Suppose $f,g$ are polynomials in $\Bbb K^n$ where $\Bbb K$ is a number field (namely $\Bbb Q\subset \Bbb K$). If $fg$ is a homogeneous polynomial then how to (slickly) prove both $f$ and $g$ are homogeneous?
Well, one hopeful way is to group the terms of $f$ by their total orders and do the same for $g$, then their respective highest order terms should multiply into terms whose orders are strictly greater than terms multiplied from other groups. Is there any other slick approach?
On
You can use any field $F$. In the sequel, $\mathbf{x}$ stands for $(x_1,\dots,x_n)$. Write \begin{align} f(\mathbf{x})&=f_0(\mathbf{x})+f_1(\mathbf{x})+\dots+f_r(\mathbf{x})\\ g(\mathbf{x})&=g_0(\mathbf{x})+g_1(\mathbf{x})+\dots+g_s(\mathbf{x}) \end{align} where $f_k$ and $g_k$ are homogeneous of degree $k$. Suppose $fg$ is homogeneous of degree $n$. Then, by definition, $$ f(t\mathbf{x})g(t\mathbf{x})=t^nf(\mathbf{x})g(\mathbf{x}) $$ where $t$ is an indeterminate over $F[\mathbf{x}]$. Thus we have $$ t^n \biggl(\,\sum_{h}f_h(\mathbf{x})\biggr) \biggl(\,\sum_{k}g_k(\mathbf{x})\biggr) = \biggl(\,\sum_{h}t^hf_h(\mathbf{x})\biggr) \biggl(\,\sum_{k}t^kg_k(\mathbf{x})\biggr) $$ Equate the coefficients.
It's maybe easier to see the converse.
We can decompose any polynomial into its homogeneous parts: For $f, g \neq 0$, write $$f = f_a + f_{a + 1} + \cdots + f_{b - 1} + f_b , \qquad g_c + g_{c + 1} + \cdots + g_{d - 1} + g_d,$$ where $f_i$, $g_i$ are homogeneous of degree $i$ (and these representations are minimal in the sense that $f_a, f_b, g_c, g_d$ are nonzero). The lowest- and highest- degree homogeneous parts of $fg$ have degree $a + c$ and $b + d$, but if, e.g., $f$ is not homogeneous, then $a < b$, and so $a + c < b + d$, i.e., $fg$ is not homogeneous.