Splitting field of $f$ as smallest field extension containing all BUT ONE zero of $f$

Question

I'm just working with splitting fields and I have to prove something which I don't understand.

Let $L$ be a splitting field of the polynomial $f$ over $K$ and $f = \prod_{i=1}^n(X-\alpha_i)$.

Prove: $L = K(\alpha_1, …, \alpha_{n-1})$ (so one alpha less!)

By definition we have $L = K(\alpha_1, …, \alpha_{n-1}, \alpha_{n}) \subset K(\alpha_1, …, \alpha_{n-1})$. That means we have to show that $\alpha_n \in K(\alpha_1, …, \alpha_{n-1})$, but I don't know why that should be the case?

Answer 1

We have that over $\;L':=K(\alpha_1,...,\alpha_{n-1})\;$ :

$$f(x)=\prod_{i=1}^{n-1} (X-\alpha_i)\cdot g(x)\;,\;\;g(x)\in L'[x]$$

but it must be that $\;g(x)=X-\alpha_n\;$ , so we've no option but to deduce that $\;\alpha_n\in L'\;$ and we're done.

Answer 2

An alternative point of view:

Write $f(X) = X^n+a_{n-1}X^{n-1}+\dots+a_0$, where $a_i \in K$ by definition. Then $\sum_{i=1}^{n} \alpha_i=-a_{n-1}$ so clearly $\alpha_n=-a_{n-1}-\sum_{i=0}^{n-1}\alpha_i$ , which gives us what we want.