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If we consider a map of the form $f(a,b)=x^2+ax+b$, then $f(a,b)=0$ only has iterated roots along the curve $b=\left(\frac{a}{2}\right)^2$. The set $\left\{(a,b)\in\mathbb{R}^2:b=\left(\frac{a}{2}\right)^2\right\}$ has ($\mathbb{R}^2$)Lebesgue measure zero. I would like to know if there is any generalization of this for polynomials of arbitrary degree (I know it is true up to $n=4$). I.e.,

  1. Does polynomials with real coefficients with iterated roots have Lebesgue measure zero (in the same sense as above, considering the coefficient set)?

  2. Is there any sufficiency condition that guarantees that no roots (or no real roots) are repeated?

2 Answers2

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Yes, the set of polynomials with repeated roots always has measure zero. This is because it is cut out by a polynomial condition on the coefficients, namely the condition that the discriminant vanishes, and the zero set of a polynomial function always has measure zero.

Qiaochu Yuan
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  • @dfsEss3: this is an answer to question 2. The necessary and sufficient condition is that the discriminant is nonzero. – Qiaochu Yuan Oct 11 '22 at 07:28
  • Yes, I saw that and that is why removed the comment about this, thank you! –  Oct 11 '22 at 08:03
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There is an informal and intuitive proof that yes, the set of degree $n$ (monic) polynomials will have measure $0$ in $\mathbb{R}^n$, based on basic counting.

Here is the idea: a degree $n$ (monic) polynomial is uniquely determined by it's $n$ roots, and vice-versa. Now, count. How many (monic) polynomials have exactly $n-1$ simple roots? Well, I can freely place $n-1$ roots, and the last one has $n-1$ options (place it on top of any of the former roots). Thus, I have $n-1$ real degrees of freedom, and $n-1$ discreet degrees of freedom. This obviously represents disjoint union of $n-1$ copies of $\mathbb{R}^{(n-1)}$, which I will write as $(n-1)*\mathbb{R}^{n-1}$. How many (monic) polynomials have exactly $n-2$ simple roots? Same argument, $(n-2)^2*\mathbb{R}^{n-2}$. In total, you will have $$(n-1)*\mathbb{R}^{n-1} \cup (n-2)^2*\mathbb{R}^{n-2} \cup \dots \cup \mathbb{R} = \cup_{i=1}^{n-1}(n-i)^i\mathbb{R}^{n-i},$$ where all the unions are disjoint. Each of those sets has measure $0$ in $\mathbb{R}^n$ and they are disjoint, so the total measure is $0$.

Intuitively, throw $n$ darts in $\mathbb{R}^n$, you clearly have $0$ probability of darts coinciding.

*Keep in mind that since the coefficients are real, all complex roots appear in pair with their conjugates, so a pair of complex roots still uniquely determines two real degrees of freedom.

Carl Chaanin
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  • I think this argument can be made rigorous using Sard's lemma. Maybe a simpler argument is available though. – Qiaochu Yuan Oct 11 '22 at 07:29
  • Thanks for the feedback, I'll look into it. I never formally studies measure theory so I'm not familiar with most theorems in it, which is why I kept my argument on the informal side. I also think it provides more clarity than theorem citing, and is more amenable to open up new and interesting questions. – Carl Chaanin Oct 11 '22 at 07:36