• Support PF! Buy your school textbooks, materials and every day products Here!

Finding inverse in polynomial factor ring

  • #1
784
11

Homework Statement


find the inverse of r in R = F[x]/<h>.
r = 1 + t - t^2
F = Z_7 (integers modulo 7), h = x^3 + x^2 -1

Homework Equations


None

The Attempt at a Solution


The polynomial on bottom is of degree 3, so R will look like:
R = {a + bt + ct^2 | a,b,c are elements of z_7 and x^3 = 1 - ^2}

To solve this problem I realized that the inverse must obviously have the form of some element in R, so I set up:
(a + bt + ct^2)(1 + t - t^2) = 1

then I multiplied it all out whilst continuously substituting for t^3 and then solving for coefficients where the constant coefficient should equal 1 and the other two should equal 0.

I did all of this and got the constant coefficient to be zero and nonzero answers for the other two >.<. I checked my calculations and can't find an error (doesn't necessarily mean there isn't one...), is something wrong with the way I set up the problem? is my substitution for x^3 correct?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618

Homework Statement


find the inverse of r in R = F[x]/<h>.
r = 1 + t - t^2
F = Z_7 (integers modulo 7), h = x^3 + x^2 -1

Homework Equations


None

The Attempt at a Solution


The polynomial on bottom is of degree 3, so R will look like:
R = {a + bt + ct^2 | a,b,c are elements of z_7 and x^3 = 1 - ^2}

To solve this problem I realized that the inverse must obviously have the form of some element in R, so I set up:
(a + bt + ct^2)(1 + t - t^2) = 1

then I multiplied it all out whilst continuously substituting for t^3 and then solving for coefficients where the constant coefficient should equal 1 and the other two should equal 0.

I did all of this and got the constant coefficient to be zero and nonzero answers for the other two >.<. I checked my calculations and can't find an error (doesn't necessarily mean there isn't one...), is something wrong with the way I set up the problem? is my substitution for x^3 correct?
The usual way to do this is to use the Extended Euclidean Algorithm to explicitly write Bezout's identity. gcd(h,r)=a*h+b*r. Then mod both sides by h. Does that sound familiar? It is kind of a tedious calculation and it's easy to make a mistake. What did you do with t^4?
 
Last edited:

Related Threads on Finding inverse in polynomial factor ring

  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
9
Views
2K
Replies
1
Views
2K
Replies
5
Views
3K
  • Last Post
Replies
13
Views
28K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
872
Replies
8
Views
3K
  • Last Post
Replies
1
Views
871
Top