Z/p [x] consists of polynomials in x with coefficients in a field Z/p.(adsbygoogle = window.adsbygoogle || []).push({});

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 Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Is there a faster way to do this?

**Physics Forums | Science Articles, Homework Help, Discussion**