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I am currently looking at Lemma $\mathbf{24}$ from Dathan Ault-McCoy's note 'On Bézout's Theorem'. I will state it below and show the proof presented, commenting my understanding and doubts along the way.

Lemma 24. Let $f \in k[x,y]$ be a nonzero homogeneous polynomial in two variables with degree $n$. Then $f$ factors as $$ f(x,y) = \lambda y^t \prod_{j=1}^s (x-\alpha_i y), $$ for some $\lambda, \alpha_j \in k$ with $\lambda \neq 0$ and $s+t = n$.

Proof. Let $s$ be the $x$ degree of $f$ and let $t = n-s$. Then the term of $f$ with highest degree will be of the form $\lambda y^tx^s.$ Furthermore, since $f$ is homogenous, every other term will have $y$ degree of, at least, $t$. So, we can factor $f$ as $$ f(x,y) = \lambda y^tf'(x,y) $$ where $f'$ is homogeneous of degree $s$ and its degree on $x$ is also $s$. Hence $f'(x,1)$ is a polynomial of one variable that is monic and of degree $s$. Since $k$ is algebrically closed, we can factor this as $$ f'(x,1) = \prod_{j=1}^s (x-\alpha_j), $$ where $\alpha_j$ are the roots of $f'(x,1)$. [UNTIL HERE I UNDERSTAND EVERYTHING]

Now comes the part I don't understand: the author states: $\color{red}{\text{Then by homogeneity}}$

$$ \color{red}{f'(x,y) = y^t f\left( \frac{x}{y},1 \right) = y^t\prod_{j=1}^s \left( \frac{x}{y} -\alpha_j \right) = \prod_{j=1}^s (x - \alpha_j y).} $$

I really can't understand the logic in the equation in red. Perhaps I am missing something really obvious but I can't see it.

Note that after this, the result follows directly.

Thanks for any help in advance.

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    $\begingroup$ Do you know what homogeneity means? Here, one has $f’(zx,zy)=z^tf’(x,y)$. $\endgroup$
    – Aphelli
    Commented Jun 21, 2023 at 20:08
  • $\begingroup$ @Aphelli Thanks for your comment. I know that for a homegeneous polynomial $f(x_1, \dots, x_n)$ of degree $n$ we have that $$ f(\lambda x_1, \lambda x_2, \dots , \lambda x_n ) = \lambda^n f(x_1, \dots, x_n).$$ Is this just a direct application of this? I can't see where $t$ comes from since the degree of $f'(x,y)$ is $s$, perhaps I am just being dumb $\endgroup$
    – xyz
    Commented Jun 21, 2023 at 20:22
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    $\begingroup$ That's a typo - is the $s$ versus $t$ thing the only thing you're concerned about, or is there something else? (PS: it would be nice if you said explicitly what your concern was in the post - knowing it's "why $t$" is much more helpful, because I didn't see that typo the first time I read your post and was inclined to ask you the same question as the first comment.) $\endgroup$
    – KReiser
    Commented Jun 21, 2023 at 20:27
  • $\begingroup$ @KReiser So it should be $s$ instead of $t$ in the statement of the Lemma? Even knowing this, I can't see how to reach the equality in red using the statement Aphelli provided in the comments (or what I stated). I am sorry about any confusion caused in the question. $\endgroup$
    – xyz
    Commented Jun 21, 2023 at 20:31
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    $\begingroup$ Sorry, mis-typed, got caught scrolling. I'll write up an answer in a second so I can give it more care. $\endgroup$
    – KReiser
    Commented Jun 21, 2023 at 20:37

1 Answer 1

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The goal is to move from $f'(x,1) = \prod_{j=1}^s (x-\alpha_j)$ to $f'(x,y)=\prod_{j=1}^s (x-\alpha_jy)$. One hopefully clearer way to do this is to recognize that $f'(\frac{x}{y},1) = \prod_{j=1}^s (\frac{x}{y}-\alpha_j)$, then use the fact that $f'$ is homogeneous of degree $s$ to write $y^s f'(\frac{x}{y},1)=f'(x,y)$ and $y^s \prod_{j=1}^s (\frac{x}{y}-\alpha_j) = \prod_{j=1}^s y(\frac{x}{y}-\alpha_j)= \prod_{j=1}^s (x-\alpha_jy)$.


It appears as if there are a few issues in the red text. Here's the offending line from the post: $$ f'(x,y) = y^t f\left( \frac{x}{y},1 \right) = y^t\prod_{j=1}^s \left( \frac{x}{y} -\alpha_j \right) = \prod_{j=1}^s (x - \alpha_j y) $$

First, the factor of $y$ occuring in the third term should be $y^s$ if it's going to be multiplied with $\prod_{j=1}^s \left( \frac{x}{y} -\alpha_j \right)$ to give $\prod_{j=1}^s (x - \alpha_j y)$. Next, there's no need to bring $f$ in to this, we've already defined $f'$, so I'd prefer to write the second term as $y^sf'(\frac{x}{y},1)$. (You can write something involving $f$ in there if you want, but it seems less clear to me and more likely to be a typo.)

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  • $\begingroup$ @xyz you're making an error with your multiplication there: if you want to pull a factor of $y$ in to each of the $s$ terms in the product, you'll need a $y^s$ outside the product symbol. Try an example: $y^3\prod_{j=1}^3 (\frac{x}{y}-j)=y^3(\frac{x}{y}-1)(\frac{x}{y}-2)(\frac{x}{y}-3) = (x-y)(x-2y)(x-3y)$. $\endgroup$
    – KReiser
    Commented Jun 21, 2023 at 21:16
  • $\begingroup$ You're right. I saw your edit before your comment and understood it (that's why I deleted the previous comment). Thank you for your help and carefull explanation. $\endgroup$
    – xyz
    Commented Jun 21, 2023 at 21:18

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