MHB How to Use Euclidean Division in $\mathbb{Z}_7$?

Click For Summary
In the discussion on applying Euclidean division in $\mathbb{Z}_7$, a polynomial $[2]x^5+x^4+x^3+[3]x^2+[2]x+[2]$ is divided by $[3]x^2+[2]x+[3]$. The divisor is made monic by multiplying by $[5]$, resulting in $x^2+[3]x+[1]$. The quotient obtained is $[2]x^3+[2]x^2+[1]$ with a remainder of $[6]x+[1]$, which is then made monic, leading to further division yielding a final quotient of $x+[4]$ and a remainder of $[5]$. The conclusion drawn is that the two polynomials are coprime over $\mathbb{Z}_7$.
evinda
Gold Member
MHB
Messages
3,741
Reaction score
0
Hello! :D
I am given the following exercise:
In $\mathbb{Z}_7$ apply the Euclidean division, dividing $[2]x^5+x^4+x^3+[3]x^2+[2]x+[2]$ by $[3]x^2+[2]x+[3]$.
That's what I have done:
$$\text{ the units of } \mathbb{Z}_7 \text{ are: } \{ 1,2,3,4,5,6\}$$
We want $[3]x^2+[2]x+[3]$ to be monic,so we multiply it by $[5]$ ($[3] \cdot [5]=1 \pmod 7$) and we get: $[5] \cdot ([3]x^2+[2]x+[3])=x^2+[3]x+[1] $.

Then ,dividing $[2]x^5+x^4+x^3+[3]x^2+[2]x+[2]$ by $x^2+[3]x+[1] $ I got, that the quotient is equal to $[2]x^3+[2]x^2+[1]$ and the remainder: $[6]x+[1]$.
Then, to make $[6]x+[1]$ monic,I multiplied it by $[6]$ : $[6]([6]x+[1])=x+[6]$ and dividing $x^2+[3]x+[1]$ by $x+[6]$,I found that the quotient is equal to $x+[4]$ and that the remainder is $[5]$.
Could you tell me if it is right or if I have done somethig wrong? (Blush)
 
Physics news on Phys.org
I agree with your calculation. So the conclusion is that those two polynomials are coprime over $\mathbb{Z}_7$.
 
Opalg said:
I agree with your calculation. So the conclusion is that those two polynomials are coprime over $\mathbb{Z}_7$.

Great! (Clapping) Yes,since the greatest common divisor will be equal to $1$ (Nod) Thank you! :)
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 1 ·
Replies
1
Views
823
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 26 ·
Replies
26
Views
809
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 21 ·
Replies
21
Views
1K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 0 ·
Replies
0
Views
795
Replies
13
Views
3K