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I'm trying to follow a proof of the Bezout's theorem via Euclid's generalized algorithm on pdf1 and pdf2, specially the first one. I'm having trouble understanding homogeneity of some polynomials.

Let $f,g \in k[x_1,x_2,x_3]$ be homogeneous polynomials, apply Euclid's generalized algorithm on them as members of $k(x_1,x_2)[x_3]$ and you get $a,b \in k(x_1,x_2)[x_3]$ with $af + bg = 1$ (Gauss Lemma). Clearing the denominators of $a$ and $b$ we have $pf + qg = r \in k[x_1,x_2]$.

Now, the author say we can assume $p,q$ and thus $r$ as homogeneous, and though it seems to be clear, I can't really understand why. Thanks in advance.

Gustavo Labegalini
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2 Answers2

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Unfortunately your affirmation is false. Take $f= x_1 + x_3$, $g = x_1 - x_3$, $a= b=\frac{1}{2x_1} $; then $p=q=1$ and $af + bg = 1 \in K[x_1,x_2,x_3] $. ($a_2 \nmid pf$).

If I understand correctly your question, given $ f,g \in K[x_1, x_2,x_3]_h $ coprimes, you need $p,q \in K[x_1, x_2, x_3]_h\setminus\{0\} $ such that $pf+qg\in K[x_1, x_2]_h $. By doing an appropiate change of coordinates, we can assume that $f,g \notin K[x_1, x_2]_h $.

By Gauss lemma, there exist $a,b \in K(x_1,x_2)[x_3]$ such that $ af+ bg = 1 $... (*), after clearing denominators we obtain $p, q \in K[x_1, x_2, x_3]\setminus \{0\} $ satisfying $pf + q g \in K[x_1, x_2]$. Now, consider $p = p_0 + \ldots + p_n$, $q = q_0 + \ldots + q_m$, sum of their homogeneous components ($f_i, q_i $ are homogeneous of degree $i$). Choose any $i,j\leq \max\{n,m\}$ that satisfy $i + \deg f = j + \deg g=:d $. Then $r_d = p_{d-i} f + q_{d-j}g $, as desired ($p_{d-i} =0$ implies $g \in K[x_1,x_2]_h$, resp. $q_{d-j}$).

I am willing to accept any correction :)

Abraham Rojas Vega
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After some whatsapp discussion, I may have reached an answer.

Say $a = \frac{p}{a_2}$ and $b = \frac{q}{b_2}$ are already on they're irreducible form, the sum $af + bg \in k[x_1,x_2,x_3]$. Now if $a_2 \nmid p f$ or $b_2 \nmid q g$ we would not have that, (since terms with $a_2 b_2$ denominator wouldn't vanish)$^{\sharp}$, thus:

$$ a_2 \mid p f \qquad \land \qquad b_2 \mid q g $$

By the hypothesis of irreducibility on the fractions, $a_2 \nmid p$ or $b_2 \nmid q$, since $k[x_1,x_2,x_3]$ is a UFD, $a_2 \mid f$ and $b_2 \mid g$, then they are homogeneous, and so is $r = a_2 b_2$ they're multiplication. Then so must be $p$ and $q$ because if one of them were not, $r = pf + qg$ would not be homogeneous.

Gustavo Labegalini
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