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# Abstarct-Algebra

In Abstarct-Algebra , some of the proofs of polynomial rings are below.

Polynomial Rings

Definition of Polynomial Ring :

Let R be a ring . An infinite sequence $$(a_{0}, a_{1}, a_{2},....a_{n}, ......)$$ of elements of R, with at most a finite number of non-zero terms, is called a Polynomial over the ring R.

Q) If $$f(x),g(x)\epsilon F(x)$$ and g(x)|f(x) then prove that $$(f(x))\subset (g(x))$$ .

Proof:

g(x)|f(x) $$\Rightarrow$$ f(x)=a(x)g(x), for some a(x)$$\epsilon$$F(x) .

Let h(x)$$\epsilon$$((f(x)) then h(x)=b(x)f(x) , where b(x) is a

polynomial function in F(x).

h(x)=b(x)f(x)

=b(x)a(x)g(x)

=c(x)g(x)   , where c(x)=b(x)a(x)$$\epsilon$$F(x).

h(x) $$\epsilon$$ (g(x))

If  h(x)  $$\epsilon$$  (f(x)) $$\Rightarrow$$   h(x)$$\epsilon$$(g(x))

Hence (f(x))$$\subset$$(g(x)).

Q2)   If $$f(x),g(x)\epsilon F(x)$$ and f(x), g(x) are co-primes and f(x)|h(x),          g(x)|h(x) then prove that f(x)g(x)|h(x).

Proof :   f(x), g(x) are co-primes means no common factor other than 1.

f(X)|h(x) $$\Rightarrow$$ h(x)=a(x)f(x)  where a(x) $$\epsilon$$F(x).

If  g(x)|a(x)f(x) $$\Rightarrow$$ g(x)|a(x) as g(x) does not divide f(x).

g(x)|a(x)  $$\Rightarrow$$ a(x)=b(x)g(x) , b(x) $$\epsilon$$F(x).

but  we have h(x) = a(x)f(x)

h(x) = b(x)g(x)f(x)

Therefore f(x)g(x)|h(x) .

Hence it is proved.

What we learned in the branch of Abstarct-Algebra , Polynomial Rings:

• If $$f(x),g(x)\epsilon F(x)$$ and g(x)|f(x) then  $$(f(x))\subset (g(x))$$
• If $$f(x),g(x)\epsilon F(x)$$ and f(x), g(x) are co-primes and f(x)|h(x),          g(x)|h(x) then  f(x)g(x)|h(x).
• Two functions are said to be Co-Primes if they have only 1 as their common factors, i.e., no common zeros .

In the following page it has the definition of Irreducible Polynomial and its examples .