# 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?