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I have a set of polynomials $P_t(z)= z^n+ a_{n-1}(t)z^{n-1}+\cdots+ a_0(t)$ which depends on a real parameter $t \in [a,b]$ and where $a_{n-1}(t),\ldots, a_0(t)$ are real continuous functions.

May I say that there exists a continuous map $\theta(t)$ such that $\theta(t)$ is a root of $P_t$ (for all $t$)?

I mean, I know that there exists a continuous dependence of the roots of a polynomial with respect to the coefficients and that the Viète map descends to a homeomorphism $w:C^n/S_n\to C^n$, but, can I 'choose' a root? Or I need the axiom of choice to affirm that there exists a map $C^n/S_n\to C^n$? In that case, may I get a such map to be continuous?

Any bibliography reference for all this?

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Finding a function is easy, even without the AoC: any ordering of the set of complex numbers will be a well-ordering on the roots of $P_t(z)$ for each $t$. Choose the smallest root. Of course, this need not be continuous. –  Hurkyl Jun 8 '12 at 10:52
    
I have to wonder, why are you asking? –  Asaf Karagila Jun 8 '12 at 11:11
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Great question.

If the roots are all always real the answer is yes (this comes from the fact that in $\mathbb R$ you can order the roots from the lowest to the highest).

If the roots or the coefficients may be complex, the answer is in general negative. Take for example the polynomial $t^2-z \in \mathbb{C}[t]$, with $z \in \mathbb C$.

However there is a deep theorem (by Kato) that may help you: it states that if the roots of your polynomial depend only on a real parameter $t \in \mathbb {R}$ than you have $N$ continuous functions that describe the roots.

Anyway, I suggest you to give a look to Kato, Perturbation theory for linear operators, Springer (Theorem 5.2, pag 109 in my edition).

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$\theta(t) = t^2$ is a continuous root of $t^2 - z$. If you meant $t - z^2$, then $\theta(t) = \begin{cases} \sqrt{t} & t \geq 0 \\ i \sqrt{-t} & t \leq 0 \end{cases}$ is. –  Hurkyl Jun 8 '12 at 10:56
    
You cannot define a continuous function $\omega: \mathbb C \to \mathbb C$ such that $\omega^2(z)=z$ for every $z \in \mathbb C$. –  Romeo Jun 8 '12 at 11:00
    
But you can define a continuous function on $[a,b] \subseteq \mathbb{R}$. –  Hurkyl Jun 8 '12 at 11:00
    
Yes, I agree with you, but I meant $z \in \mathbb C$ (I should have written that, sorry). –  Romeo Jun 8 '12 at 11:04
    
But does Kato's theorem use the axiom of choice? This is a key issue to the question. –  Asaf Karagila Jun 8 '12 at 11:08
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