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)
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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)
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Can anyone show me (formally) why the last statement ie " , for otherwise deg deg g(x) which would makje equation (3) impossible " follows?
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 r_{1}(x) ≠ r_{2}(x), then deg(r_{1} - r_{2}) = max{deg(r_{1}),deg(r_{2})} > -∞
(the two remainders can't BOTH be zero, or else they would be equal).
since deg(r_{1}) < deg(g), and deg(r_{2}) < deg(g), it follows that -∞ < deg(r_{1} - r_{2}) < deg(g).
now deg((q_{1}-q_{2})(g)) = deg(q_{1} - q_{2}) + 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 q_{1} - q_{2} = 0.
however, if q_{1} - q_{2} = 0, we get:
(q_{1}(x) - q_{2}(x))(g(x)) = (0)(g(x)) = 0 (0 here means the 0-polynomial, and not the field element)
and since (q_{1}(x) - q_{2}(x))(g(x)) = r_{1}(x) - r_{2}(x), if q_{1} = q_{2}, r_{1} = r_{2}.
since this violates our assumption that r_{1} ≠ r_{2}, we can rule out this niggling doubt, and say with confidence:
deg((q_{1}-q_{2})(g)) ≥ deg(g) > deg(r_{1} - r_{2}),
which is a contradiction, since (q_{1}-q_{2})(g) = r_{1} - r_{2}
(equal polynomials have the SAME degree).
since r_{1} ≠ r_{2} leads to a contradiction, it must be the case that this cannot happen, so in point of fact it must be that r_{1} = r_{2}.
since F[x] is an integral domain (this is important!!!), from (q_{1} - q_{2})(g) = 0, and g ≠ 0, w conclude q_{1} - q_{2} = 0, that is:
q_{1} = q_{2}.
************
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 "x^{n}" 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})(x^{n}) = x^{m+n}
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 (q_{1} - q_{2})(g) = 0, and g ≠ 0, w conclude q_{1} - q_{2} = 0, that is:
q_{1} = q_{2}.'
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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