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I am just beginning to learn algebraic geometry. An exercise in Reid, p. 24 is to prove that if $Q(x,y,z)$ is a quadratic form over a field $k$ with at least 4 elements, and $Q$ vanishes on the zero set of the conic $\{xz=y^2\}\subset \mathbb{P}^2_k$, then $Q=\lambda (xz-y^2)$. I have solved the problem, I assume in the way the author intended, but in trying to integrate the solution with my previous knowledge, I feel that I am missing something. Can you help me get a sense of "the real story"?

The solution I have: $\{xz=y^2\}\subset \mathbb{P}^2_k$ is parametrized by $(u:v)\in \mathbb{P}^1_k \mapsto (u^2:uv:v^2)\in \mathbb{P}^2_k$. Thus $Q(u^2,uv,v^2)$ vanishes for all choices of $u,v$. This is a homogeneous polynomial in two variables, so I can interpret its zero set as a set of elements in $\mathbb{P}^1_k$. It is degree 4 so unless it is identically zero as a polynomial, it cannot have more than $4$ zeros in $\mathbb{P}^1_k$. But $\mathbb{P}^1_k$ has at least 5 points, (at least) 4 for the elements of $k$ and also $\infty$. Since $Q(u^2,uv,v^2)$ vanishes on all of them, it must be zero as a polynomial. Now an elementary calculation with generic coefficients for $Q$ shows it has the desired form. (Auxiliary question: is this correct?)

This kind of reasoning in terms of projective space is new to me so I sought an alternative solution in the hopes that it would help me understand this solution better.

ATTEMPTED alternative solution: Since $Q$ and $xz-y^2$ are low degree and I want to show $(xz-y^2)|Q$, divide $Q$ by $xz-y^2$ in the ring $k(z,y)[x]$ to obtain a remainder $R\in k(z,y)$ since $xz-y^2$ is degree 1 in $x$. Clear the denominator in $R$ to obtain an element of $k[z,y]$. Essentially this amounts to replacing $zx$ with $y^2$ throughout $z^2Q$. Thus $R$ is a homogeneous degree at most 4 polynomial in $y$ and $z$, and by construction it vanishes whenever $xz=y^2$.

I want to conclude that $R$ is zero as a polynomial, but I can't do what I did last time because I do not know that $R$ vanishes for every $(y:z)\in \mathbb{P}^1_k$. In particular, I do not know if it vanishes on $(1:0)$, although it does vanish on $(u:1)$, all $u\in k$ (i.e. on a copy of $\mathbb{A}^1_k\subset \mathbb{P}^1_k$), because let $x=u^2$. Thus, e.g. if the field is $\mathbb{F}_4$, I only know that the associated inhomogeneous polynomial to $R$ has four zeros, and this means I do not know it is the zero polynomial.

It seems to me that morally, what is going on is that I'm attempting to parametrize $xz=y^2$ by taking $(y:z)\in\mathbb{P}^1$, solving $xz=y^2$ for $x$, and then mapping $(y:z)\in\mathbb{P}^1_k \mapsto (x:y:z)\in\mathbb{P}^2_k$, but this isn't working because if $(y:z)=(1:0)$ then there is no $x$ solving $xz=y^2$. But I remain confused. In particular:

Questions:

1) Is there a way to do this without referring to projective space? We are proving a statement about low-degree polynomials in 3 variables. I feel like I should be able to do this with very simple commutative algebra. (If $k$ is algebraically closed, then the result is immediate from Hilbert's Nullstellensatz, but this feels like way too much power and doesn't work if $k$ is, e.g., $\mathbb{F}_4$.) Can you see such a solution?

2) Can you help me understand better "why" the second (attempted) solution above didn't work? (Apologies for the vague question. I am going for "moral / conceptual" satisfaction, a slippery thing. I will appreciate any attempt to help.) Is there a way to complete that solution using projective or non-projective ideas?

Thanks. Apologies in advance for the imprecision of the questions.

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up vote 1 down vote accepted

Your first solution is correct. The second one also, but you have to go further. You have $$z^2Q(x,y,z)=(xz-y^2)F(y,z)+R(y,z).$$ Over $\mathbb F_4$, your hypothesis implies that $R(y,z)=c(y^4-yz^3)$ for some constant $c$. Then $F(y,0)=cy^2$. But the term $cxzy^2$ doesn't show up in $R(y,z)$ neither in $z^2Q(x,y,z)$, so $c=0$ and you are done.

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Yes, got it, this works. So if I want to avoid talking about projective space, the proof splits into two cases based on whether $k=\mathbb{F}_4$ or not? Any thoughts on why this happens? (I.e. why did using projective ideas make the proof more economical?) –  Ben Blum-Smith Nov 15 '11 at 15:45
    
I think this is because with the projective space $P^2$, you don't privilege any point, while with second method, you need a special treatement for the point $x=1,y=z=0$. –  user18119 Nov 15 '11 at 23:40
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