Show that there existe two polynomials $q$ and $r$ in $K[x]$ such that $f=qg+r$ 
For a field $K$, we set $K[x]$ for the set of polynomials with the
  variable $x$ and with coefficient in $K$. The notion of degree will be
  taken as usual, that is, the greatest degree of $x$ that appears in a
  polynomial. Given $f$ and $g$ in $K[x]$, where $g \not= 0$, show that
  there existe two polynomials $q$ and $r$ in $K[x]$ such that $f=qg+r$
  (division with remainder).

I would like to solve the problem myself, but I don't know how to begin. Are there someone who could give me a good hint to begin.
 A: Look at the proof for integers. In this case, one looks at the set $I=\{a-bq\mid q\in\mathbb{Z}\}$, and uses the fact that $I^+=I\cap\mathbb{Z}_{\geq 0}$ is nonempty and bounded below (hence, has a smallest element).
For $K[x]$, consider the set $I=\{f-gq\mid q\in K[x]\}$. Note that if $0\in I$ you are done. Otherwise consider $I^+=\{\deg(r)\mid r\in I\}$ and proceed like you would for integers. 
Theorem: Let $a,b\in\mathbb{Z}$, $b\neq 0$. Then there exist unique $q,r\in \mathbb{Z}$ such that
$$a=bq+r\;\;\;0\leq r<|b|.$$
Proof: We may assume that $b>0$ (after proving the statement for $b>0$, we have $a=(-b)q+r=b(-q)+r$ for $b<0$).
As above, we see that $I^+$ is nonempty and bounded below, so $I^+$ contains a least element $r$. Since $r\in I$, $r=a-bq$ for some $q$, so we get
$$a=bq+r$$
We now want to see that $0\leq r <b$. If not, say $k=r-b\geq 0$, then $k=a-b(q+1)\in I^+$ contradicting the minimality of $r$. Hence, $0\leq r<b$ as desired.
For uniqueness, assume $bq+r=a=bt+s$ where $0\leq r,s<b$. Then $b(q-t)=s-r$, so $b|(s-r)$. But, $-b<s-r<b$, forcing $s-r=0$. Thus, $s=r$ and $q=t$.
A: hint : prove that $K[x]$ is an euclidian domain
