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I have come up with this problem but cannot seem to come up with a rigorous proof for it.

Fix $n$ (a positive integer) and $m \in \mathbb{Z} \cap (0, n]$. Prove that there are

  • Uncountably many
    • polynomials of degree $n$ with real coefficients with exatcly $m$ distinct real roots, and
    • polynomials of degree $n$ with real coefficients with exatcly $m$ distinct complex roots.
  • Countably infinitely many
    • polynomials of degree $n$ with integer coefficients with with exatcly $m$ distinct real roots, and
    • polynomials of degree $n$ with integer coefficients with exatcly $m$ distinct complex roots.

I tried applying FTA but that only enables me to prove that every given polynomial has $n$ roots, included repeated roots, and what I am interested in are distinct roots. Plus, I am trying to show there are infinitely many polynomials satisfying the property, not that a particular one satisfies it.

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  • $\begingroup$ What you have tried to do? I suggest you to show your attempt...if you have tried $\endgroup$
    – pawel
    Commented Apr 10, 2021 at 20:27
  • $\begingroup$ I tried applying FTA but that only enables me to prove that every given polynomial has $n$ roots, included repeated roots, and what I am interested in are distinct roots. Plus, I am trying to show there are infinitely many polynomials satisfying the property, not that a particular one satisfies it. $\endgroup$
    – Run27.35
    Commented Apr 10, 2021 at 20:31

1 Answer 1

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  • There are uncountable many polynomials with real coefficient, in particular, the polynomials with real coefficients and $m$ distinct roots are at most uncountable. At the same time there are at least uncountable polynomials with $m$ distinct roots: $$p_\lambda=x(x-\lambda)\cdots(x-\lambda^{m-1})$$for any $\mathbb R\ni\lambda\neq0,1,-1$.
  • The real polynomials with $m$ district real roots form a subset of the set of real polynomials with $m$ distinct complex roots, so the latter is uncountable as well (actually at least uncountable, then at most uncountable for the same reason as above).

The case with integral coefficients is similar (clearly are at least countable, and at most countable for a similar reason as above).

In general if you want to prove that a certain set has cardinality $x$, you show that it has at most cardinality $x$ (usually by defining a 1-1 map from the set in question to one having cardinality $x$) and at least cardinality $x$ (usually by defining a 1-1 map from a set of cardinality $x$ to the set in question) . For example, in the first point there at at most uncountable many polynomials because each polynomial of degree $n$ is determined by the $n+1$-tuple of its coefficients, which is an element of $\mathbb R^{n+1}$, which is uncountable.

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