Some confusion of the concepts about polynomials

In summary, the conversation discusses the concept of prime elements in the sets P[X] and P[[X]]. It is clarified that in P[[X]], the concept of degree is not applicable and the index of the first nonzero coefficient, denoted as w(f), is introduced to determine the prime element. The definition of an irreducible element is also mentioned and it is stated that in these sets, every irreducible element is also prime. The conversation also includes a proof that X is the only irreducible element in P[[X]].
  • #1
boombaby
130
0
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?
 
Physics news on Phys.org
  • #2
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
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?
 

Related to Some confusion of the concepts about polynomials

1. What is a polynomial?

A polynomial is a mathematical expression consisting of variables and coefficients that are combined using addition, subtraction, and multiplication operations. It can also include powers and exponents of variables, but not division by a variable.

2. What are the different types of polynomials?

The three main types of polynomials are monomials, binomials, and trinomials. Monomials have one term, binomials have two terms, and trinomials have three terms.

3. How do I simplify a polynomial?

To simplify a polynomial, you need to combine like terms and then use the rules of addition, subtraction, and multiplication to simplify further. You can also use the distributive property to simplify polynomials.

4. What is the degree of a polynomial?

The degree of a polynomial is the highest exponent of the variable in the expression. For example, the polynomial 3x^2 + 5x + 2 has a degree of 2.

5. How do I solve a polynomial equation?

To solve a polynomial equation, you need to set the equation equal to 0 and then use algebraic methods such as factoring, the quadratic formula, or synthetic division to find the values of the variable that make the equation true.

Similar threads

  • Calculus and Beyond Homework Help
Replies
13
Views
2K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
573
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
  • Math Proof Training and Practice
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
480
Back
Top