Is there a faster way to do this?

  • Context: Graduate 
  • Thread starter Thread starter e12514
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on the process of finding the inverse of a polynomial in the ring Z/p[x] when modded out by an irreducible polynomial i(x). The method described involves tedious calculations that often lead to errors, particularly when the prime p is large. The user seeks a more efficient method for verifying that (Z/p[x])/I is a field, emphasizing the need for a deeper understanding of concepts like principal ideal domains (PIDs) and maximal ideals, which are crucial for simplifying these calculations.

PREREQUISITES
  • Understanding of polynomial rings, specifically Z/p[x]
  • Knowledge of irreducible polynomials and their properties
  • Familiarity with concepts of ideals, particularly prime and maximal ideals
  • Basic principles of abstract algebra, including fields and division rings
NEXT STEPS
  • Study the properties of principal ideal domains (PIDs) in detail
  • Learn about efficient algorithms for finding polynomial inverses in finite fields
  • Explore the concept of field extensions and their applications in abstract algebra
  • Investigate computational tools for polynomial arithmetic in Z/p[x]
USEFUL FOR

Students and researchers in abstract algebra, particularly those studying polynomial rings and field theory, as well as anyone looking to improve their efficiency in polynomial computations within finite fields.

e12514
Messages
27
Reaction score
0
Z/p [x] consists of polynomials in x with coefficients in a field Z/p.
I is an ideal of Z/p [x] generated by some irreducible polynomial i(x) in Z/p [x]
(let's say I is generated by an irreducible polynomial i(x) of degree 3 in Z/p [x])

The factor (or quotient)
(Z/p [x])/I = (Z/p [x])/<i(x)> is a field.
To verify that it is,
* (Z/p [x])/<i(x)> is commutative because Z/p [x] is. (ok)
* i(x) is irreducible so (Z/p [x])/<i(x)> has no divisors of zero. (ok)
* (Z/p [x])/<i(x)> needs to be a division ring, that is, every nonzero element needs to be invertible. (needs to be verified)

My (tedious) method is,
WLOG let f(x) = x^2 + ax + b be a polynomial in Z/p [x]
need to find some iverse of f(x) (call it g(x)) in Z/p [x] such that
(f(x).g(x)) + <i(x)> = 1 + <i(x)>

which can be done by multiplying out f(x).g(x) and then dividing by i(x) to get the remainder (very long) which would be of degree 2 (one less than deg(i(x)) )
and then solving the coefficient of x^2 and x to be congruent to 0 mod p
and the constant term to be congruent to 1 mod p
(three equations and three unkowns)

and not only does that take very long, if p is not small then in the end I often end up making a mistake somewhere so my calculated "inverse" of f(x) is incorrect
(end up having f(x).g(x) + <i(x)> different to 1 + <i(x)> )

Is there a better/cleaner/easier/more efficient way to do these sorts of things rather than having to go through all that mess?
(Can take p to be a specific prime, say p=5, or 7, or whatever.)
 
Physics news on Phys.org
Z/p is a field, so (Z/p)[x] is a principal ideal domain. If i(x) is irreducible over Z/p, then <i(x)> is a prime ideal and consequently (because we're in a PID) a maximal ideal. And what happens when you mod out by a maximal ideal?
 
Then the factor is a field! Yay, fantastic! Thanks!
This shows that I need to do a lot of extra reading on stuff that is not taught in order to fully understand the material.. I've never heard of a pid before, unless it's lying on some fresh page that I've never looked at in our notes, or it's not covered at my level - first course in abstract algebra... (I had to look up maximal and prime ideals too...)
 

Similar threads

Replies
48
Views
6K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 24 ·
Replies
24
Views
5K
  • · Replies 28 ·
Replies
28
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K