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An algebraic set (not necessairly a variety) $X \subseteq \mathbb{A}^2$ defined by a polynomial of degree $2$ is called a conic.

The problem is:

Show that any irreducible conic is isomorphic either to $Z(y-x^2)$ or to $Z(xy-1)$ after an affine change of coordinates in $\mathbb{A}^2$.

However, with this definition of conic, I only can conclude that if the polynomial is irreducible, then the conic is irreducible (because $f$ irreducible $\Rightarrow Z(f)=X$ is irreducible)

What can I do to prove that if a conic is irreducible, then the polynomial that defines the conic is irreducible?

And for the main problem, how can I do the affine change of coordinates?

Thanks

Leafar
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    Hint : think to linear algebra. You can associate to each conic a matrix $A$ such that the equation of the conic is $(x,y)A(x,y)^T = 0$ ... Secondly if you work on an algebraic closed field don't forget that you can factor $x^2 + y^2$ for example ... –  Mar 22 '15 at 14:18
  • Isn't your vector $(x,y)$ supposed to be $(x,y,1)$? I'm reading from http://en.wikipedia.org/wiki/Matrix_representation_of_conic_sections to remember the linear algebra – Leafar Mar 22 '15 at 14:41
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    If you have a degree two polynomial which is reducible into factors then the factors can have degree one. Using the correspondence between zero sets of product of polynomials and the union of the zero set of each factor you get reducibility of the conic iff reducibility of the polynomials. Adding to the above hint: write down the most general degree 2 equation in two variables and then try to use substitutions for example given $x^{2}+ y^{2}$ a change of variables $u = x+ iy, v= x-iy$ yields $uv = 0$... – DBS Mar 22 '15 at 14:41
  • But, writing the most general degree 2 equation in two variables: $ax^2+bxy+cx^2+dx+ey+f=0$; how can I get a thing like $x^2+y^2$ if we have the $a$ and the $c$? And by the way thanks for clarify the first doubt – Leafar Mar 22 '15 at 15:01
  • oh yeah, I forgot to put the 1 you´re right. –  Mar 22 '15 at 15:14
  • I have some notes of linear algebra that prove that a conic can be reduced in 3 forms: $Jx^2+Ky^2=L$, $x^2=Ly$ or $y^2=Lx$ – Leafar Mar 22 '15 at 18:04
  • However, we must not forget that the conic is irreducible! – Leafar Mar 22 '15 at 18:05
  • @DBS Can you explain better the idea? – Leafar Mar 22 '15 at 18:17
  • Ok, omfg, after searching several pages on google, I think I found the solution of this problem, and it's hard and full of change of coordinates, which I couldn't make by myself. I'll post the solution later (after understanding it) – Leafar Mar 22 '15 at 19:00

2 Answers2

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Dragos Oprea's solution from problem 3 on solution set 2 to the course Fall 2012: Algebraic Geometry (Math 203A) at UC San Diego:

Let $F(x,y)=ax^2+by^2+cxy+dx+ey+f$. We will show that after a linear/affine change of coordinates the conic can be written as $XY − 1 = 0$ or $Y − X^2 = 0$. We discuss the following cases.

Case1. If $a=b=0$, then $F(x,y)=cxy+dx+ey+f =c(xy+ \frac{d}{c}x+ \frac{e}{c}y) +f =c(x+ \frac{e}{c})(y+ \frac{d}{c}) +\tilde{f}$ If $\tilde{f} = 0$, $F$ is reducible, which is not allowed. Therefore $\tilde{f}\neq 0$.

Let $X=−\frac{c}{\tilde{f}} (x+\frac{e}{c})$ and $Y =y+\frac{d}{c}$, so that $F =−\tilde{f}XY +\tilde{f}$. Thus $F (x, y) = 0$ implies $XY−1=0$.

After an affine change of coordinates, the conic $Z(F)$ becomes $XY−1=0$.

Case 2. If either $a$ or $b$ is not $0$, without loss of generality, we may assume $a \neq 0$. Then, $F=ax^2 +by^2 +cxy+dx+ey+f=a(x+\frac{c}{2a}y)^2 +\tilde{b}y^2 +dx+ey+f$ Let $x_1=\sqrt{a}(x+ \frac{c}{2a}y)$ and $y_1=y$, (choose any one of the square roots). There exist constants $\tilde{d}$, $\tilde{e}$, $\tilde{f}$ such that $F=x_1^2+by_1^2+\tilde{d}x_1+\tilde{e}y_1+\tilde{f}= (x_1+\frac{\tilde{d}}{2})^2+by_1^2+\tilde{e} y_1+\tilde{\tilde{f}}$. Let $x_2 = x_1 + \frac{\tilde{d}}{2}$.

Subcase (i): If $b = 0$, $F = x_2^2 +\tilde{e} y_1 + \tilde{\tilde{f}}$. We claim $\tilde{e}\neq 0$ because otherwise $F =(x_2 +i \sqrt{\tilde{\tilde{f}}})(x_2 −i \sqrt{\tilde{\tilde{f}}})$ is reducible. Let $X=x_2$, $Y =−(\tilde{e}y_1+\tilde{\tilde{f}})$. Then $F = X^2 − Y$.

Subcase (ii): If $b\neq 0$, let $y_2=\sqrt{b}y_1+\frac{\tilde{e}}{2\sqrt{b}}$ so that $F = x^2 + y^2 + g$. Letting $X=\sqrt{−g}(x_2+iy_2)$ and $Y=\sqrt{−g}(x_2−iy_2)$, we have $F = −g(XY − 1)$.

Therefore, the conic $Z(F)$ can be written in the form $XY −1=0$ after an affine change of coordinates.

Jan-Magnus Økland
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To the last question: Saying "the" polynomial might be a little misleading, since $Z(f) = Z(f^2)$ and so on. But actually this is the only thing that can go wrong: if you write down a prime factorization $f = p_1^{m_1} \cdots p_r^{m_r}$ then $Z(f) = Z(p_1) \cup \cdots \cup Z(p_r)$ is the decomposition of $Z(f)$ into irreducible components. Check this! So if a hypersurface like this is irreducible then it really is the zero set of an irreducible polynomial. [It's harder to show that a subvariety of $\mathbb{A}^n$ of dimension $n - 1$ must be of this form; see (1.13) of Hartshorne.]

But this brings up an issue: under your definition it seems that $Z(x^2)$ is an irreducible conic, and that's probably not something you want.

Regarding the problem overall: It's somehow less scary than it looks. It's very common to suggest moving to $\mathbb{P}^2$ and working with quadratic forms but one can just charge ahead and complete the square wherever possible. If I can think of a more intermediate hint I'll update this post, but do try it.

Hoot
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