Number Theory: An Introduction to Mathematics

3 Polynomials 97

It is easily verified that, for all polynomialsf,g,

|f+g|≤max{|f|,|g|}, |fg|=|f||g|.

Since|f|≥0, with equality if and only iff=O, the last property implies thatR[t]is
an integral domain. Thus we can define divisibility inR[t], as explained in Section 1.
The set of all polynomials of the form(a 0 , 0 , 0 ,...)is a subdomain isomorphic
toR. By identifying this set withR, we may regardRas embedded inR[t]. The only
units inR[t] are the units inR,since1=efimplies 1=|e||f|and hence|e|=1.
If we putt=( 0 , 1 , 0 , 0 ,...),then

t^2 =tt=( 0 , 0 , 1 , 0 ,...), t^3 =tt^2 =( 0 , 0 , 0 , 1 ,...),....

Hence if the polynomialf =(a 0 ,a 1 ,a 2 ,...)has degreen, then it can be uniquely
expressed in the form

f=a 0 +a 1 t+···+antn (an= 0 ).

We refer to the elementsa 0 ,a 1 ,...,anofRas thecoefficientsoff. In particular,a 0
is theconstantcoefficient andanthehighestcoefficient. We say thatfismonicif its
highest coefficientan=1.
If also

g=b 0 +b 1 t+···+bmtm (bm= 0 ),

then the sum and product assume their familiar forms:

f+g=(a 0 +b 0 )+(a 1 +b 1 )t+(a 2 +b 2 )t^2 +···,

fg=a 0 b 0 +(a 0 b 1 +a 1 b 0 )t+(a 0 b 2 +a 1 b 1 +a 2 b 0 )t^2 +···.

Suppose now thatR=Kis a field, and let

f=a 0 +a 1 t+···+antn(an= 0 ),

g=b 0 +b 1 t+···+bmtm(bm= 0 )

be any two nonzero elements ofK[t]. If|g|<|f|,i.e.ifm<n,theng=qf+r,
withq=Oandr=g. Suppose on the other hand that|f|≤|g|.Then

g=an−^1 bmtm−nf+g†,

whereg†∈K[t]and|g†|<|g|.If|f|≤|g†|, the process can be repeated withg†in
place ofg. Continuing in this way, we obtainq,r∈K[t] such that

g=qf+r, |r|<|f|.

Moreover,qandrare uniquely determined, since if also

g=q 1 f+r 1 , |r 1 |<|f|,