[proofplan]
We prove both assertions directly from the monomial formula for homogenization. First, we expand an affine polynomial $f$ into monomials and observe that setting $x_0=1$ removes exactly the powers of $x_0$ introduced during homogenization. For the converse, we expand the [homogeneous polynomial](/page/Homogeneous%20Polynomial) $G$ by separating the exponent of $x_0$ from the exponents of $x_1,\dots,x_n$; dehomogenization sets the factor $x_0^r$ and every remaining $x_0$ to $1$, and homogenizing back to the prescribed degree $\deg G$ restores the missing powers of $x_0$.
[/proofplan]
[step:Write the affine polynomial in multi-index form and dehomogenize its homogenization]
Let $\alpha=(\alpha_1,\dots,\alpha_n)\in\mathbb Z_{\ge 0}^n$ be a multi-index, and define
\begin{align*}
|\alpha|:=\alpha_1+\cdots+\alpha_n.
\end{align*}
For such $\alpha$, write
\begin{align*}
x^\alpha:=x_1^{\alpha_1}\cdots x_n^{\alpha_n}\in A.
\end{align*}
Since $f\in A$ is nonzero, it has a finite monomial expansion
\begin{align*}
f=\sum_{\alpha\in E} c_\alpha x^\alpha,
\end{align*}
where $E\subset\mathbb Z_{\ge 0}^n$ is finite, each $c_\alpha\in k\setminus\{0\}$, and
\begin{align*}
d=\deg f=\max_{\alpha\in E}|\alpha|.
\end{align*}
By the definition of homogenization to total degree $d$,
\begin{align*}
f^h=\sum_{\alpha\in E} c_\alpha x_0^{d-|\alpha|}x^\alpha.
\end{align*}
Therefore dehomogenizing by substituting $x_0=1$ gives
\begin{align*}
(f^h)^{\mathrm{deh}}=\sum_{\alpha\in E} c_\alpha 1^{d-|\alpha|}x^\alpha.
\end{align*}
Since $1^{d-|\alpha|}=1$ for every $\alpha\in E$, this becomes
\begin{align*}
(f^h)^{\mathrm{deh}}=\sum_{\alpha\in E} c_\alpha x^\alpha=f.
\end{align*}
[guided]
The definition of homogenization is designed to add just enough powers of the new variable $x_0$ so that every monomial has the same total degree. Let $\alpha=(\alpha_1,\dots,\alpha_n)\in\mathbb Z_{\ge 0}^n$ be a multi-index, define
\begin{align*}
|\alpha|:=\alpha_1+\cdots+\alpha_n,
\end{align*}
and write
\begin{align*}
x^\alpha:=x_1^{\alpha_1}\cdots x_n^{\alpha_n}.
\end{align*}
Because $f$ is a nonzero polynomial in $A=k[x_1,\dots,x_n]$, it has a finite monomial expansion
\begin{align*}
f=\sum_{\alpha\in E} c_\alpha x^\alpha,
\end{align*}
where $E\subset\mathbb Z_{\ge 0}^n$ is finite and each coefficient $c_\alpha$ is nonzero. The degree of $f$ is
\begin{align*}
d=\max_{\alpha\in E}|\alpha|.
\end{align*}
Thus, for each monomial $c_\alpha x^\alpha$, the exponent $d-|\alpha|$ is a nonnegative integer. Homogenization to degree $d$ is therefore
\begin{align*}
f^h=\sum_{\alpha\in E} c_\alpha x_0^{d-|\alpha|}x^\alpha.
\end{align*}
Now dehomogenization on the chart $D_+(x_0)$ means substituting $x_0=1$. Applying this substitution to the displayed formula gives
\begin{align*}
(f^h)^{\mathrm{deh}}=\sum_{\alpha\in E} c_\alpha 1^{d-|\alpha|}x^\alpha.
\end{align*}
Every factor $1^{d-|\alpha|}$ equals $1$, including the case $d-|\alpha|=0$. Hence
\begin{align*}
(f^h)^{\mathrm{deh}}=\sum_{\alpha\in E} c_\alpha x^\alpha=f.
\end{align*}
This proves that dehomogenizing the usual degree-$d$ homogenization recovers the original affine polynomial.
