Let $J \subseteq \mathbb{K}[x_o,\ldots,x_n]$ be a homogeneous ideal.
I am struggling to 'prove' (more to understand) this statement in my Algebraic Geometry book:
Every homogeneous element of $J$ can be written in the form $x_0^d f^h$ (where $f^h$ denotes the homogenization of $f$ with respect to $x_0$) for some $d \in \mathbb{N}$ and for some $f \in \mathbb{K}[x_1,\ldots,x_n]$.
We let $f$ be a homogeneous polynomial in $J$. Then, $f$ is a sum of monomials $f_k$ and those monomials are all of the same degree $deg(f)$. What can I do now?
Other fact: In general, $(f+g)^h$ is in general not equal to $f^h + g^h$
Thanks!
Let $g(x_0,\ldots,x_n) \in J$ be a homogeneous polynomial.
Denote $f(x_1,\ldots,x_n) := g(1,x_1,\ldots,x_n)$ the de-homogenization of $g$.
Then, $f^h(x_0,\ldots,x_n) = g(x_0,\ldots,x_n)$.
But $f^h(x_0,\ldots,x_n) = x_0^{deg(f)}f(\frac{x_1}{x_0},\cdots,\frac{x_n}{x_0})$.
So we conclude that $g(x_0,\ldots,x_n) = x_0^{deg(f)}f(\frac{x_1}{x_0},\cdots,\frac{x_n}{x_0})$, which is what we had to prove.
[Credit to @Hoot]