Proof of Lemma 2,2 in the book Elliptic Curves, Number Theory and Cryptography by Lawrence Washington.

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!

Solution 1:

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 $m-k$.