Finding inverse in polynomial factor ring

In summary, the problem involves finding the inverse of r in a ring R, where F is the integers modulo 7 and h is a polynomial of degree 3. The solution involves using the Extended Euclidean Algorithm to write Bezout's identity and then solving for coefficients using substitution. However, there may be an error in the calculation, particularly with t^4.
  • #1
PsychonautQQ
784
10

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?
 
Physics news on Phys.org
  • #2
PsychonautQQ said:

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:

FAQ: Finding inverse in polynomial factor ring

What is an inverse in a polynomial factor ring?

An inverse in a polynomial factor ring is an element that, when multiplied by another element, results in the multiplicative identity element of the ring. In simpler terms, it is a number that, when multiplied by another number, gives a result of 1. This is similar to how the number 5 is the inverse of 1/5, as 5 * 1/5 = 1.

Why is finding the inverse important in polynomial factor rings?

Finding the inverse is important in polynomial factor rings because it allows us to divide polynomials and solve equations in these rings. It also helps us determine if a polynomial is irreducible, meaning it cannot be factored into simpler polynomials.

How do you find the inverse in a polynomial factor ring?

To find the inverse in a polynomial factor ring, we use the Extended Euclidean Algorithm. This algorithm involves finding the greatest common divisor of the polynomial and the modulus, and then using the coefficients of that common divisor to find the inverse. There are also other methods, such as the Bezout's Identity method, that can be used to find the inverse.

Can every polynomial in a factor ring have an inverse?

No, not every polynomial in a factor ring has an inverse. Only polynomials that are relatively prime to the modulus have an inverse. In other words, the greatest common divisor of the polynomial and the modulus must be 1 in order for the polynomial to have an inverse. This is why it is important to check if a polynomial is irreducible before attempting to find its inverse.

What are some real-world applications of finding the inverse in polynomial factor rings?

One real-world application of finding the inverse in polynomial factor rings is in error correction codes used in computer networks. These codes use polynomial factor rings to encode and decode data, and finding the inverse is crucial in the decoding process. It is also used in cryptography, specifically in the RSA algorithm, which involves finding inverses in large polynomial factor rings to encrypt and decrypt messages.

Similar threads

Back
Top