Satz 1. Sowohl die Summe wie das Produkt zweier Polynome über <<N>>
oder <<Q>> ist ein Polynom über <<N>> bzw. <<Q>>

Beweis. Sei <MATH>f(x)=Sum{i=0,...,m:a_i x^i}</MATH> und <MATH>g(x)=Sum{i=0,...,n:b_i x^i}</MATH>.
Dann ist</P>

<MATHP>f(x)+g(x)=Sum{i=0,...,max(m,n):(a_i +b_i )x^i}</MATHP>

<P>und</P>

<MATHP>f(x)g(x)=Sum{i=0,...,m+n:c_i x^i}</MATHP>

<P>mit</P>

<MATHP>c_i =Sum{j=0,...,i:a_j b_i-j}</MATHP>

<P>Dabei haben die Koeffizienten <MATH>a_i</MATH> mit <MATH>i<0</MATH>
oder <MATH>i>m</MATH> der Wert 0; analog für die <MATH>n_i</MATH>.</P>

Satz 2. Das Produkt zweier primitiver Polynome über <<N>> ist ein primitives Polynom
über <<N>>.

Beweis. 

(1) Polynomials can be multiplied to yield polynomials.

(2) Polynomials can be divided, yielding a quotient polynomial
    and a remainder polynomial with a degree less than the degree of
    the divisor polynomial, i.e.

    For f(x), g(x) there is q(x) and r(x) with
    f(x) = g(x) * q(x) + r(x) and degree(r) < degree(g)

(3) If f(z) = 0, then f(x) is divisible by (x-z).
    (Proof: Divide according to (2) and see what happens when x = z)

(4) If h(x) = f(x) * g(x) has all coefficients divisible by some prime
    p, then already one of f and g has had.
    (Proof: what are the coefficients of h? If there are coefficients
    of both f and g that are not divisable by p, then one of the
    coefficients is also not divisible by p.)

(5) From (3) and (4), one can conclude that if f(z) = 0 for rational
    z = p/q, then f(x) is divisible by (xq - p) yielding a polynomial
    with integer coefficients.

(6) From (1) and (5) you can infer the theorem.