I found this result on some notes but I don't understand the steps. Let $k$ be an algebraically closed field.
Every degree 2 homogeneous polynomial is a product of two homogeneous polynomials of degree $1$
Consider the degree $2$ homogeneous polynomial $g(x,y)=ax^2+bxy+cy^2$, then $$g(x,y)=a\bigg[\bigg(x+\frac{b}{2a}y\bigg)^2-\bigg(\frac{b^2-4ac}{4a^2}\bigg) y^2\bigg].$$
Question 1. How do I get this?
Since $k$ is algebraically closed take any $\beta\in k$ with $\beta^2=(b^2-4ac)/4a^2$.
Then we have $$g(x,y)=a(x+(b/2a+\beta)y)(x+(b/2a-\beta)y)$$
Question 2. How do I get this?
Now let $f(x,y)=ax^2+bxy+cy^2+dx+ey+f\in k[x,y]$ be an irreducible quadratic polynomial.
Let $X:=\alpha_1 x+\beta_1 y$ and $Y:=\alpha_2x + \beta_2 y$ be the homogeneous polynomials of degree $1$.
Case 1: $X$ and $Y$ are $k-$linearly dependent
I can't figure out how to get what I'm going to write: transforming the term $dx+ey+f$ with respect to $X$ and $Y$, the quadratucììic polynomial $f(x,y)$ takes the form $$f(x,y)=a_0 X^2+b_0 X+c_0 Y+ d_0.\tag1$$ Since $f(x,y)$ is irreducible $c_0\ne 0$ and hence $X$ and $b_0X+c_0Y+d_0$ are $k-$linearly independent.
Summary of what I would like to know: How do I get $(1)$? What does it mean that two polynomials are linearly dependent and where do I use this hypothesis?
Case 2: $X$ and $Y$ are $k-$linearly independent
In this case $f(x,y)$ takes the form $$f(x,y)=a_0 XY+b_0X+c_0Y+d_0\tag 2$$
Summary of what I would like to know: How do I get $(2)$? What does it mean that two polynomials are linearly independent and where do I use this hypothesis?
Best Answer
Question 1
You can just check the expression expanding all the parentheses. However, I guess that you would like to know how to do this in "direct motion". It seems like this formula for $g(x,y)$ results from the desire to write the function $g$ as a parabola with the explicit centre. For this we need just to complete quadratic and linear in $x$ terms to the full square: $$ ax^2+bxy+cy^2=a(x^2+2x\left(\frac{b}{2a}y\right)+\left(\frac{b}{2a}y\right)^2)-a\left(\frac{b}{2a}y\right)^2+cy^2=\\ =a(x+\left(\frac{b}{2a}y\right))^2-a(\frac{b^2}{4a^2}-\frac{c}{a})y^2=\\ =a\left((x+\frac{b}{2a}y)^2-(\frac{b^2-4ac}{4a^2})y^2\right) $$
Question 2
The field is closed, so the equation $x^2-\frac{b^2-4ac}{4a^2}=0$ must have a solution $\beta$ by definition of the closed field. Then $$ g(x,y)=a\left((x+\frac{b}{2a}y)^2-\beta^2y^2\right) $$ Now we can use the usual square difference formula $a^2-b^2=(a-b)(a+b)$ $$ g(x,y)=a(x+\frac{b}{2a}y-\beta y)(x+\frac{b}{2a}y+\beta y) $$
Case 1
Two polynomials $X,Y$ are linearly dependent means that there are $c_1,c_2 \in k\setminus \{0\}$ such that $$ c_1 X+c_2 Y =0. $$ At this point, I can only guess what the point of the statement is. I guess that $X$ and $Y$ are not arbitrary polynomials, but the polynomials from questions 1 and 2 for the quadratic part of $f(x,y)$. $f$ is an irreducible quadratic polynomial. Thus its quadratic part is not zero, which means that both $X$ and $Y$ are not zero, and we can write $$ f(x,y)=XY+dx+ey+f. $$ Now let's use the fact that those polynomials are linearly dependent and non zero $$ c_1X+c_2Y=0\Rightarrow Y=-\frac{c_1}{c_2}X=a_0 X. $$ Thus we have $$ f=a_0X^2+dx+ey+f. $$
Here we should notice that since X and Y are essentially the same polynomials up to some constant factor, it seems like in your equation (1), you should have at least one small y. Otherwise, you would get reducible $f(x,y)$. I assume that $Y$ in (1) is really $y$. Then the statement is indeed true. We can take $b_0 = \frac{d}{\alpha_1}$, then $$ f(x,y)=a_0X^2+b_0(\alpha_1 x+\alpha_2 y- \alpha_2 y)+ey+f=\\ =a_0X^2+b_0X+(e-b_0\alpha_2)y+f=a_0X^2+b_0X+c_0y+f $$ We see that $c_0 \neq 0$, otherwise $f(x,y)$ would be reducible. $X$ and $b_0 X+c_0 y +f$ must be linearly independent, because if those would be dependent we would be able to write $b_0 X+c_0 y +f=kX$ and get reducible polynomial.
Case 2
To be linearly independent means not to be dependent. Thus, it means that we can't find $c_1,c_2 \in k\setminus \{ 0\}$ such that $c_1 X+c_2 Y=0$. Thus, we again have $$ f(x,y)=XY+dx+ey+f. $$ However, we should notice that since $X$ and $Y$ are independent we can express $x$ and $y$ in terms of $X$ and $Y$. We can do it as follows. $X=\alpha_1 x+\beta_1y$ and $Y=\alpha_2 x+\beta_2 y$. Let's assume that $\beta_2 \neq 0$, the case when $\beta_2=0$ will be considered in a second. If $\beta_2 \neq 0$ we can write $X-\frac{\beta_1}{\beta_2}Y=(\alpha_1-\frac{\alpha_2}{\beta_2})x$. Since $X$ and $Y$ are independent $\alpha_1-\frac{\alpha_2}{\beta_2} \neq 0$, otherwise we would find $c_1$ and $c_2$ form the definition. Thus $x=(\alpha_1-\frac{\alpha_2}{\beta_2})^{-1}(X-\frac{\beta_1}{\beta_2}Y)$. In the same manner you can show that there is an expression for $y$ through the $X$ and $Y$. In the case when $\beta_2=0$ everything is even simpler because $x=\frac{1}{\alpha_2}Y$ and using it we can express $y$ as well. Therefore, using independence of $X$ and $Y$ we can show that there are some coefficients $a_1,b_1,a_2,b_2$ such that $x=a_1X+b_1Y$ and $y=a_2X+b_2Y$. Now we can substitute this into expression for $f(x,y)$ $$ f(x,y)=XY+d(a_1X+b_1Y)+e(a_2X+b_2Y)+f=XY+b_0X+c_0Y+f. $$
comment
I guess that I didn't consider all the possible cases above. As an example, when I took $b_0=\frac{d}{\alpha_1}$ I didn't check $\alpha_1=0$ case. However, it is just an additional technical job. Also, I guess that it could be useful to assume that $\mathrm{char} k \neq 2$