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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $B[x]$?

share|cite|improve this question

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.

share|cite|improve this answer
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

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.