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!
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 $B$.