# 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!

• 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) Dec 28, 2021 at 1:32
• Thank you so much!! It helps to have a precise reference correctly stated. I am grateful!! Dec 29, 2021 at 2:05
• 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! Dec 29, 2021 at 2:17

• 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 $$m-k$$.