In a field, if we require that all polynomials have at least one root, then it's algebraically closed and all polynomials factor completely. In a *ring*, the same requirement implies that it's an ACF, because linear polynomials need a root and this ensures multiplicative inverses. This raises the question, is there some other way a ring can be "almost" algebraically closed, without turning completely into a field?


So my question: Is there a ring, in which all polynomials of degree $\ge 2$ have a root, that is not also a field?


Naturally we can't require that all polynomials of degree $\ge 2$ have as many roots as their degree, otherwise one of the two roots of $$ax^2 - (a-1)x + 1 = (x-1)(ax-1)$$ would give us an inverse of $a$.



Ring where nonlinear polynomials have at least one root
Rings in which every monic polynomial has a root, are called *absolutely integrally closed*.


Such rings are not necessarily a field: For example the ring of all algebraic integers (that is the ring all roots of monic polynomials with rational integral coefficients) is absolutely integrally closed, but no field.


For more Information on absolutely integrally closed rings see <https://stacks.math.columbia.edu/tag/0DCK>. See also Section 4.7 of the book "Swanson, Huneke: Integral Closure of Ideals, Rings and Modules" (LMS, Lecture Notes Series 336), online: [https://www.math.purdue.edu/~iswanso/book/SwansonHuneke.pdf](https://www.math.purdue.edu/%7Eiswanso/book/SwansonHuneke.pdf) for results on the absolute integral closure of a reduced ring with finitely many minimal primes.


