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Let $f \in \Bbb C[x_1, ..., x_n]$ be a homogeneous polynomial of degree $d$, and assume that $f=gh$, where $g,h \in \Bbb C[x_1, ..., x_n]$. Is it true that $g$ has to be a homogeneous polynomial of degree $k \leq d$.

I tried to write $$f(\lambda x_1, ..., \lambda x_n) = \lambda^d f(x_1, \dots, x_n) = g(\lambda x_1, ..., \lambda x_n)h(\lambda x_1, ..., \lambda x_n) = \lambda^d g(x_1, \dots, x_n) h(x_1, \dots, x_n)$$

for all $\lambda, x_i \in \Bbb C, \lambda \neq 0$. What can I do next?

Alphonse
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  • If you write $g=g_r+g_{r+1}+\cdots+g_m$, where $g_i$ is homogeneous of degree $i$, what does $g(\lambda x_1,\ldots,\lambda x_n)$ look like in terms of the $g_i$s? – Mohan Mar 20 '17 at 17:19

1 Answers1

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Decompose $g$ and $h$ into homogeneous components, $g = \sum g_i$ and $h = \sum h_j$ with $\deg g_i = d_i$ and $\deg h_j = e_j$. It follows that $f = \sum _{i,j} g_i h_j $ and, since $f$ is homogeneous, it follows that the products $f_i g_j$ must all have the same total degree, namely $\deg f$. This means that $d_i + e_j = \deg f$ for all $i,j$. Choosing $i \ne k$ it follows that $d_i + e_j = \deg f = d_k + e_j$, whence it follows that $d_i = d_k$, i.e. the components of $g$ have all the same degree, i.e. $g$ is homogeneous. The same argument applied to $h$ shows that $h$ is homogeneous too.

Alex M.
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