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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?


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The word you're looking for in the first question is prime subfield, e.g. the subfield generated by 1. But I think your second question is the one you're actually interested in. –  Qiaochu Yuan Apr 10 '10 at 23:48
To expand a bit on Qiaochu's comment, the property you describe implies the field contains an algebraic closure of its prime subfield, but there are many such fields that are not algebraically closed, namely transcendental extensions of algebraically closed fields. You can probably find more information in an undergraduate abstract algebra text. –  S. Carnahan Apr 11 '10 at 0:52
Thanks Qiaochu; I don't actually work on fields much... this is part of a "real world" example demonstrating the limitations of a certain kind of logic. You say that the field with all the roots I mention is the "subfield generated by 1" -- I wasn't aware that "generated by" included passing from polynomials to their roots. If this is the case, what is the name for the operation which closes only under the field operations $\langle +,*,-,1,0,{}^{-1}\rangle$? Thanks, –  Adam Apr 12 '10 at 0:05

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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).

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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.

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I think this can be shown quite directly without Galois Theory. –  Mike Benfield Apr 11 '10 at 15:08

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