Existence of a homogeneous system of parameters In the book on invariant theory by Sturmfels, it is stated that every finitely generated graded $\mathbb{C}$-algebra $R$, of finite Krull dimension $n$, has a homogeneous system of parameters. Moreover, it is said that this homogeneous system of parameters always will be algebraically independent over $\mathbb{C}$. 
There are two problems I did not manage to solve. It is clear that by Noether's normalization lemma there always exists $r_1,\ldots,r_n$ in $R$ such that $R$ is finitely generated as a module over $\mathbb{C}[r_1,\ldots,r_n]$. Although it seems very clear to me, I can not find a rigid argument why these $r_1,\ldots, r_n$ should be algebraically independent. I would like to avoid Nakayama's lemma, because I totally have the feeling it should be possible to proof without invoking it.
Secondly, I do not understand how one can conclude that there always exists a homogeneous system of parameters, by the existence of the $r_1,\ldots,r_n$ as in Noether's normalization lemma.
Any help is welcome! 
Note: A homogeneous system of parameters for $R$ is a set of homogeneous elements $\theta_1,\ldots, \theta_n$ such that $R$ as a module over the subring $\mathbb{C}[\theta_1,\ldots,\theta_n]$ is finitely generated. 
 A: First question: Why are the $r_i$'s algebraically independent?
Algebraic independence is sometimes taken as part of the definition of a system of parameters guaranteed by the Noether normalization lemma, but since you are working from a statement of the theorem that does not make this explicit, let me show you why finiteness of $R$ over $\mathbb{C}[r_1,\dots,r_n]$ does actually imply the $r_i$'s are algebraically independent. The key thing is that $n$ is both the number of $r_i$'s and also the Krull dimension of $R$.
Let $r_1,\dots,r_n$ be the elements guaranteed by the Noether normalization lemma such that $R$ is finitely generated as a module over $S:=\mathbb{C}[r_1,\dots,r_n]\subset R$. By assumption, $R$ has Krull dimension $n$. Since finiteness of $R$ over $S$ implies $R$ is integral over $S$ (see Proposition 5.1 in Atiyah-MacDonald), and integrality preserves dimension (proof in appendix below), it follows that $S$ has Krull dimension $n$. Since it is already generated by the $n$ elements $r_1,\dots,r_n$, they must be algebraically independent. We can see this as follows.
Let $\mathbb{C}[X_1,\dots,X_n]$ be the polynomial ring. There is a canonical surjection $\phi:\mathbb{C}[X_1,\dots,X_n]\rightarrow S$ obtained by mapping $X_i\mapsto r_i$ for each $i=1,\dots,n$. Any nontrivial algebraic relation between the $r_i$'s would show up as a nontrivial element of the kernel of $\phi$; thus our job is to show that the kernel of $\phi$ is trivial. Suppose for a contradiction that there existed a nonzero polynomial $F\in \ker\phi$. Let $\mathfrak{p}_0\subset\dots\subset \mathfrak{p}_d$ be a maximal chain of prime ideals of $S$. The fact that $S$ is dimension $n$ means that $d=n$. But on the other hand, $\phi^{-1}(\mathfrak{p}_0)\supset \ker \phi\ni F$, so the inclusion $(0)\subset \phi^{-1}(\mathfrak{p}_0)$ is strict. It follows that $(0)\subset \phi^{-1}(\mathfrak{p}_0)\subset\dots\phi^{-1}(\mathfrak{p}_d)$ is a chain of primes of length $d+1$ in the polynomial ring $\mathbb{C}[X_1,\dots,X_n]$, which has Krull dimension $n$, and therefore $d+1\leq n$. This contradiction proves the nonexistence of $F$, so $\phi$ is injective, and this means the $r_i$'s are algebraically independent.
Second question: Why can the $r_i$'s be taken to be homogeneous?
Strong versions of the Noether normalization lemma exist which explicitly say that if the ring is graded then you can take the $r_i$'s to be homogeneous. For example, Theorem 13.3 in Eisenbud's book, which is even stronger than you need. (See also Lemma 13.2, where much of the work of the proof is done.) The statement is this:

Theorem 13.3 (Noether Normalization). Let $R$ be an affine ring of dimension $d$ over a field $k$. If $I_1\subset\dots\subset I_m$ is a chain of ideals of $R$ with $\dim I_j = d_j$ and $d_1>d_2>\dots>d_m\geq 0$, then $R$ contains a polynomial ring $S=k[x_1,\dots,x_d]$ in such a way that $R$ is a finitely generated $S$-module and
$$ I_j\cap S = (x_{d_i+1},\dots,x_d), \;\forall j=1,\dots,m.$$
If the ideals $I_j$ are homogeneous, then the $x_i$ may be chosen to be homogeneous. If $k$ is infinite, and $R$ is generated over $k$ by $y_1,\dots, y_r$, then for $j\leq d_m$ the element $x_j$ may be chosen to be a $k$-linear combination of the $y_i$.

To apply this statement in your situation, present your $R$ as $T/I$ where $T=\mathbb{C}[X_1,\dots,X_\ell]$ is a polynomial ring and $I$ is a homogeneous ideal. This is possible because $R$ is a finitely generated graded algebra over $\mathbb{C}$; note that the $X_i$'s might need to be assigned degrees other than 1, but this is okay because tracing through Eisenbud's proof shows that the theorem stated here doesn't require a standard grading on the polynomial ring. Then apply Theorem 13.3 to the ring $T$ with the ideal $I_1=I$ (and $m=1$). The conclusion of the theorem supplies you with a subring $S'=\mathbb{C}[R_1,\dots,R_\ell]$, with each $R_i\in T$ homogeneous, such that $I\cap S' = (R_{n+1},\dots,R_\ell)$ and $T$ is finite over $S'$. Thus the canonical homomorphism $\pi:T\rightarrow T/I=R$ maps $R_1,\dots,R_n$ to homogeneous elements $r_1,\dots,r_n\in R$, and $R_{n+1},\dots,R_\ell$ are mapped to zero. Let $S\subset R$ be the image of $S'$ under $\pi$. Then $S=\mathbb{C}[r_1,\dots,r_n]$, and $T/I=R$ is finite over $S'/(S'\cap I) = S$ because it is generated as an $S$-module by the images of $S'$-module generators for $T$ under $\pi$.
Appendix: proof that integrality preserves dimension
Let $R/S$ be an integral extension of commutative, unital rings. Take a maximal chain of prime ideals in $R$ and intersect with $S$ to produce a chain of primes of equal length in $S$; they are distinct by the incomparability theorem [Corollary 5.9 in Atiyah-MacDonald]; thus $\dim R\leq \dim S$. Take a maximal chain of prime ideals in $S$ and lift the bottom one to $R$ by the lying-over theorem [Theorem 5.10 in A-M]; then lift the rest of the chain to $R$ by the going-up theorem [Theorem 5.11 in A-M]; thus $\dim R\geq \dim S$.
