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May I ask how to show that the property of being algebraically closed is reflected by elementary extensions?

The reason that I want to show that is to prove the following:

Prove:

If $p(x)=x^{n+1}+a_nx^n+\cdots+a_1x+a_0$ is a polynomial such that $\{a_0,\cdots,a_n\}\subset\mathbb{Q}$, then $p(x)=0$ has a solution in $F$.

Sincere thanks for any help!

Sorry, I missed out info abt F: Assume that the structure $(F,0,1,+,\cdot)$ is a countable elementary submodel of the complex field $(\mathbb{C},0,1,+,\cdot)$.

yoyostein
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    What is $F$? And what do you mean by "reflected by elementary extensions"? – Alex Kruckman Apr 25 '13 at 18:22
  • Thanks!

    Assume that the structure $(F,0,1,+,\cdot)$ is a countable elementary submodel of the complex field $(\mathbb{C},0,1,+,\cdot)$.

     – yoyostein Apr 25 '13 at 18:34
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    Perhaps it's only me, but I'm not quite sure many know what is a "countably elementary submodel of the complex field"...In fact, the countable thing pretty much confuses me. – DonAntonio Apr 25 '13 at 18:41
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    @DonAntonio This is basic terminology from mathematical logic. The question as now written makes perfectly good sense. – Alex Kruckman Apr 25 '13 at 19:17
  • Nobody claimed otherwise, @AlexKruckman : I, though, don't know that terminology. – DonAntonio Apr 25 '13 at 19:32
  • Note that your statement is quite a bit weaker than algebraic closedness: ${\bf Q}^{\textrm {alg}}(\pi)$ satisfies it, but isn't algebraically closed (or an elementary submodel of the complex field). – tomasz Apr 27 '13 at 00:28

2 Answers2

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What the field $F$ is has not been described. But the following should take care of your problem: any elementary extension $F$ of an algebraically closed field $K$ is algebraically closed.

For any fixed $n$, let $\phi_n$ be the formula $$\forall w_0\forall w_1\cdots\forall w_n \exists x\left(x^{n+1}+w_nx^n+\cdots+w_0=0\right).$$ (We are being a bit sloppy: $x^k$ is an abbreviation.)

Since $\phi_n$ is true in $K$, it is true in $F$.

André Nicolas
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  • Thanks! I wonder if is it true that any elementary extention of a non-algebraically closed field is also not algebraically closed? – yoyostein Apr 25 '13 at 18:34
  • @yoyostein You're right: if $F$ is not algebraically closed, then it will fail to have a root for some polynomial (of degree $n$, say), and then it will satisfy $\exists w_0,\dots,w_{n-1} \lnot \exists x , x^n + w_{n-1}x^{n-1} + \dots + w_0 = 0$, and so will any elementarily equivalent field. – Alex Kruckman Apr 25 '13 at 19:19
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Andre Nicolas's answer is quite correct, but it proves something stronger than the actual statement of your problem (which amounts to proving that $F$ contains the algebraic closure of $\mathbb{Q}$) and I think that misled you at first.

To prove the weaker statement, you don't need to quantify over the coefficients. You can fix the polynomial $p(x)$ for the course of your proof. You need to represent the equation $p(x) = 0$ in the language, $\cal L$, say, defined by the signature you are using for the fields, namely $(0, 1, +, \cdot)$. To do this multiply through by the least common multiple of the coefficients to get a polynomial with integer coefficients. Now (as you have not included negation in the signature), move terms with negative coefficients to the right-hand side of the equation. You now have an equation $q(x) = r(x)$ say, equivalent to $p(x) = 0$ in which $q(x)$ and $r(x)$ are polynomials with positive integer coefficients. Now you can write $q(x)$ and $r(x)$ in $\cal L$, by expressing the positive integer constants in the form $1 + 1 + 1 + \ldots +1$. With $q$ and $r$ expressed in this form, $\exists x(q(x) = r(x))$ is a sentence in $\cal L$ and, as $F$ is an elementary submodel of $\mathbb{C}$, it is valid in $F$ iff it is valid in $\mathbb{C}$ (which, of course, it will be, as your original $p(x)$ is monic of degree greater than $0$).

Rob Arthan
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