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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.

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I think you made a typo in b). If $T$ is the integral closure of $R$ in $S$, then you should have $R \subset T \subset S$. –  Rankeya Apr 11 '12 at 18:11
    
Thank you! Fixed. –  Eric Gregor Apr 11 '12 at 18:14

1 Answer 1

up vote 1 down vote accepted

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.

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Why does it suffice to show that this is true for one polynomial? Unless I made a mistake I have already shown that this is true. Where do my (a) and (b) fit in this answer? –  Eric Gregor Apr 11 '12 at 18:38
    
I see what you're getting at, maybe. Are you suggesting that (a) is unnecessary to prove (b)? I think the integral case might be equivalent to the non-integral case provided our $f$ and $g$ both split in some integral ring extension. –  Eric Gregor Apr 11 '12 at 18:45
    
Sorry, I posted something that it seems you already knew. But, this is sufficient. For if $f,g \in S[x]$ with $fg \in T[x]$, then you have some extension $S \subset S'$, such that in $S'[x]$, $f,g$ split into linear factors. So in $S'[x]$, write $f(x)= (x-a_1)...(x-a_m)$, and $g(x) = (x- b_1)...(x-b_n)$. Then $a_i, b_j$ are roots of $fg$, hence integral over $T$. Then the coefficients of $f$, $g$ are integral over $T$. But, then the coefficients of $f$, $g$ must be in $T$, since $T$ is integrally closed in $S$. –  Rankeya Apr 11 '12 at 18:46
    
So, yeah. I think you don't need a) to prove b). –  Rankeya Apr 11 '12 at 18:48
    
Basically my answer is bogus, because it does not answer a), which I think is what you want an answer to. –  Rankeya Apr 11 '12 at 18:49

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