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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 ????

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  • $\begingroup$ What are your thoughts? You can start by writing $A,B$ as sums of homogeneous parts, and then... $\endgroup$
    – KReiser
    Dec 28, 2020 at 18:05
  • $\begingroup$ 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 ? . $\endgroup$
    – L. ZWALD
    Dec 28, 2020 at 18:29
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    $\begingroup$ 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. $\endgroup$
    – KReiser
    Dec 28, 2020 at 18:37

1 Answer 1

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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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  • $\begingroup$ Thank you very much KReiser. I am OK with your response. $\endgroup$
    – L. ZWALD
    Jan 5, 2021 at 14:45
  • $\begingroup$ @L.ZWALD After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. $\endgroup$
    – KReiser
    Jan 6, 2021 at 5:11
  • $\begingroup$ This is done. Yhank you. $\endgroup$
    – L. ZWALD
    Jan 7, 2021 at 9:36

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