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In section 3.1 (3rd paragraph on page 4) in this paper, I cannot understand why $Q$ and $R$ are homogeneous: enter image description here

(Given $A$, $B$ are homogenous. Capital letters denote homogenous polynmials.)

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Thanks Zev Chonoles, for making it look nice. – Biswarup Ray Jun 26 '12 at 7:40

The trick is to realize that all terms of $q$ (or $r$) are homogeneous of the same degree BUT you have to introduce a grading with negative degrees in $K(y,z)$.

For example, if you want to divide $yz^2+x^3$ by $xz+y^2$ you first use the euclidean algorithm to obtain $$ yz^2+x^3= (\frac {y}{x}z -\frac {y^3}{x^2}) \cdot (xz+y^2) + (x^3+ \frac {y^5}{x^2} ) $$
Notice that every term of $r=x^3+ \frac {y^5}{x^2}$ above has degree $3$, but that $\frac {y^5}{x^2}$ has degree $3=5-2$ because $\frac {1}{x^2}$ has negative degree $-2$ . Similarly every term of $q=\frac {y}{x}z -\frac {y^3}{x^2}$ has degree $1$.

Getting rid of denominators will just multiply all terms of $q$ and $r$ by a homogeneous polynomial $H$ of degree $d$ and will thus conserve homogeneity of $q$ and $r$ .
In our example you multiply throughout by $x^2$ which has degree $2$ and obtain the final equality $$ x^2yz^2+x^5= (xyz-y^3) \cdot (xz+y^2) + (x^5+ y^5 ) $$

where $Q=xyz -y^3$ is homogeneous of degree $1+2=3$ and $R=x^5+ y^5 $ is homogeneous of degree $3+2=5$, as required.

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