"Reducing" spliting fields

Question

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!

Answer 1

Note that $f=(X-K_1)\dots(X-K_n)$. The field $F=\Bbb Q(K_1,\dots,K_{n-1})$ contains the roots $K_1,\dots,K_{n-1}$ and hence the polynomial $(X-K_1)\dots(X-K_{n-1})$ is defined over $F$. We may therefore divide $f$ by $(X-K_1)\dots(X-K_{n-1})$ and get a polynomial over $F$ again. But $$\frac{f}{(X-K_1)\dots(X-K_{n-1})}=X-K_n$$ hence $K_n\in F$.

Answer 2

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 $(x-K_1)\cdots (x-K_{n-1})$.

Answer 3

The sum of the roots $K_1+\dots+K_{n-1}+K_n$ is $-a_{n-1}\in Q\subseteq Q[K_1,\dots, K_{n-1}]$.

Also $K_1+\dots+K_{n-1} \in Q[K_1,\dots, K_{n-1}]$.

So $K_n\in Q[K_1,\dots, K_{n-1}]$.