roots of polynomials with multiple-of-unity coefficients implies algebraically closed?

Question

Suppose a field has roots of all polynomials whose coefficients are 0, 0+1, 0+1+1, 0+1+1+1, 0+1+1+1+1, etc or additive inverses thereof. Is such a field necessarily algebraically closed?

The algebraic numbers are produced by starting with the rationals and closing under roots of polynomials as described above (ie integer coefficients). The algebraic numbers are algebraically closed -- any polynomial with any coefficients in the algebraic numbers has a root (not just polynomials with integer coefficients). Is this phenomenon specific to the algebraic numbers, or is it true for fields in general?

Thanks,

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Answer 1

It is certainly not a general property of fields. Consider for instance the rational function field $\mathbb{C}(t)$. This field is far from algebraically closed: for instance, it does not contain any $n$th roots of $t$ for $n > 1$, but every polynomial with integer coefficients already has a root, since the complex numbers are contained in the ground field.

Here is a related question: for which fields $F$ of characteristic $0$ does the compositum of $F$ with an algebraic closure of $\mathbb{Q}$ give an algebraic closure of $F$? This is true for instance when $F$ is the real numbers (rather obviously) or is any $p$-adic field (by Krasner's Lemma).

Answer 2

A positive answer to a related question which may very well been behind your question is that if $L$ is an algebraic extension of the field $K$ and every nonconstant polynomial over $K$ has a root in $L$, then $L$ is algebraically closed. The proof of this result is an exercise in Galois theory.