Sunday, 16 October 2016

abstract algebra - Factor Rings over Finite Fields



Given a polynomial ring over a field F[x], I can factor, for example, the ideal generated by an irreducible polynomial ax2+bx+c: $F[x]/\left$, and guarantee that this factor ring is also a field.



My question concerns the structure of this factor ring. For example, if I consider the factor ring $Z_p[x] / \left$ for some irreducible polynomial ax2+bx+c, I can guarantee, for example, that this field has p2 elements.




I am unsure why this is the case. My understanding is that the coset representitives of this factor ring are possible remainders by division by ax2+bx+c. Is this the right idea, and how would I know that two different remainders aren't in the same coset? Thanks.


Answer



Indeed the elements of the factor ring Fp[x]/ax2+bx2+c can be represented by the remainders by division by ax2+bx+c. This is true because we can divide polynomials in Fp[x] by ax2+bx+c with remainder. What this means is that




For every polynomial fFp[x] there exist unique q,rFp[x] with degr<2 such that
f=q(ax2+bx+c)+r.




This equality shows that f and r are in the same coset of ax2+bx+c, and hence they are mapped to the same element of the factor ring Fp[x]/ax2+bx+c. Hence the image of

f in the factor ring is represented by r, and so every element of the factor ring is represented by a linear polynomial.



To see that no two linear polynomials represent the same element of Fp[x]/ax2+bx+c, it suffices to note that the remainder r in (1) is unique for every fFp[x], meaning in particular that every linear polynomial is represented only by itself.



Alternatively, if two linear polynomials r and r represent the same coset of ax2+bx+c in the factor ring, then rr is a multiple of ax2+bx+c. Because degrr<deg(ax2+bx+c) it follows that rr=0.


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