abstract algebra - Question about a proof about finite normal extensions

In my book they show that if $K \subset L$ is a finite normal extension, then $L$ is the splitting field for some polynomial $f(X) \in K[X]$.

They do so as follows:

Suppose $a_1, ... ,a_n$ is a basis for $L$ as vector space over $K$, hence $L = K(a_1,\ldots,a_n)$. Now let $f_i$ be the minimal polynomial of $a_i$. Since $a_i$ is a root of $f_i$ and since $f_i$ is irreducible, $f_i$ splits completely over $L$, hence $f = f_1\cdots f_n$ also splits completely over $L$. Thus $L$ is the splitting field of $f(X)$.

Now my question. My definition in my book says that $L$ is a splitting field of $f(X)$ over $K$, if

• $f(x) = a(X-\lambda_1)^{m_1}\ldots(X-\lambda_q)^{m_q}$ where $a \in K^*, m_i \in \mathbb{N}$


• $L = K(\lambda_1,\ldots,\lambda_q)$

Now in the proof when $f(X)$ splits into linear factors in $L[x]$ it could have more roots than just $a_1,\ldots,a_n$, hence according to the definition the splitting field would equal to $K(a_1,\ldots,a_n,\lambda_1,\ldots,\lambda_p)$, where $\lambda_1,\ldots,\lambda_p$ are the remaining roots of $f$. Now, I wonder whether my reasoning is correct:

$$L = K(a_1,\ldots,a_n) \subseteq K(a_1,\ldots,a_n,\lambda_1,\ldots,\lambda_p)\subseteq L$$

hence $L$ is the splitting field.

Answer

The proof shows precisely that $K(a_1,\ldots,a_n)=K(a_1,\ldots,a_n,\lambda_1,\ldots,\lambda_p)$, which follows from the assumption that $L/K$ is normal. If you read the proof carefully, that's exactly what it says: since $L/K$ is normal, and one root of $f_i$ is in $L$, they are all in $L$.

Maybe, it's easier to parse if you assume that $L=K(\alpha_1)$. Then, if $f$ is the minimal polynomial of $\alpha_1$ and if $\alpha_2,\ldots,\alpha_r$ are the remaining roots of $f$, then $L$ being normal implies that $K(\alpha_1) = K(\alpha_1,\ldots,\alpha_r)$.