"Reducing" spliting fields Let $f(X)=X^n+a_{n-1}X^{n-1}+\ldots a_{1}X+a_{0} \in \Bbb Q[X]$, irreducible in $\Bbb Q[X].$ Let $K_{1},K_{2},\ldots,K_{n}$ be the $n$ roots of $f(X)$ in $\mathbb{C}$. Why $\Bbb Q[K_{1},K_{2},\ldots,K_{n}] = \Bbb Q[K_{1},K_{2}, \ldots, K_{n-1}]?$
I think that i need to use the multiplicative property of dimensions of field extensions but i dont know if i'm right. Any help is appreciated!

Well, you have $f=(x-K_1)\cdots (x-K_{n-1})(x-K_n)$ over $\Bbb C$.
Thus if you have the first $n-1$ roots $K_1,\ldots,K_{n-1}$, the remaining root is given by division of $f$ into $(x-K_1)\cdots (x-K_{n-1})$.
