Discriminant is zero iff $f\in K[X]$ has repeated roots

Question

I have to prove the statement in the title. Proving from right to left is easy. Now from left to right: $D=(\alpha_1-\alpha_2)^2(\alpha_1-\alpha_3)^2\cdots(\alpha_{n-1}-\alpha_n)^2$ where $\alpha_i$ are the roots of $f$ in the splitting field. Now $D=0$ implies that somewhere in the product we have a zero, so for some $i$ and $j$ we have that $\alpha_i=\alpha_j$. This $\alpha_i$ doesn't have to be in $K$. So we can't conclude deg(gcd($f,f'$))>0.From here I don't know what to do. Can someone give the proof?

Answer 1

As you say $\alpha_i$ and $\alpha_j$ are not in $K$, they are in a algebraic closure $\bar{K}$ of $K$. So ${\sf gcd}_{\bar{K}}(f,f')$ is nonconstant. But the euclidean algorithm shows that ${\sf gcd}_{\bar{K}}(f,f')={\sf gcd}_{K}(f,f')$, so you are done.