Two questions about integral "splitting ring" extensions We have a ring $R$, commutative, and $f_1,\dots,f_n$ polynomials in $R[x]$ monic, with $\deg f_i\ge 1$. It is straightforward to show that there is a ring extension $R\subset S$ such that $S$ contains all of the roots of the set of polynomials (the polynomials split in $S$). Basically one uses the result that if $x$ is integral over $R$, then $R[x]$ is a finitely generated as an $R$-module, and then induct.
I would like to
a) Show that there is an integral ring extension $T$ over $R$ such that every monic polynomial (non-constant) splits in $R$.
and use a) to show
b) That if we have $R\subset T\subset S$, with $T$ the integral closure of $R$ in $S$, then for any $f,g\in S[x]$ monic, with $fg\in T[x]$, then $f$ and $g$ are each in $T[x]$. [No assumptions about the integrality of the ring $R$.]
If we could assume that $R$ were integral it would be easier, since we could think of things in the field of fractions. Any help for how to proceed would be much appreciated.
 A: It suffices to show for b) that 

given a ring $R$ and a monic $f(x) \in R[x]$, there is some ring $D$, with $R \subset D$ such that $f(x)$ split into linear factors in $D[x]$. 

Proceed by induction on the degree of $f(x)$. The case where $\deg(f(x)) = 1$ is trivial. So, let $\deg(f(x)) =n$, where $1 \leq n$. We have a natural injection $R \rightarrow R[T]/(f(T))$, so we can view $R$ as a subring of $R' =R[T]/(f(T))$. Hence, $R[x]$ is a subring of $R'[x]$. 
Now, $f(x)$ has a root in $R'$. So, we have in $R'[x]$, $f(x) = (x - r)f'(x)$, where $
\deg(f'(x)) = n-1$. Then by induction $f'(x)$ splits into linear factors in some extension $D$ of $R'$. In that extension $f(x)$ clearly splits into linear factors.
A: First note that if $R$ is the product of algebraically closed field then chose $T=R$ and you get a)
If you furthermore assume that $R$ has the multiplicative identity and is reduced/semiprime then you can show a). 
Proof:
Since $R$ is reduced, $R$ can be embedded (ring monomorphism) into product of fields (in fact this is a characterisation of reducedness for commutative unitary rings). The product of fields can be given by the product of algebraic closure of the quotient fields of the integral domain $R/\mathfrak{p}$  for each prime ideal $\mathfrak{p}$ of $R$. Let us call this product of fields $A$. Take $T$ to be the integral closure of $R$ in $T$ and you have a). 
You can, in fact, even do more. You can generalize algebraic closureness by a further assumption: a needed property is essential extension of rings. For this I refer you to the paper of Mel Hochster: "Totally integrally closed rings and extremal spaces", Pac. Journal of Math. 1970, Vol. 32, No. 3, p.767-779. Note: here he calls essential extension, tight extension. The definition is much more palatable in the category of commutative unitary rings (you can generalize the definition to any category).
