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Let $A$ be an integral domain. Assume that any monic polynomial (different from $1$) with coefficients in $A$ has a root in $A$. Does it follow that $A$ is a field (necessarily algebraically closed)?

If we assume that any non constant polynomial $f \in A[X]$ has a root in $A$ (an "algebraically closed ring"), then $A$ is a field, for $aX-1$ has a root in $A$ for any $a \in A \setminus \{0\}$. But what happens if the assumption is only made on monic polynomials?

I think that my condition is equivalent to say that $A$ has no integral extension, except $A$ itself.

I didn't find any obvious counter-examples. I tried to show that any maximal ideal had to be $0$, without success.

Thank you for your help!

Watson
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No, it might be the ring of integers of an algebraically closed field, for example (as Eric Wofsey notes in the comments, this idea can be generalized). The integral closure $B$ of $\Bbb Z$ in $\overline{\Bbb Q}$ satisfies this property. Any nonconstant polynomial $f\in B[x]$ necessarily has a root in $\overline{\Bbb Q}$. If we assume $f$ is monic, then because $B$ is integrally closed, the root of $f$ must lie in $B$.

Stahl
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    More generally, you can replace $\overline{\mathbb{Q}}$ with any algebraically closed field $K$ and $\mathbb{Z}$ with any subring of $K$ that is not a field. – Eric Wofsey Dec 04 '16 at 21:16
  • Thank you! Do you know if actually all the examples must arise as integral closure of a subring $D$ of an algebraically closed field $K$ (such that $D$ is not a field)? – Watson Dec 04 '16 at 21:19
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    @Watson: Sure, rather trivially: just let $D=A$ and $K$ be an algebraically closed field containing $A$. Your condition says that $D$ is already integrally closed in $K$. (In fact, you can take $K$ to just be the field of fractions of $A$, since your condition implies the field of fractions is algebraically closed.) – Eric Wofsey Dec 04 '16 at 21:27