3

If $l_1$ and $l_2$ are linear forms in $\mathbb{C}[x, y]$, then $l_1^2 + l_2^2$ is a square if and only if one of the $l_i = 0$ or they scalar multiples of each other.

What is the analogous statement for three linear forms $l_1$, $l_2$, $l_3$ in $\mathbb{C}[x, y, z]$? Clearly the sum $l_1^2 + l_2^2 + l_3^2$ is a square if the nonzero $l_i$ are scalar multiples of each other, but is this the only possibility for it to be a square?

user44413
  • 171
  • 1
    Not sure I get this fully but what happens if you consider $(\frac{x+1}{\sqrt{2}})^2 + (\frac{x-1}{\sqrt{2}})^2 + i^2 = x^2$ ? – Aydin Gerek Nov 07 '16 at 20:15
  • @AydinGerek, those are not linear forms. A linear form is homogeneous linear polynomial, so $l_i = ax + by + cz$, where $a, b, c \in \mathbb{C}$. – user44413 Nov 07 '16 at 21:18
  • 1
    I see. How do you feel about $(\frac{x+y}{\sqrt{2}})^2 + (\frac{x-y}{\sqrt{2}})^2 + (iy)^2 = x^2$ then? – Aydin Gerek Nov 07 '16 at 21:33
  • @AydinGerek, Excellent! Then it is clearly possible to write a square as a sum of three linear forms that are not multiples of each other. – user44413 Nov 07 '16 at 22:04

1 Answers1

2

The analogous statement does not hold. Even in $\mathbb{C}[x,y]$ it is possible for the squares of three linear forms to add up to another square and for these linear forms not to be scalar multiples of each other. For an example consider:

$$ \left(\frac{x+y}{\sqrt{2}}\right)^2 + \left(\frac{x-y}{\sqrt{2}}\right)^2 + (iy)^2 = x^2$$

Aydin Gerek
  • 440
  • A related question which I cannot immediately answer is whether the statement with two squares holds in $\mathbb{C}[x,y,z]$ or indeed in polynomial rings with even greater number of variables. – Aydin Gerek Nov 08 '16 at 06:36
  • 1
    Suppose there are two squares such that their sum is a square: $l_1^2(x,y,z,\dots)+l_2^2(x,y,z,\dots)=p^2(x,y,z,\dots)$. Then $l_1^2=(p-l_2)(p+l_2)$. Then it follows that $l_1^2=0$ (ans, consequently, $l_1=0$) on the union of the sets defined by $p-l_2=0$ and $p+l_2=0$. What are these sets? (Hyper)planes. Can a union of two hyperplanes itself be a hyperplane? Yeah, if they coincide. – Ivan Neretin Nov 08 '16 at 08:44