This is Exercise 7.30 of Gathmann's 2021 notes of Algebraic Geometry.
A conic in $\mathbb{P}^2$ over an algebraically closed field $K$ of characteristic not equal to $2$ is an irreducible quadric curve, which is known to be the zero locus of an irreducible homogeneous polynomial of degree $2$.
The set of homogeneous polynomials of degree $2$ in $K[x_0,x_1,x_2]$ is of the form
$$\{c_0x_0^2+c_1x_1^2+c_2x_2^2+c_3x_0x_1+c_4x_0x_2+c_5x_1x_2\mid c_0,\cdots, c_5\in K, \exists i\in \{0,\cdots,5\} \text{ s.t. }c_i\neq 0 \}. $$
By sending $(c_0,\cdots, c_5)$ to $[y_0:\cdots :y_5]\in \mathbb{P}^5$, the conics in $\mathbb{P}^2$ are identified with a subset $U$ in $\mathbb{P}^5$.
My question is: The exercise asserts that $U$ is open in $\mathbb{P}^5$, but I don't see a clue how to show this.
I think all I need is a general method to tell that if a homogeneous polynomial of degree $2$ is irreducible using its coefficients, which might translate the reducible ones to the zero locus of a homogeneous polynomial. But I just cannot come up with any of such things. Could anyone help me a little bit?
Thanks in advance for any help.
Best Answer
Let me give two ways of doing this, first following the suggestion of Albanese above.
Any reducible conic is just the product of two linear forms. So it is precisely the image of $\mathbb{P}^2\times \mathbb{P}^2\to\mathbb{P}^5$ given by $(l_1,l_2)\mapsto l_1l_2$, where the $\mathbb{P}^2$ parametrizes linear forms. The image is closed (do you know why?) and thus the complement is open.
The other argument is to think of quadrics as bilinear forms. Then it corresponds to a symmetric $3\times 3$ matrix ( the entries can be thought of as the coordinates of $\mathbb{P}^5$) and $U$ is defined as where this matrix is nonsingular and thus open.