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I want to proof the following lemma:

Given a polynomial $P \in F[X]$ the number of distinct roots is d = \deg(P) - \deg(\gcd(P,P')).

I see that if $z_1, \dots, z_n$ are the roots and $\mu_1, \dots, \mu_n$ are the multiplicities then \gcd(P,P') = (X-z_1)^{\mu_1-1} \cdots (X-z_n)^{\mu_n-1}.

Now if $P$ has exactly $\deg(P)$ roots that's fine and the lemma holds. But I do not see why it holds if there are less then $\deg(P)$ roots.

Any ideas?

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    Good to know! Thanks!2011-11-15

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It doesn't.

Take $F=\mathbb{R}$, $P(x)=x^2+1$. Then P'(x) = 2x, \gcd(P,P') = \gcd(x^2+1,2x) = 1, but $P(x)$ has zero roots in $\mathbb{R}$, not $2-0=2$ as the claim holds.

What is true is that the number of distinct roots in an algebraic closure of $F$ is equal to the given formula, and your argument gives the result.

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    @joachim: No, because polynomials with integer coefficients may have *rational* roots; e.g., $f(x) = 2x-3$ has $\deg(f)=1$, $\deg(\gcd(f,f')) = 0$, but $f(x)$ has *no* roots in $\mathbb{A}$. It will work for monic polynomials with integer coefficients, by considering them as polynomials over $\mathbb{Q}$ and then noting that any roots in $\overline{\mathbb{Q}}$ will actually lie in $\mathbb{A}$. Also for polynomials that are constant multiples of monic polynomials with integers in $\mathbb{Z}$, of course.2011-11-15