# a linear combination of polynomials is a homogeneous polynomials implies that each polynomial is homogeneous

Let $$F$$ and $$G$$ be relatively prime homogeneous polynomials in $$k[X,Y,Z]$$ and $$A,B\in k[X,Y,Z]$$ such that $$AF+BG$$ is a homogeneous polynomial. How to show that $$A$$ and $$B$$ are homogeneous polynomials ?

I am interested in this question since in the proof of Bezout Theorem of Fulton's book

http://www.math.lsa.umich.edu/~wfulton/CurveBook.pdf

page 58, it is asserted the following.

Let $$\pi:k[X,Y,Z]_d\longrightarrow k[X,Y,Z]_d/(F,G)$$ be the canonical map where $$k[X,Y,Z]_d$$ is the set of homogeneous polynomials of degree $$d$$. Let $$\varphi:k[X,Y,Z]_{d-m}\times k[X,Y,Z]_{d-n} \longrightarrow k[X,Y,Z]_d$$ defined by $$\varphi(A,B)=AF+BG$$ where $$m=deg(F)$$ and $$n=deg(G)$$. Fulton asserts that $$Im(\varphi)=Ker(\pi)$$.

However, $$Ker(\pi)$$ is the set of $$AF+BG$$ such that $$AF+BG$$ is homogeneous of degree $$d$$ with $$A$$, $$B$$ in $$k[X,Y,Z]$$. In order to show that it is $$Im(\varphi)$$, one has to show that $$A$$ and $$B$$ are homogeneous. The counter-example in the previous comment of K.Reiser seems to show that the assertion of Fulton is wrong ????

• What are your thoughts? You can start by writing $A,B$ as sums of homogeneous parts, and then... Dec 28, 2020 at 18:05
• Is the following reasoning correct ? $A=\sum_i A_i$ where $A_i$ is a monomial of degree $d_i$, $B=\sum_j B_j$ where $B_j$ is a monomial of degree $e_j$, $F=\sum_a F_a$ where $F_a$ is a monomial of degree $m$, $G=\sum_b G_b$ where $G_b$ is a monomial of degree $n$. $AF+BG$ is the sum of monomials $A_iF_a$ and $B_jG_b$. Since $AF+BG$ is homogeneous, each of these monomials has the same degree. Thus each monomials of $A$ get the same degree and each monomials of $B$ get the same degree. Finally, $A$ and $B$ are homogeneous. Correct ? . Dec 28, 2020 at 18:29
• No, because $F=x$, $G=y$, $A=y+1$, $B=-x+1$ provides a counterexample. (Should have seen this when I wrote the first comment!) Still, that sort of attempt should be edited in to your post, not placed in the comments; you should also add a bit about why you're interested in the problem. Dec 28, 2020 at 18:37

It is not true that if $$F,G$$ are relatively prime homogeneous polynomials in $$k[x,y,z]$$ and $$A,B\in k[x,y,z]$$ are such that $$AF+BG$$ is homogeneous, then $$A,B$$ are homogeneous: consider $$F=x$$, $$G=y$$, $$A=y+1$$, $$B=-x+1$$.
This is fine in your situation, though: the goal is to say that if $$P$$ is a polynomial in $$k[x,y,z]_d$$ which lies in $$(F,G)$$, then there exist $$A\in k[x,y,z]_{d-m}$$ and $$B\in k[x,y,z]_{d-n}$$ so that $$P=AF+BG$$. (Note that this is not what you've written!) We can start by using the fact that $$P\in (F,G)$$ to write $$P=CF+DG$$ for some polynomials $$C,D\in k[x,y,z]$$. Now write $$C=\sum C_i$$ and $$D=\sum D_i$$ where $$C_i,D_i$$ are homogeneous of degree $$i$$. This gives that $$P= \left(\sum C_iF\right)+\left(\sum D_iG\right).$$ Therefore the degree-$$e$$ homogeneous part of $$P$$ is equal to $$C_{e-m}F+D_{e-n}G$$, and by the fact that $$P$$ is homogeneous of degree $$d$$, we have that $$C_{e-m}F+D_{e-n}G=0$$ when $$e\neq d$$ and $$C_{e-m}F+D_{e-n}G=P$$ when $$e=d$$.
So $$P=C_{e-m}F+D_{e-n}G=P$$ and therefore if $$P\in \ker(\pi)$$, $$P$$ is in the image of $$\varphi$$ as $$\varphi(C_{e-m},D_{e-n})$$ and the claim is proven.
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