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Let $p=X^2-Y \in \Bbb R[X,Y]$ and let $g \in \Bbb R[X,Y]$. Show that $g$ can be written as $g=pq+r$ for $q \in \Bbb R[X,Y]$ and $r \in \Bbb R[X]$.
Hint: $g$ can be written as $\sum_{i=0}^n f_iY^i$, where $f_i \in \Bbb R[X]$. Use induction on $n$.

I don't quite know how to approach this problem with the given hint. Doesn't this follow from the fact that the remainder must have degree in $Y$ less than the degree of $X^2-Y$ and so it would need to have degree $0$ on $Y$ implying that it's a polynomial only on $X$?

How would this induction approach work here?

dxiv
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Gregori
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2 Answers2

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Hint : show first that for every $i\geq 0$, $Y^i$ can be written $Y^i=pq_i+r_i$ where $q_i\in {\mathbb R}[X,Y]$ and $r_i\in{\mathbb R}[X]$. Then, you can deduce $g=\sum_{i=0}^n f_i Y^i = \bigg(\sum_{i=0}^nf_iq_i\bigg)p+\bigg(\sum_{i=0}^nf_ir_i\bigg)$.

Ewan Delanoy
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Consider the bivariate polynomial $\,h(X,Y)=g(X,Y)-g(X,X^2)\,$, then $\,h(X,X^2)=0\,$ so $\,Y-X^2 \,\mid h(X,Y)\,$ per Does there exist a formal statement of the Multivariable Factor Theorem.

It follows that $\,h(X,Y) = (Y-X^2)q(X)\,$, therefore $\,g(X,Y) = \underbrace{h(X,Y)}_{(Y-X^2)q(X)} + \underbrace{g(X,X^2)}_{r(X)}\,$.

dxiv
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