Some confusion of the concepts about polynomials

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The discussion clarifies the distinction between polynomials in P[X] and formal power series in P[[X]], focusing on the concept of prime elements. It establishes that while prime elements in P[X] are defined by their degree, this concept does not apply to P[[X]], where the index of the first nonzero coefficient, w(f), is used instead. The conversation demonstrates that the element X is irreducible in P[[X]] by showing that if it were not, it would lead to a contradiction regarding the values of w(g) and w(h). Additionally, it confirms that X is the only irreducible element in P[[X]], as any higher power of X cannot be irreducible. The argument concludes that the reasoning presented is valid and supports the claim.
boombaby
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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?
 
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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.
 
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?
 
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