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I am having problems understanding this proof. I will state the lemma and the proof and then ask my question. This is on Page 22 of the Second Edition of the book. I have simply repeated the Lemma and the proof verbatim from the book for the reader's convenience.

LEMMA 2.2 Let $G(u,v)$ be a non-zero homogeneous polynomial and let $(u_0 : v_0)\in{\bf{P}}^{1}(K).$ Then, there exists an integer $k\geq{0}$ and a polynomial $H(u,v)$ with $H(u_0 , v_0)\neq{0}$ such that $$G(u,v) = (v_0{u}-{u_0}v)^{k}H(u,v).$$

PROOF. Suppose ${v_0}\neq{0}.$ Let $m$ be the degree of $G$. Let $g(u) = G(u,v_0 ).$ By factoring out as large a power of $(u-u_0 )$ as possible, we can write $$g(u) = (u-{u_0})^k{h(u)}$$ for some $k$ and for some polynomial $h$ of degree $m-k$ with $h(u_0 )\neq{0}$. Let $$H(u,v) =(v^{m-k}/{v_0}^m)h(u{v_0}/v),$$ so $H(u,v)$ is homogeneous of degree $m-k$. Then, $$G(u,v)=({\frac{v}{v_0}})^mg({\frac{uv_0}{v}})={\frac{v^{m-k}}{v^{m}_0}}({v_0}u-{u_0}v)^kh({\frac{uv_0}{v}})$$ $$=({v_0}u-{u_0}v)^kH(u,v).$$ as desired. If $v_0 =0,$ then ${u_0}\neq{0}.$ Reversing the roles of $u$ and $v$ yields the proof in this case.

This proof seems to assume (unless I am missing something) that if $G(u,v)$ is homogeneous of degree $m$, and if $v_0\neq{0},$ then $g(u) = G(u,v_0)$ must be a polynomial in the single variable $u$ of degree $m$. I am having difficulty convincing myself that this statement is always true. Consider for example $$G(u,v)=uv.$$ This polynomial is homogeneous of degree $2$. But if $v_0\neq{0},$ then $$g(u) = G(u,v_0 ) = u{v_0}$$ has degree $1$ as a polynomial in $u$. What am I missing here? Thanks in advance for indulging me!

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    $\begingroup$ I've fixed a couple errors in your question. The first one ($v$ instead of $u$) is really understandable that you'd miss it, so rarely will anyone hold it against you. The second ("homogenous" instead of "homogeneous"), however, may result in your leaving a bad impression on some mathematicians/scientists, so make sure to use "homogeneous" (you'll almost never need to use "homogenous" as it's an older term from biology, but its existence means your standard spell check will not catch it) $\endgroup$ Dec 28, 2021 at 1:32
  • $\begingroup$ Thank you so much!! It helps to have a precise reference correctly stated. I am grateful!! $\endgroup$
    – student
    Dec 29, 2021 at 2:05
  • $\begingroup$ I didn't even realize that the correct term is "homogeneous". That there was a difference between the two words is something I seem not to have taken in. Thanks for alerting me. I will remember! $\endgroup$
    – student
    Dec 29, 2021 at 2:17

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I agree that the proof technically assumes what you're written about the degrees, and this is an error. It's a rather harmless error as to the structure of the proof, as it doesn't really matter.

Here's the fix (which curiously doesn't seem to be in the official errata for the 2nd edition):

  • Replace "for some polynomial $h$ of degree $m-k$ with $h(u_0)\neq 0$" with "for some polynomial $h$ of degree at most $m-k$ with $h(u_0)\neq 0$"

Explanation: The only place where the degree of $h$ matters is when we write $$H(u,v) =(v^{m-k}/{v_0}^m)h(u{v_0}/v).$$ We want $H$ to be a polynomial so the factor $v^{m-k}$ must clear the denominator in $h(uv_0/v)$. This will happen as long as the degree of $h$ is at most $m-k$.

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  • $\begingroup$ It struck me that this proof of the associative law is probably skipped over by many readers. There are other proofs-using Bezout's Theorem or The Riemann-Roch Theorem. So people may not work through it carefully. $\endgroup$
    – student
    Dec 29, 2021 at 3:25
  • $\begingroup$ @student I couldn't be sure if it's just people skimming and that's why it's not in the current errata. If you look at the errata for the first edition, however, there were many corrections to this proof, so someone was very invested in correcting it. I don't have access to a first edition to see, but it's also possible that this [relatively] minor error was created in the flurry of edits that followed. $\endgroup$ Dec 29, 2021 at 6:05
  • $\begingroup$ Possible! It’s a lovely concrete book! Am planning to read in addition other accounts like Knapp’s excellent book in Princeton Mathematical Notes series. He seems to give a lot of intuition! $\endgroup$
    – student
    Dec 30, 2021 at 1:47

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