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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 $a$.

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    $\begingroup$ Another interesting variant would be to restrict to monic polynomials, so that the inverse trick no longer works. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Feb 4, 2018 at 6:08
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    $\begingroup$ Doesn't $\ ax^2-1=0\ $ give an inverse to each $a\neq 0$? $\endgroup$
    – abx
    Commented Feb 4, 2018 at 6:50
  • $\begingroup$ @abx, you're totally right! I'm sorry, this question was silly then. The monic question is still interesting though. $\endgroup$
    – Alex Meiburg
    Commented Feb 4, 2018 at 7:02
  • $\begingroup$ @AlexMeiburg please edit the question (including title). abx made a comment of his remark because it's not worth an answer, so it can just ask the new one (with monic) and add a remark reflecting the previous comments. $\endgroup$
    – YCor
    Commented Feb 4, 2018 at 13:46
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    $\begingroup$ I note that the condition that all monic polynomials have roots is preserved under quotients. Thus, e.g., if $R$ is the non-field ring from @tj_ 's comment, and $I$ is a non-prime ideal of $R$, then the ring $R/I$ also satisfies the condition, and it is not even an integral domain. $\endgroup$
    – Emil Jeřábek
    Commented Feb 4, 2018 at 20:26

2 Answers 2

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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 for results on the absolute integral closure of a reduced ring with finitely many minimal primes.

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  • $\begingroup$ @Brandenburg: Fixed the link (maybe you can just delete our comments) $\endgroup$
    – tj_
    Commented Dec 29, 2022 at 23:46
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I think that your condition implies that a monic degree $n$ polynomial has $n$ roots (counting multiplicity). Maybe you want to consider the weaker condition

  • a degree $n$ polynomial is a product of $n$ linear factors.

For a commutative ring with an identity the familiar result holds with the familiar proof:

Any polynomial $f(x)$ (even if not monic) we have that $r$ is a root , $f(r)=0$, exactly when there is a polynomial $g(x)$ with $f(x)=(x-r)g(x).$

So saying that every polynomial has at least one root implies that a degree $n$ polynomial is a product of $n$ linear factors.

NOTES:

  • I don't know that there are any non-field examples. I'm just trying to salvage the question.

  • Over the integers, $f(x)=6x^2-5x+1=(2x-1)(3x-1)$ although the polynomial has no roots.

  • That factorization holds $\bmod 30$ as well, though we might prefer to write it as $6x^2+25x+1.$ That makes it look as though there are no roots $\bmod 30.$

    However when $x=17,$ we have $(2x-1)(3x-1)=33\cdot 50 =0 \bmod 30.$

    And when $x=23,$ we have $(2x-1)(3x-1)=45\cdot 68 =0 \bmod 30.$

    And , indeed, $$(x-17)(6x-23)=(6x-17)(x-23)=6x^2-5x+1$$

  • Again $\bmod 30,$ we have $$x^2-19=(x-7)(x+7)=(x-7)(x-23)$$ and also $$x^2-19=(x-13)(x+13)=(x-13)(x-17)$$ Of course $x^2-1$ also has four roots, $\pm 1,\pm 11.$ But that seems less mysterious.

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  • $\begingroup$ Could you please elaborate why this holds for any commutative ring with 1? Usually, for polynom division one needs $R[X]$ to be an Euclidean domain (but I guess one can weaken it to be a gcd domain). $\endgroup$
    – M.G.
    Commented Feb 4, 2018 at 13:40
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    $\begingroup$ @July: Let $f(x)=a_nx^n + \cdots + a_1x+a_0$. Then $f(x)=f(x)-f(r) = a_n(x^n-r^n) + \cdots + a_1(x-r)=(x-r)\sum_{k=1}^n\sum_{i=0}^{k-1}r^{k-1-i}x^i$. $\endgroup$
    – tj_
    Commented Feb 4, 2018 at 17:13
  • $\begingroup$ @tj_: oh, I see, I was thinking too complicated. Thanks! $\endgroup$
    – M.G.
    Commented Feb 4, 2018 at 17:21
  • $\begingroup$ I agree that monic polynomials having one to imply n roots, but how is being a product of n linear factors any weaker? $\endgroup$
    – Alex Meiburg
    Commented Feb 4, 2018 at 19:00
  • $\begingroup$ I added some notes to cover that and other things. $\endgroup$
    – Aaron Meyerowitz
    Commented Feb 5, 2018 at 3:28

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