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

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! :)
 
##\textbf{Exercise 10}:## I came across the following solution online: Questions: 1. When the author states in "that ring (not sure if he is referring to ##R## or ##R/\mathfrak{p}##, but I am guessing the later) ##x_n x_{n+1}=0## for all odd $n$ and ##x_{n+1}## is invertible, so that ##x_n=0##" 2. How does ##x_nx_{n+1}=0## implies that ##x_{n+1}## is invertible and ##x_n=0##. I mean if the quotient ring ##R/\mathfrak{p}## is an integral domain, and ##x_{n+1}## is invertible then...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top