Modulo $x^2 + y^2 - 1$, is every homogeneous polynomial that is a square of a polynomial, necessarily of sum of squares of homogeneous polynomials?

Question

I am hoping this question is alright for Math Overflow. I didn't get a definitive solution in Math Stack Exchange.

Let $f(x, y) \in \mathbb{R}[x, y]$ be a homogeneous polynomial with real coefficients in $2$ determinates $x , y$. Suppose that $$f(x, y) \equiv g(x, y)^2 \pmod{x^2 + y^2 - 1}$$ for some polynomial $g(x, y)$, where $g(x, y)$ is not necessarily homogeneous.

Is it true that there exists homogeneous polynomials $h_1(x, y), \dots, h_n(x, y)$ such that $$f(x, y) \equiv h_1(x, y)^2 + \cdots + h_n(x, y)^2 \pmod{x^2 + y^2 - 1}?$$

Answer 1

Yes. Will Sawin's answer uses a topological fact about the unit circle in $\mathbb{R}^2$. Here is an answer replacing $\mathbb{R}$ with any field of characteristic not $2$.

First, working modulo $x^2+y^2-1$, we can multiply terms of $g$ with powers of $x^2+y^2$, so we can assume $g = g_{2k} + g_{2k+1}$ for some $k$, where each $g_i$ is homogeneous of degree $i$. For convenience write these as $g_0,g_1$.

Observe that

$$ f \equiv (g_0 + g_1)^2 \equiv (g_0^2(x^2+y^2) + g_1^2) + 2 g_0 g_1 \pmod{x^2+y^2-1}. $$

There is a polynomial $h$ so that

$$ f + (x^2+y^2-1)h = (g_0^2(x^2+y^2) + g_1^2) + 2 g_0 g_1 . $$

Write $h = h_0 + h_1$ where $h_0$ is all the terms of even degree and $h_1$ is all the terms of odd degree (these are not necessarily homogeneous).

The first case is if $f$ is homogeneous of odd degree. In this case

$$ f + (x^2+y^2-1)h_1 = 2 g_0 g_1, \qquad (x^2+y^2-1)h_0 = g_0^2(x^2+y^2) + g_1^2 $$

In the second equation the right hand side is homogeneous. The only way for the left hand side to be homogeneous is to have $h_0 = 0$. Then $g_0^2(x^2+y^2) + g_1^2 = 0$, but $x^2+y^2$ is not a perfect square (or negative of a perfect square) outside of characteristic $2$. So it must be $g_0=g_1=0$, and then $f \equiv 0$.

If $f$ is homogeneous of even degree then

$$ f + (x^2+y^2-1)h_0 = g_0^2(x^2+y^2) + g_1^2, \qquad (x^2+y^2-1)h_1 = 2 g_0 g_1 $$

Since $x^2+y^2-1$ is irreducible outside characteristic $2$ (in characteristic $2$ it's equal to $(x+y+1)^2$) it must be either $g_0 \equiv 0$ or $g_1 \equiv 0$ modulo $x^2+y^2-1$.

So then $f \equiv g_1^2$ or $f \equiv g_0^2$. These are the same polynomials that Will Sawin's answer ends with.

Answer 2

Yes. In fact, you can take $n=1$.

Let $g = \sum_{i=0}^d g_i$ with $g_i$ homogeneous of degree $i$. Then $$ \begin{aligned} g^2 &= \bigg( \sum_{k=0}^{ \lfloor d/2\rfloor} g_{d-2k} + \sum_{k=0}^{ \lfloor d-1/2 \rfloor } g_{d-1-2k} \bigg)^2 \\ &= \bigg( \sum_{k=0}^{ \lfloor d/2\rfloor} g_{d-2k} \bigg)^2 + \bigg(\sum_{k=0}^{ \lfloor d-1/2 \rfloor } g_{d-1-2k} \bigg)^2 + 2 \bigg( \sum_{k=0}^{ \lfloor d/2\rfloor} g_{d-2k}\bigg )\bigg( \sum_{k=0}^{ \lfloor d-1/2 \rfloor } g_{d-1-2k} \bigg). \end{aligned} $$

Now $f(x,y) \mapsto f (-x,-y)$ is an involution of $\mathbb R[x,y]$ for which homogeneous polynomials are eigenfunctions, with eigenvalue $1$ if the degree is even or eigenvalue $-1$ if the degree is odd.

If $f$ has odd degree then $f(-x,-y) = -f(x,y)$, so $f$ takes a negative value on the unit circle unless it's identically zero on the unit circle. The first possibility contradicts that $f$ is congruent to a square mod $x^2+y^2-1$ and the second possibility makes $f$ a multiple of $x^2+y^2-1$ so the problem is trivial. So we may assume $f$ has even degree.

Furthermore $(x^2+y^2-1)$ is invariant, so the space of multiples of $x^2+y^2-1$ is an invariant subspace. Thus
$$g^2 = \bigg( \sum_{k=0}^{ \lfloor d/2\rfloor} g_{d-2k} \bigg)^2 + \bigg(\sum_{k=0}^{ \lfloor d-1/2 \rfloor } g_{d-1-2k} \bigg)^2 + 2 \bigg( \sum_{k=0}^{ \lfloor d/2\rfloor} g_{d-2k}\bigg )\bigg( \sum_{k=0}^{ \lfloor d-1/2 \rfloor } g_{d-1-2k} \bigg)$$ is an eigenfunction (mod $x^2+y^2-1$ ) with eigenvalue $1$. Thus $$2 \bigg( \sum_{k=0}^{ \lfloor d/2\rfloor} g_{d-2k} \bigg)\bigg( \sum_{k=0}^{ \lfloor d-1/2 \rfloor } g_{d-1-2k} \bigg)$$ is divisible by $x^2+y^2-1$. Since $x^2+y^2-1$ is an irreducible polynomial, one of the two terms is divisible by $x^2+y^2-1$, so $f$ is congruent mod $x^2+y^2-1$ to either $$\bigg( \sum_{k=0}^{ \lfloor d/2\rfloor} g_{d-2k}\bigg )^2 \quad\text{or}\quad \bigg(\sum_{k=0}^{ \lfloor d-1/2 \rfloor } g_{d-1-2k}\bigg )^2 .$$ But both of these are congruent to the square of a homogeneous polynomial, i.e. $$\bigg( \sum_{k=0}^{ \lfloor d/2\rfloor} g_{d-2k}(x^2+y^2)^k \bigg)^2 \quad\text{or}\quad \bigg(\sum_{k=0}^{ \lfloor d-1/2 \rfloor } g_{d-1-2k} (x^2+y^2)^k\bigg)^2 $$