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It follows from the fundamental theorem of algebra that, if $P(x)\in\mathbb R[x]$ is a nonzero polynomial of degree $n$, then $P(x)$ has at most $n$ distinct roots in $\mathbb C$ (to be precise, it has exactly $n$ roots if counted with their multiplicity).

But do almost all (in the sense of measure theory) polynomials of degree $n$ have exactly $n$ distinct roots in $\mathbb C$?

If we consider polynomials of degree $2$ it's clearly the case, since the set of all $(a,b,c)\in\mathbb R^3$ such that $b^2-4ac=0$ has Lebesgue measure equal to $0$ in $\mathbb R^3$. But of course this reasoning cannot be applied in general. So how can this concept be generalized? Does the statement hold for any degree?

Thanks in advance.

Edit: solved, thanks to the commenters. It was actually much simpler than I thought.

NtLake
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    Consider: the measure of an $n-1$ dimensional subspace in $n$ dimensional space is zero. – JMoravitz Apr 25 '24 at 12:14
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    For each fixed degree you can generalize what you did for degree $2$. Polynomials of any degree have a discriminant, which is a polynomial in the coefficients. (the higher the degree, the more complicated it is, but this doesn't make a difference) – Mark Apr 25 '24 at 12:14
  • @Mark got it, thanks. – NtLake Apr 25 '24 at 13:19
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    Please either answer your own question or delete it, so that it does not linger on the unanswered queue. – Ethan Bolker Apr 25 '24 at 14:11

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As mentioned in the comments, the space of polynomials having $<n$ roots has a dimension smaller than $n$, so they have measure zero.

ultralegend5385
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