A zero for a homogeneous polynomial is a zero for the associated inhomogeneous polynomial I am trying to prove a simple statement from Reid, Undergraduate Algebraic Geometry, pg 16.
Let $F(U,V)$ be a nonzero homogeneous polynomial of degree $d$:
$$F(U,V)=a_dU^d+a_{d-1}U^{d-1}V+...+a_0V^d.$$
We can always find an associated inhomogenous polynomial in 1 variable
$$f(u)=a_du^d+a_{d-1}u^{d-1}+...+a_0.$$
And we have the following statement:
$$f(\alpha)=0\Leftrightarrow(u-\alpha)|f(u)\Leftrightarrow (U-\alpha V)|F(U,V)\Leftrightarrow F(\alpha,1)=0$$
I want to prove each of these arrows as an exercise since it exemplifies some of the things I find confusing about more some more complicated situations. I would prefer the method of proof to be as basic as possible (illustrated by what I have done so far). I have 2 of the 6 directions done.
Arrow 1: $f(\alpha)=0\Leftrightarrow(u-\alpha)|f(u)$
$\Rightarrow$: Help!
$\Leftarrow$: If $(u-\alpha)|f(u)$ then $f(u)/(u-\alpha)=n\in\mathbb{Z}$, so $f(u)=n(u-\alpha)$ and $f(\alpha)=0$.
Arrow 2: $(u-\alpha)|f(u)\Leftrightarrow (U-\alpha V)|F(U,V)$
$\Rightarrow$: Help!
$\Leftarrow$: Help! If I just define $f(U)=F(U,1)$ I can do this: 
$$\frac{F(U,V)}{U-\alpha V}=n\in\mathbb{Z}\rightarrow\frac{F(U,1)}{U-\alpha}=\frac{f(U)}{U-\alpha}=n$$
But I'm guessing that's not his intention. 
Arrow 3: $(U-\alpha V)|F(U,V)\Leftrightarrow F(\alpha,1)=0$
$\Rightarrow$: $F(U,V)/(U-\alpha V)=n\in\mathbb{Z}$ so $F(\alpha,1)=n(\alpha-\alpha)=0$.
$\Leftarrow$: Help!
 A: First off, I don't think what you've got so far is correct. For example, for arrow one, if $f(u)=(u-\alpha)^2,$ then certainly $f(u)/(u-\alpha)=u-\alpha\notin\mathbb Z.$ A similar thing happens in the other parts. 
Arrow 1
Presumably, Reid is working over an algebraically closed field. Thus $f(u)$ decomposes into a product of linear factors, and $\alpha$ is a root if and only if $u-\alpha$ is a factor (up to a scalar). 
(Edit: Of course, as Asal Beag Dubh points out, the first equivalence is true more generally, so long as we are assuming $\alpha$ and the coefficients of $f$ lie in the same field (otherwise we could have something like $f(i)=0$ for $f(u)=u^2+1\in\mathbb R[u]$).)
Arrow 3
I'm skipping to the last equivalence, since $F(u,1)=f(u)$ by definition. Thus $\alpha$ is a root of $f(u)$ if and only if it is a root of $F(u,1).$ So instead of proving the third equivalence directly, it is simpler to notice that it will follow from the second, along with this remark.
Arrow 2
If $u-\alpha$ divides $f(u),$ say $f(u)=(u-\alpha)g(u),$ then we can use the fact that $F(U,V)$ is what we get by replacing $u=U/V$ and multiplying everything by $V^{\deg F}.$ That is, $$F(U,V)=V^{\deg F}(U/V-\alpha)g(U/V)=(U-\alpha V)(V^{\deg F -1}g(U/V))=(U-\alpha V)G(U,V)$$ where $G$ is the homogenization of $g.$
On the other hand, if $U-\alpha V$ divides $F(U,V),$ say $F(U,V)=(U-\alpha V)G(U,V),$ then clearly $f(u)=F(u,1)$ is divisible by $u-\alpha,$ which we see by evaluating both sides of the equality at $(u,1).$
