Flatness and the Family of Projective Plane Curves I've been working through Chapter 6: Flat Families of Eisenbud's Commutative Algebra: With a View Toward Algebraic Geometry, and I have gotten myself stuck on Exercise 6.6, which deals with the family of projective plane curves. The question reads as follows (with some paraphrasing to keep it short):

Consider $k\mathbb{P^3}$, and fix a degree $d$. For each multi-index $\alpha = (a_0,a_1,a_2)$ with $a_0+a_1+a_2 = d$ let $x_\alpha$ be an indeterminate. Set $R = k[\{x_\alpha\}]$, and let $S$ be the $R$-algebra: $$ S = \frac{R[y_0,y_1,y_2]}{\left(\sum_\alpha x_\alpha y_0^{a_0}y_1^{a_1}y_2^{a_2}\right)}.$$ Show that:
$(a)$ if we invert any $x_\alpha$, then the family becomes flat. That is, the $R[x_\alpha^{-1}]$-algebra $S[x_\alpha^{-1}]$ is flat, by showing that it is a free module (it is not finitely generated).
$(b)$ $S$ is an integral domain, and moreover that it contains $R$, so that it is torsion-free as a $R$-module.
$(c)$ $S$ is actually not flat over $R$, by proving and using the following two facts:

*

*If $S$ is a flat module over a ring $R$ and $R \to T$ is any map of rings, then $S\otimes_R T$ is flat over $T$.


*There is a map of rings $R = k[\{x_\alpha\}] \to k[t] = T$, such that $$T \otimes_R S = \frac{k[t,y_0,y_1,y_2]}{(ty_0^d)}$$ is not a flat $T$-algebra.

I understand geometrically that this corresponds to all projective curves of degree $d$, and that the $x_\alpha$ represent the "coefficients" of the curves. I also know that the fiber of any prime $P$ (essentially picking out a specific set of constants to get an irreducible curve) is a polynomial ring over $\text{Frac}(P)$ modulo the equation of degree $d$ that we choose.
I can prove part $(b)$ without too much trouble, but I am very much stuck on part $(a)$. I have tried a few obvious aproaches such as trying to enumerate a basis, but I am getting confused with the structure of the localization. I would very much appreciate a hint for part $(a)$ and possibly part $(c)$.
Thanks in advance.
 A: Ok so I have figured out this problem completely. For part $(a)$, simply look at the generating set given by all monomials of any degree, that is, $\{1,y_0,y_1,y_2,y_0y_1, y_1y_2,y_0y_2,...\}$. A more explicit decomposition turns $S$ into a graded module, with each homogeneous component $S_k$ having all monomials of degree $k$ as its generating set.
Now returning to our large generating set, we need to show that if we invert some $x_\beta$, then this generating set becomes a basis. If we denote $y_0^{a_0} y_1^{a_1} y_2^{a_2}$ by $y^\alpha$ for the multi-index $\alpha = (a_0,a_1,a_2)$, in $S$: $$\sum_\alpha x_\alpha y^\alpha = 0.$$ 
So pre-localization, this generating set fails to be a basis. However, in $S[x_\beta^{-1}]$, we can use the the fact that $x_\beta$ is a unit to obtain
$$- x_\beta^{-1} \left( \sum_{\alpha\neq \beta} x_\alpha y^\alpha \right) = y^\beta.$$ All this tells us is that in the localization, one of the generating elements is linearly dependent on the others, and so should be omitted from the generating set. Since there are no other relations between the other generators, we get unique representation, and so this shows that $S[x_\beta^{-1}]$ is free.
For part $(b)$, simply use the graded decomposition stated above, and note that $S_0 = R$. Showing that $\sum_\alpha x_\alpha y^\alpha$ is irreducible is easy (indeed, you are taking the sum of all monomials of fixed degree, nothing can factor out), and sufficient to show that $S$ is an integral domain.
Part $(c)$ follows by contradiction from the two listed facts, the first is proven by noting that for an injection of $T$-modules $N \to N'$, we have 
$$(S\otimes_R T) \otimes_T N \to (S \otimes_R T)\otimes_TN', $$ which by associativity, gives $$S \otimes_R N \to S\otimes_R N'.$$ Since $S$ is flat and $R \to T$ is a ring morphism (so $N$, $N'$ can be regarded as $R$-modules), have we that the above is an injection. So the tensor product is flat. The second fact is just a choice about where to send the $x_\alpha$ plus a computation.
