3+1)x^{phi(3)}+(3*4-1)x+3*3$$
$$2x+1-x=0 ( mod 3)$$
$$x=-1 ( mod 3)$$
So we have a solution in mod 3. Since there is no solution in mod 4, we have no solution in mod 12.
However, suppose we did have a solution in mod 4, that our solutions were:
$x=p ( mod 4)$
$x=q ( mod 3 )$
Then you could use the Chinese Remainder theorem to find solutions for $Z_{12}[x]$
answered Nov 21 '18 at 2:27
TurlocTheRed
838311
answered Nov 21 '18 at 2:27
TurlocTheRed
838311
answered Nov 21 '18 at 2:27
TurlocTheRed
838311
answered Nov 21 '18 at 2:27
TurlocTheRed
838311
838311
add a comment |
add a comment |
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It is not over a field so, factor theorem does not hold
– Sorfosh
Nov 19 '18 at 2:14
Looks like you want $mathbb{Z}/12mathbb{Z}$, rather than $mathbb{Z}_{12}=varprojlim(mathbb{Z}/12^nmathbb{Z})$.
– user10354138
Nov 19 '18 at 2:23