Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.