Possibly wrong proof - about homogeneous polynomials.

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.

• Do you know what homogeneity means? Here, one has $f’(zx,zy)=z^tf’(x,y)$. Commented Jun 21, 2023 at 20:08
• @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
– xyz
Commented Jun 21, 2023 at 20:22
• 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.) Commented Jun 21, 2023 at 20:27
• @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.
– xyz
Commented Jun 21, 2023 at 20:31
• Sorry, mis-typed, got caught scrolling. I'll write up an answer in a second so I can give it more care. Commented Jun 21, 2023 at 20:37

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.)
• @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: $y^3\prod_{j=1}^3 (\frac{x}{y}-j)=y^3(\frac{x}{y}-1)(\frac{x}{y}-2)(\frac{x}{y}-3) = (x-y)(x-2y)(x-3y)$. Commented Jun 21, 2023 at 21:16