[/guided]
[/step]
[step:Expand the homogeneous factor $G$ with the $x_0$ exponent determined by total degree]
Set $D:=\deg G$. Since $G\in S$ is homogeneous of total degree $D$, there is a finite set $E\subset\mathbb Z_{\ge 0}^n$ and coefficients $c_\alpha\in k\setminus\{0\}$ such that
\begin{align*}
G=\sum_{\alpha\in E} c_\alpha x_0^{D-|\alpha|}x^\alpha.
\end{align*}
Indeed, each monomial of $G$ has the form
\begin{align*}
c x_0^{\beta_0}x_1^{\beta_1}\cdots x_n^{\beta_n}
\end{align*}
with
\begin{align*}
\beta_0+\beta_1+\cdots+\beta_n=D,
\end{align*}
so after setting $\alpha=(\beta_1,\dots,\beta_n)$ one has $\beta_0=D-|\alpha|$. The uniqueness of monomial expansion in the [polynomial ring](/page/Polynomial%20Ring) $S$ justifies collecting the corresponding coefficients as the $c_\alpha$.
[/step]
[step:Dehomogenize $F=x_0^rG$ and homogenize back to degree $\deg G$]
Using $F=x_0^rG$ and the definition of dehomogenization,
\begin{align*}
F^{\mathrm{deh}}=F(1,x_1,\dots,x_n)=1^rG(1,x_1,\dots,x_n)=G(1,x_1,\dots,x_n).
\end{align*}
Substituting the expansion of $G$ from the previous step gives
\begin{align*}
F^{\mathrm{deh}}=\sum_{\alpha\in E} c_\alpha x^\alpha.
\end{align*}
Now homogenize this affine polynomial to the prescribed total degree $D=\deg G$. By the defining formula for degree-$D$ homogenization,
\begin{align*}
(F^{\mathrm{deh}})^h=\sum_{\alpha\in E} c_\alpha x_0^{D-|\alpha|}x^\alpha.
\end{align*}
The right-hand side is exactly the monomial expansion of $G$, so
\begin{align*}
(F^{\mathrm{deh}})^h=G.
\end{align*}
Multiplying by $x_0^r$ in $S$ yields
\begin{align*}
x_0^r(F^{\mathrm{deh}})^h=x_0^rG=F.
\end{align*}
This proves both the converse identity and the final factorization.
[guided]
The factor $x_0^r$ has no effect after dehomogenization because the chart $D_+(x_0)$ sets $x_0=1$. Starting from the factorization $F=x_0^rG$, the definition of dehomogenization gives
\begin{align*}
F^{\mathrm{deh}}=F(1,x_1,\dots,x_n)=1^rG(1,x_1,\dots,x_n)=G(1,x_1,\dots,x_n).
\end{align*}
From the previous step, the homogeneous polynomial $G$ of total degree $D$ has the expansion
\begin{align*}
G=\sum_{\alpha\in E} c_\alpha x_0^{D-|\alpha|}x^\alpha.
\end{align*}
Substituting $x_0=1$ into this expansion removes exactly the powers of $x_0$, so
\begin{align*}
F^{\mathrm{deh}}=\sum_{\alpha\in E} c_\alpha x^\alpha.
\end{align*}
Now we homogenize this affine polynomial to the specified total degree $D=\deg G$, not necessarily to its ordinary affine degree. This distinction is important because some highest-degree affine terms may be absent after dehomogenization. By the degree-$D$ homogenization formula, every monomial $c_\alpha x^\alpha$ is multiplied by $x_0^{D-|\alpha|}$, and hence
\begin{align*}
(F^{\mathrm{deh}})^h=\sum_{\alpha\in E} c_\alpha x_0^{D-|\alpha|}x^\alpha.
\end{align*}
The right-hand side is the already established monomial expansion of $G$, so
\begin{align*}
(F^{\mathrm{deh}})^h=G.
\end{align*}
Multiplying this equality by $x_0^r$ in the polynomial ring $S$ gives
\begin{align*}
x_0^r(F^{\mathrm{deh}})^h=x_0^rG=F.
\end{align*}
This proves the converse identity and the stated factorization.
[/guided]
[/step]
