An irreducible polynomial in a subfield of $\mathbb{C}$ has no multiple roots in $\mathbb{C}$ 
Let $K\subset \mathbb{C}$ be a subfield and $f\in K[t]$ an irreducible polynomial. Show that $f$ has no multiple roots in $\mathbb{C}$.

If I understand this question correctly, I must show that there is no $a \in \mathbb{C}$ such that $(t-a)^n|f$ in $F[t]$ with $n>1$. So suppose $(t-a)^2|f$ and $f=(t-a)^2h$. Then we have $f'=2(t-a)h+(t-a)^2h' \Rightarrow (t-a)|f'$ so $\gcd(f,f'$) is not constant. Therefore $f$ is divisible by some square of non-constant polynomial in $F[t]$, which is a contradiction. Is my argument correct? Thank you.
 A: Given a root $\alpha$ of $f(x)$, $f(x)$ (divided by his director coefficient but who cares, we have coefficients in a field) is the minimal polynomial of $\alpha$. Thus there cannot be another polynomial with lower degree such that $p(\alpha)=0$.
If $\alpha$ is a double root for $f(x)$, then $f'(x)$ is again a polynomial in $K[x]$ and of course $f'(\alpha)=0$. But his degree is lower then $f(x)$. We conclude that $f'(x) \equiv 0$. Since we are working in a subfield of $\mathbb C$, thus a field of characteristic $0$, we know that this implies $f(x)\equiv 0$ , absurd.
A: We can assume $f$ has degree $\ge 2$. Suppose that $f$ and $f'$ are relatively prime as polynomial over $K$. Then (Bezout) there exist polynomials $u(t)$ and $v(t)$, with coefficients in $K$ such that $uf+vf'=1$. 
But then $f$ and $f'$ cannot have a common root in any field extension of $K$, for such a root would have to be a root of the polynomial $1$. In particular, they cannot have a common root in $\mathbb{C}$. This contradicts the assumption that $f$ has a double root in $\mathbb{C}$.
So $f$ and $f'$ are not relatively prime as polynomials over $K$. Let $g=\gcd(f,f')$. Then $g$ is a polynomial over $K$ of degree $\ge 1$ and less than the degree of $f$, and $g$ divides $f$. This contradicts the irreducibility of $f$ over $K$.
