Thread: Rings of Polynomials _ Division ALgorithm

I am reading Papantonopoulou - Algebra, Ch 8, Rings of Polynomials

Theorem 8.2.2 (Division Algorithm) reads as follows: (see attachment for full proof)

Let F be a field and f(x) and g(x) elements of F[x], with g(x) 0 . Then there exist unique elements Q(x) and r(x) of F[x] such that

(1) f(x) = q(x)g(x) + r(x)
(2) r(x) = 0 or deg r(x) deg g(x)

================================================== ================================

I am having trouble following the uniqueness argument which runs as follows:

Suppose we have:

f(x) =

f(x) =

where or deg deg g(x).


Subtracting the two equations give we get

0 =

Hence

(3)

We must have , for otherwise deg deg g(x) which would makje equation (3) impossible .... etc etc

(see the proof on the attachment)
================================================== =========================================

Can anyone show me (formally) why the last statement ie " , for otherwise deg deg g(x) which would makje equation (3) impossible " follows?

Originally Posted by Bernhard

Suppose we have:

f(x) =

f(x) =

where or deg deg g(x).

First, it should say .

Originally Posted by Bernhard

Subtracting the two equations give we get

0 =

Hence

(3)

We must have , for otherwise deg deg g(x) which would makje equation (3) impossible

In general, if , then . So, if , then . On the other hand, and , so .

in general, deg(f+g) = max{deg(f),deg(g)} (the 0-polynomial sort of goobers things up. for this reason deg(0) is often taken to be -∞).

also deg(fg) = deg(f) + deg(g) (the 0-polynomial REALLY messes with this formula. but see above).

from the second rule we get deg(-f) = deg((-1)(f)) = deg(-1) + deg(f) = 0 + deg(f) = deg(f).

now in the proof at hand: if r₁(x) ≠ r₂(x), then deg(r₁ - r₂) = max{deg(r₁),deg(r₂)} > -∞

(the two remainders can't BOTH be zero, or else they would be equal).

since deg(r₁) < deg(g), and deg(r₂) < deg(g), it follows that -∞ < deg(r₁ - r₂) < deg(g).

now deg((q₁-q₂)(g)) = deg(q₁ - q₂) + deg(g).

we'd like to show that this is greater than or equal to deg(g), but there is one wrinkle: we might have q₁ - q₂ = 0.

however, if q₁ - q₂ = 0, we get:

(q₁(x) - q₂(x))(g(x)) = (0)(g(x)) = 0 (0 here means the 0-polynomial, and not the field element)

and since (q₁(x) - q₂(x))(g(x)) = r₁(x) - r₂(x), if q₁ = q₂, r₁ = r₂.

since this violates our assumption that r₁ ≠ r₂, we can rule out this niggling doubt, and say with confidence:

deg((q₁-q₂)(g)) ≥ deg(g) > deg(r₁ - r₂),

which is a contradiction, since (q₁-q₂)(g) = r₁ - r₂

(equal polynomials have the SAME degree).

since r₁ ≠ r₂ leads to a contradiction, it must be the case that this cannot happen, so in point of fact it must be that r₁ = r₂.

since F[x] is an integral domain (this is important!!!), from (q₁ - q₂)(g) = 0, and g ≠ 0, w conclude q₁ - q₂ = 0, that is:

q₁ = q₂.

************

perhaps you recall when you were looking at "group rings" over a field: F[G]. what polynomials are, are "monoid rings", of the monoid , over the ring (we in general, for most things only need the ring properties of the field F) F, .

in other words, polynomials in F[x] are F-linear combinations of natural numbers (we identity the natural number n with the "xⁿ" term, or if you like, the n-th coordinate (term) in a finite sequence of field elements).

more acccurately, we are identifying the monoid (N,+) with the sub-monoid (<x>,*) of (F[x],*). this is why:

(x^m)(xⁿ) = x^m+n

Thanks for a very helpful post!

Will now work through your post in detail.

Peter

Yes, followed all that and now understand the proof thanks to the clarity of your post.

Thank you by the way for the point:

"since F[x] is an integral domain (this is important!!!), from (q₁ - q₂)(g) = 0, and g ≠ 0, w conclude q₁ - q₂ = 0, that is:

q₁ = q₂.'

I had completely missed this point - and yes, it is very important!

Now will look at polynomials and monoid rings - thanks for that interesting lead ...

Peter

I don't know much about this.But I have learned something useful.







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