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As for an arbitrary field $K$, we know that its algebraic closure always exists and it is unique up to an isomorphism. However, when we talk about integral closure of some commutative ring $A$, we are always given $A$ as some subring of a larger ring $B$ and its closure is defined to be all the elements of $B$ integral over $A$.

It seems like, due to the lack of cancellation law for multiplication, there doesn't seem to be a natural choice for even "simple extensions" by a root of a polynomial, hence making the concept rather meaningless. It still seems, however, possible to arbitrary append $A$ with a root of a polynomial in $A[x]$ and get some "integral extension" of $A$, albeit it not being a natural choice. For example, $\overline{\mathbb{Q}}$ in $\mathbb{C}$ might be considered as an integral closure of $\mathbb{Z}$.

So here is my question: Given a commutative ring $A$, is there a ring $B$ such that $B$ is integral over $A$ and every polynomial in $A[x]$ somehow splits completely in $B[x]$?

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For a domain $A$ the absolute integral closure $\widetilde{A}$ always exists and is unique up to $A$-isomorphisms. It is defined to be a maximal integrally closed domain, which is an integral extension of $A$. Quite similar to the case of an algebraically closed field. One gets $\widetilde{A}$ by taking the integral closure of $A$ within an algebraic closure of the fraction field of $A$.

Your last remark seems to imply that you have the impression that polynomials $f\in A[X]$ split completely over $\widetilde{A}$. Of course this is not the case in general, and is not intended by the definition. Instead polynomials $f\in A[X]$ split over $\widetilde{A}$ into linear factors and factors without a root in $\widetilde{A}$.

The case of a ring with zero-divisors is more complicated, and I cannot summarize the situation by heart.

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Thanks for the insight. I see how linear factors might not have a root in its absolute integral closure. As for domains, it seemed like it would work out, although I should go work out the details myself. Although, my main concern was about $A$ not being a domain. I don't even know where to start. I tried to come up with a counterexample, but to no avail. – Sam Nov 28 '12 at 9:15

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