Homework Help: Some confusion of the concepts about polynomials

1. Nov 10, 2008

boombaby

I'm not reading a text in English, so I should clarify some notation first:
P[X], is the set of all polynomials.
P[[X]], is the set of formal power series, which may infinitely many nonzero coefficients.
I'm asked to prove that X is the only prime(?correct term?) element in P[[X]]. By saying 'only', I mean uX and X is considered to be the same, if u is the unit in P.( of course every nonzero element in P is the unit)

The prime element f in P[X] is defined to be so that there's no other polynomial g with 0<deg(g)<deg(f) such that g|f. And I know that every f with deg(f)=1 is a prime polynomial.
But deg(f) makes no sense in P[[X]]. Then what kind of element is the prime one in P[[X]]?My book does not tell me, but introduce w(f) to me, which is defined to be the index of the first nonzero coefficient. More precisely, if f=a_n*X^n+a_(n+1)*X^(n+1)+.....then w(f)=n

for instance, how to show that 1+X is not prime?

2. Nov 10, 2008

Hurkyl

Staff Emeritus
What you describe sounds like the notion of an irreducible element. (But in these rings, every irreducible is prime, and vice versa) a is irreducible iff:
1. a is not a unit
2. If b divides a, then b is a unit or b = ua for some unit u

Note that for P[X], this is equivalent to the definition you gave.

3. Nov 10, 2008

boombaby

thanks!

let f=a_0+a_1*X+.....
so if a_0!=0,that is w(f)=0 then there is a g such that fg=1, which implies that f is a unit.(g can be found by comparing the coefficients of fg and 1)

Suppose X is not irreducible, then X=gh, where w(g)>=1 and w(h)>=1 (which means g and h are not units). Hence 1=w(X)=w(g)+w(h)>=2 and we get a contradiction.

if w(f)=s>0. then f=X^(s)*h, where h has the form h=b_0+b_1*X+....and w(h)=0.
It is sufficient to show that X^s is not irreducible, but this is obviously true if s>=2. Hence X is the only irreducible element in P[[X]].

It works, right?