3 Polynomials 103{( 1 +t)p− 1 }/t=tp−^1 +pCp− 1 tp−^2 +···+pC 2 t+psatisfies the hypotheses of Proposition 19.
For any fieldK,wedefinetheformal derivativeof a polynomialf∈K[t],
f=a 0 +a 1 t+···+antn,to be the polynomial
f′=a 1 + 2 a 2 t+···+nantn−^1.If the fieldKis ofcharacteristic0 (see Chapter I,§8), then∂(f′)=∂(f)−1.
Formal derivatives share the following properties with the derivatives of real
analysis:
(i)(f+g)′=f′+g′;
(ii)(cf)′=cf′for any c∈K;
(iii)(fg)′=f′g+fg′;
(iv)(fk)′=kfk−^1 f′for any k∈N.
The first two properties are easily established and the last two properties then need
only be verified for monomialsf=tm,g=tn.
We can use formal derivatives to determine when a polynomial issquare-free:
Proposition 20Let f be a polynomial of positive degree with coefficients from a
field K. If f is relatively prime to its formal derivative f′, then f is a product of
irreducible polynomials, no two of which differ by a constant factor. Conversely, if f
is such a product and if K has characteristic 0, then f is relatively prime to f′.
Proof Iff=g^2 hfor some polynomialsg,h∈K[t] with∂(g)>0 then, by the rules
above,
f′= 2 gg′h+g^2 h′.Henceg|f′andf,f′are not relatively prime.
On the other hand, iff=p 1 ···pmis a product of essentially distinct irreducible
polynomialspj,then
f′=p′ 1 p 2 ···pm+p 1 p′ 2 p 3 ···pm+···+p 1 ···pm− 1 p′m.If the fieldKhas characteristic 0, thenp′ 1 is of lower degree thanp 1 and is not the zero
polynomial. Thus the first term on the right is not divisible byp 1 , but all the other terms
are. Thereforep 1 f′, and hence (f′,p 1 )=1. Similarly, (f′,pj)=1for1<j≤m.
Since essentially distinct irreducible polynomials are relatively prime, it follows that
(f′,f)=1.
For example, it follows from Proposition 20 that the polynomialtn− 1 ∈K[t]is
square-free if the characteristic of the fieldKdoes not divide the positive integern.