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I'm trying to show that if $f \in \mathbb{Q}[t]$ is irreducible then all the roots of $f$ in $\mathbb{C}$ are distinct.

My first issue, am I right in thinking that the roots are distinct iff $hcf(f,f')$ is constant?

Then suppose $g$ divides $f$ then as $f$ is irreducible and $\mathbb{Q}$ is a field (so only units in $\mathbb{Q}[t]$ are constant polynomials) then $g$ must be a constant polynomial, hence we know that hcf is constant.

But I'm not sure that this is correct because I haven't used anything about $f'$ or $\mathbb{C}$ in the argument?

Thanks

EDIT

Does this argument work?

If $\alpha$ is a repeated root then $f(\alpha) = 0$ and $f'(\alpha)= 0$ so by bezouts lemma we can write: $hcf(f,f') = r(t)f(t) + s(t)f'(t)$ so then $hcf(f,f')(\alpha)= 0$ and then $(t-\alpha) | f$ which contradicts irreducibility.

But I'm still not quite sure why $\mathbb{Q}$ and $\mathbb{C}$ come in, because this argument would work over any field.

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  • $\begingroup$ Show any double (or higher) root of $f$ is also a root of $\mathrm{hcf}(f,f')$. Here is where you have to use something about $f'$. $\endgroup$
    – GEdgar
    Feb 18, 2014 at 17:58
  • $\begingroup$ So by Bezout's lemma we can write $hcf(f,f') = r(t)f(t) + s(t)f'(t)$ and then if $\alpha$ is a root of $f$ and $f'$, i.e a repeated root, then it is also a root of $hcf(f,f')$ so then $hcf(f,f')$ cannot be constant? $\endgroup$
    – Wooster
    Feb 18, 2014 at 18:57
  • $\begingroup$ You still need the proof: any repeated root of $f'$ is a root of $f$. You are (almost) right. The argument works over any fields of characteristic zero. $\endgroup$
    – GEdgar
    Feb 18, 2014 at 19:53
  • $\begingroup$ Which part of my argument relies on that? $\endgroup$
    – Wooster
    Feb 18, 2014 at 19:56

1 Answer 1

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Hint: Let $\alpha$ be a multiple root of $f$. Then $\alpha$ is a root of $f'$. "The" minimal polynomial of $\alpha$ divides $f(x)$ and $f'(x)$.

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  • $\begingroup$ Would you mind explaining what a minimal polynomial is? I haven't covered it, perhaps there is a way to do it without this? Thanks $\endgroup$
    – Wooster
    Feb 18, 2014 at 19:35
  • $\begingroup$ It is a polynomial $g(x)$ (with rational coefficients) such that $g(\alpha)=0$ and no (non-zero) polynomial with rational coefficients of degree smaller than the degree of $g$ has $\alpha$ as a root. To show that $g(x)$ divides $f$ (or $f'$) try to divide, we get $f(x)=q(x)g(x)+r(x)$ where $r(x)$ has degree less than the degree of $g$. Note that by substitution $r(\alpha)=0$. This contradicts definition of minimal polynomial unless $r$ is identically $0$, that is, $g(x)$ divides $f(x)$. $\endgroup$ Feb 18, 2014 at 19:43
  • $\begingroup$ But of course you do not need the notion of "minimal polynomial" to solve this problem. $\endgroup$
    – GEdgar
    Feb 18, 2014 at 19:51
  • $\begingroup$ Okay, yes I'm not comfortable with using it, I have made an edit to the question with a possible answer. Thank you for your help! $\endgroup$
    – Wooster
    Feb 18, 2014 at 19:52
  • $\begingroup$ In your most recent post, $t-\alpha$ divides $f$ does not directly contradict irreducibility. The coefficients of $t-\alpha$ are not necessarily rational. But you can certainly argue that $\gcd(f,f')\ne 1$, which shows some non-constant polynomial $d(x)$ divides $f'(x)$ and $f(x)$. This polynomial has degree $\le$ the degree of $f'(x)$, so it is a proper divisor of $f(x)$. $\endgroup$ Feb 18, 2014 at 20:02

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