I have to show that a splitting field over $K$ for a polynomial of deg $n$ is generated over $K$ by any $n-1$ roots of the polynomial.
I know that if $c$ is algebraic over $K$ of degree $n$ then $K(c)$ is generated by ${1,c,c^2, \dots , c^m}$ , where $m=n-1$ over $K$.I thought it will be helpful in the problem, but that's not the case.
I think I have to show that if $\alpha_1,\alpha_2, \dots , \alpha_n $ are $n$ roots of a polynomial of degree $n$ over $K$ in the splitting field $\mathcal S$ , then $\mathcal S =$ span{any of the $(n-1)$ roots of the polynomial }
Obviously $\mathcal S =$ $K(\alpha_1, \alpha_2, \dots , \alpha_n)$
Any help.Thank You.
Let $c_1,\dots,c_{n-1}$ be any $n-1$ roots of the polynomial $f$ of which we want to compute the splitting field $\mathcal{S}$. Obviously we have $K(c_1,\dots,c_{n-1})\subset \mathcal{S}$. The question is if the last root, $c_n$ is an element of $K(c_1,\dots,c_{n-1})$. But if you look at the factorization of the polynomial $f$ in $K(c_1,\dots,c_{n-1})[X]$, you can divide out any of the roots $c_1,\dots ,c_{n-1}$ so the remaining term is also a linear factor, the factor $(X-c_n)$. So $f=\prod_{i=1}^n(X-c_i)$ in $K(c_1,\dots,c_{n-1})$, hence it is the splitting field $\mathcal{S}$.