An exercise I have to make: for L a splitting field of f over K and $f = \prod_{i=1}^n(X-\alpha_{i}) \in L[X]$ prove that L = K($\alpha_{1} , ..., \alpha_{n-1}).$ Here f needs to have head coefficient equal to 1 by definition of the spittingfield. I have not found a way to prove this yet. I know f is writeable as $ f = X^n + ... + ~a_{1}X^1 +a_{0} \in K[X] $ such that $a_{0} = -\alpha_{1}~* ...*-\alpha_{n}$ . Can you give me some direction? Sorry if this turns out to be too trivial or if there is a solution somewhere here already - haven't found it yet.
Thanks.
One definition is:
Now, $f$ splits into linear factors in $L$ iff $L$ contains all the roots of $f$. Therefore, $L \supseteq K(\alpha_{1} , \dots, \alpha_{n})$. Since $f$ splits in $K(\alpha_{1} , \dots, \alpha_{n})$, we must have $L \subseteq K(\alpha_{1} , \dots, \alpha_{n})$, because $L$ is the smallest field that contains all roots of $f$. Therefore, $L = K(\alpha_{1} , \dots, \alpha_{n})$
Let $a_{n-1}$ be the coefficient of $x^{n-1}$ in $f$. Then, $\alpha_{1} + \cdots + \alpha_{n} = -a_{n-1} \in K$ and so $\alpha_{n} \in K(\alpha_{1} , \dots, \alpha_{n-1})$.
Therefore, $K(\alpha_{1} , \dots, \alpha_{n})=K(\alpha_{1} , \dots, \alpha_{n-1})$.