Characterization of homogeneous polynomials I'm studying on Lang's Algebra the formal definition of polynomial rings, in particular I'm at the chapter of several variables polynomials.
Let $A$ be a commutative ring with unity ($1$), so we have associated at $A$ the ring $A[X_1,X_2,...,X_n]$. Define $X:=(X_1,X_2,...,X_n)$ and $f(X) \in A[X]$ a polynomial in $n$ variables, and define $f^{(d)}$ to be the polynomial in which every term has degree $d$, so 
$f^{(d)}= \sum_{v_1+\cdots+v_n=d}{a_{v_1,...,v_n}X^{v_1}\cdots X^{v_n}} $ for some $a_i \in A$. 
So we can see that every polynomial in $A[X]$ of degree $n$ can be written as $f(X)= \sum_{d=0}^{n}{f^{(d)}}$
A polynomial is called homogeneous if $f(X)=f^{(d)} $.
Now, given $0\neq f(X) \in A[X]$ there is the following
Proposition: $f(X)$ is homogeneous $\iff$ given $n+1$ algebraically independent elements over $A$, say $(u,t_1,t_2,...,t_n)$, $f(ut_1,ut_2,...,ut_n)=u^df(t_1,...,t_n)$.
This is my proof.
"$\Rightarrow$" $$f^{(d)}= \sum_{v_1+\cdots+v_n=d}{a_{v_1,...,v_n}X^{v_1}\cdots X^{v_n}} = \sum_{v_1+\cdots+v_n=d}{a_{v_1,...,v_n}(ut_1)^{v_1}\cdots(ut_n)^{v_n}}=\sum_{v_1+\cdots+v_n=d}{a_{v_1,...,v_n}u^{v_1+\cdots+v_n}t_1^{v_1}\cdots t_n^{v_n}} =u^df^{(d)}.$$ Here note that the hypothesis of the algebraic independence is needed to say that those terms don't become $0$ with the evaluation homomorphism.
"$\Leftarrow$" Here I can consider $(u,X_1,...,X_n)$ and evaluating the polynomial in these variables the thesis follows.
Is my proof correct? 
 A: The direction from left to right is correct, except that you never actually need the algebraic independence for this direction: You only need to show $f(u t_1,\ldots,u t_n) = u^d f(t_1,\ldots,t_n)$, and that you have done with your calculation and by using the fact $f =f^{(d)}$.
However, the direction from right to left is actually the more difficult one. You seem to have missed this. I assume you want to try it yourself, so here's how to get started:
We are given a polynomial $f \in K[X_1,\ldots,X_n]$ and a nonnegative integer $d$ such that the property on the RHS is satisfied.
That is, for any tuple $(u,t_1,\ldots,t_n)$ of algebraically independent elements it holds that $f(ut_1,\ldots,ut_n)=u^d f(t_1,\ldots,t_n)$. We must show that $f$ is homogeneous of degree $d$, that is, $f = f^{(d)}$.
Assuming $f$ is of degree $n$, we may write $f=\sum_{d'=0}^{n} f^{(d')}$ with $f^{(d')}$ homogeneous of degree $d'$. Now let $(u,X_1,\ldots,X_n)$ be algebraically independent and substitute
$$
u^d f(X_1,\ldots,X_n)=f(uX_1,\ldots,uX_n) = \sum_{d'=0}^{n} f^{(d')}(uX_1,\ldots,uX_n) = \ldots
$$
Can you continue from here on?
