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

Beweis. Sei 
<!-- f(x)=Sum{i=0,...,m:a_i x^i} -->
<I>f</I>(<I>x</I>)&nbsp;=&nbsp;<I>a</I><SUB>0</SUB><I>x</I><SUP>0</SUP>&nbsp;+&nbsp;<I>a</I><SUB>1</SUB><I>x</I><SUP>1</SUP>&nbsp;+&nbsp;...&nbsp;+&nbsp;<I>a<SUB>m</SUB></I><I>x<SUP>m</SUP></I>
<!-- f(x)=\sum_{i=0}^{m}a_{i}x^{i} -->
 und 
<!-- g(x)=Sum{i=0,...,n:b_i x^i} -->
<I>g</I>(<I>x</I>)&nbsp;=&nbsp;<I>b</I><SUB>0</SUB><I>x</I><SUP>0</SUP>&nbsp;+&nbsp;<I>b</I><SUB>1</SUB><I>x</I><SUP>1</SUP>&nbsp;+&nbsp;...&nbsp;+&nbsp;<I>b<SUB>n</SUB></I><I>x<SUP>n</SUP></I>
<!-- g(x)=\sum_{i=0}^{n}b_{i}x^{i} -->
.
Dann ist</P>

<!-- f(x)+g(x)=Sum{i=0,...,max(m,n):(a_i +b_i )x^i} -->
<BLOCKQUOTE>
<I>f</I>(<I>x</I>)&nbsp;+&nbsp;<I>g</I>(<I>x</I>)&nbsp;=&nbsp;(<I>a</I><SUB>0</SUB>&nbsp;+&nbsp;<I>b</I><SUB>0</SUB>)<I>x</I><SUP>0</SUP>&nbsp;+&nbsp;(<I>a</I><SUB>1</SUB>&nbsp;+&nbsp;<I>b</I><SUB>1</SUB>)<I>x</I><SUP>1</SUP>&nbsp;+&nbsp;...&nbsp;+&nbsp;(<I>a</I><SUB>max(<I>m</I>,<I>n</I>)</SUB>&nbsp;+&nbsp;<I>b</I><SUB>max(<I>m</I>,<I>n</I>)</SUB>)<I>x</I><SUP>max(<I>m</I>,<I>n</I>)</SUP>
</BLOCKQUOTE>
<!-- f(x)+g(x)=\sum_{i=0}^{\max (m,n)}(a_{i}+b_{i})x^{i} -->

<P>und</P>

<!-- f(x)g(x)=Sum{i=0,...,m+n:c_i x^i} -->
<BLOCKQUOTE>
<I>f</I>(<I>x</I>)<I>g</I>(<I>x</I>)&nbsp;=&nbsp;<I>c</I><SUB>0</SUB><I>x</I><SUP>0</SUP>&nbsp;+&nbsp;<I>c</I><SUB>1</SUB><I>x</I><SUP>1</SUP>&nbsp;+&nbsp;...&nbsp;+&nbsp;<I>c</I><SUB><I>m</I>+<I>n</I></SUB><I>x</I><SUP><I>m</I>+<I>n</I></SUP>
</BLOCKQUOTE>
<!-- f(x)g(x)=\sum_{i=0}^{m+n}c_{i}x^{i} -->

<P>mit</P>

<!-- c_i =Sum{j=0,...,i:a_j b_i-j} -->
<BLOCKQUOTE>
<I>c<SUB>i</SUB></I>&nbsp;=&nbsp;<I>a</I><SUB>0</SUB><I>b<SUB>i</SUB></I>&nbsp;+&nbsp;<I>a</I><SUB>1</SUB><I>b</I><SUB><I>i</I>+1</SUB>&nbsp;+&nbsp;...&nbsp;+&nbsp;<I>a<SUB>i</SUB></I><I>b</I><SUB>2<I>i</I></SUB>
</BLOCKQUOTE>
<!-- c_{i}=\sum_{j=0}^{i}a_{j}b_{i-j} -->

<P>Dabei haben die Koeffizienten 
<!-- a_i -->
<I>a<SUB>i</SUB></I>
<!-- a_{i} -->
 mit 
<!-- i<0 -->
<I>i</I>&nbsp;&lt;&nbsp;0
<!-- i<0 -->
oder 
<!-- i>m -->
<I>i</I>&nbsp;&gt;&nbsp;<I>m</I>
<!-- i>m -->
 der Wert 0; analog für die 
<!-- n_i -->
<I>n<SUB>i</SUB></I>
<!-- n_{i} -->
.</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.