Euclidean division of polynomials Let $f, g \in \mathbb{F}[x]$ be two polynomials with $g \ne 0$. There exist $q, r \in \mathbb{F}[x]$ s.t. $f=qg + r$ and $\deg\ r < \deg\ g$.
I actually have the answer but need a bit of guidance in understanding the answer.
Proof:
We first prove the unique existence of $q, r$ such that $f = qg+r$ and $\deg\ f \geq \deg\ g$. 
Let $f = a{_m}x^m$ and $g = b{_n}x^n$ where both $a{_m}$ and $b{_n} \ne 0$.
So the base case is where $m - n = 0$. Therefore $m = n$, the quotient $q = a{_m}/b{_n}$ and $ r = f-qg$. $q$ is well defined and the coefficient $x^m$ in $r$ vanishes, and where $\deg\ r \lt m = \deg\ g$
So first question is, what do they mean when they say the coefficient of $x^m$ disappear? I understand that we defined the degree of g is greater than degree r, and I suspect they divide out or something of that sort. But can someone clarify that?
So that was the base case. The induction step is defined:
$m = n + d$ where $d \gt 0$. Let $f' = f-(a{_m}/b{_n})x^{m-n}g$. Notice $\deg\ f' \lt \deg\ f$. 
So my second question is why how did the author come to define $f' = f-(a{_m}/b{_n})x^{m-n}g$?
He then goes on:
by induction there exists $q', r$ with $\deg\ r \lt \deg\ g$ and $f'= q'g + r$. Therefore
$f = f' + (a{_m}/b{_n})x^{m-n}q=(q'+(a{_m}/b{_n})x^{m-n})g+r$. 
So at this point I have become totally lost on the induction step. Can anyone offer any guidance?
 A: This will probably make more sense if you work with actual examples. Another good thing to do is review the concepts behind long division and the division algorithm for numbers, since the idea is barely any different at all between number and polynomial settings - I stress this highly. When you do long division with numbers in base ten, you're essentially doing polynomial long division with the number as a "polynomial" in the "variable" $x=10$, since every number has a decimal representation of the form $a_n10^n+\cdots+a_2100+10a_1+a_0$, though with each $0\le a_i<10$.
Suppose you have $f(x)$ and $g(x)$. What would you personally do to find one of the "closest" (in a certain sense) polynomial multiple of $g(x)$ to that of $f(x)$? Let me give you an explicit example,
$$f(x)=3x^3+1,\quad g(x)=2x+1.$$
First, let's understand what notion of closeness we have in mind. We want the closeness of two polynomials to be determined by how many leading coefficients they share. So, to obtain a first approximation, what can we multiply $g(x)$ by so that the resulting leading coefficient matches that of $f(x)$? Why, $\color{Red}{\frac{3}{2}x^2}$, of course. We see that their difference is given not by a cubic but a quadratic:
$$f(x)-\color{Red}{\frac{3}{2}x^2}g(x)=(3x^3+1)-\left(3x^3+\frac{3}{2}x^2\right)=-\frac{3}{2}x^2+1.$$
We can go further and decompose $-\frac{3}{2}x^2+1$. Multiply $g(x)$ by $\color{Green}{-\frac{3}{4}x}$ so that its leading coefficient matches that of $-\frac{3}{2}x^2+1$. For their difference we obtain
$$\left(-\frac{3}{2}x^2+1\right)-\left(\color{Green}{-\frac{3}{4}x}g(x)\right)=\frac{3}{4}x+1.$$
Finally, we must multiply $g(x)$ by $\color{Blue}{\frac{3}{8}}$ for its leading coefficient to match $\frac{3}{4}x+1$, and
$$\left(\frac{3}{4}x+1\right)-\color{Blue}{\frac{3}{8}}g(x)=\frac{5}{8}.$$
Hence, we have
$$f(x)=\left(\color{Red}{\frac{3}{2}x^2}\color{Green}{-\frac{3}{4}x}+\color{Blue}{\frac{3}{8}}\right)g(x)+\frac{5}{8}.$$
Notice that, at each stage, we have reduced our problem of "approximating" a degree $n$ polynomial with multiples of $g(x)$ with that of approximating a degree $n-1$ polynomial with multiples of $g(x)$, and this is where induction comes into play. To go from one level of the induction to an already-assumed case, we take a difference by a "first-approximation" which reduces the degree of what we're trying to approximate. (Going from $f(x)$ to $f(x)-(a_m/b_n)x^{m-n}g(x)$, of lesser degree.)
Review!


*

*http://en.wikipedia.org/wiki/Long_division

*http://en.wikipedia.org/wiki/Division_algorithm

*http://en.wikipedia.org/wiki/Polynomial_long_division
A: Let $\rm\:\ell(f) =\:$ leading term of $\rm\:f,\:$ and $\rm\:d(f)=deg(f).\:$ $\rm\: d(g) < d(f)\:\Rightarrow\: \ell(g)\mid \ell(f),\:$ so $\rm\:\ell(f) = cx^k \ell(g).\:$ Therefore $\rm\:f' \,:=\, f\,-\, cx^k g\:$ has $\rm\  d(f') < d(f)\ $ because the subtraction cancels $\rm\,\ell(f).\,$ 
Thus an element $\rm\:r\:$ of $\rm\:R = f - g\, \Bbb F[x]\:$ of least degree must have
$\rm\:d(r) < d(g)\:$  (for otherwise, as above, we deduce $\rm\:r' \in R\:$ has $\rm\:d(r') < d(r),\:$ contra minimality of $\rm\:r\,).$
