A question on the definition of splitting field

The definition of splitting field is as follows:

Suppose that $$L$$ is a field extension of $$K$$, and that $$f ∈ K[X]$$. We say that $$f$$ splits completely over $$L$$ if there exist $$c,α_1,α_2,...,α_n ∈ L$$ such that $$f (X ) = c(X − α_1 )(X − α_2 ) · · · (X − α_n )$$ in $$L[X]$$. If moreover $$L = K(α_1, α_2, . . . , α_n)$$, then we call $$L$$ a splitting field of $$f$$ (over $$K$$).

I understand that we need all the roots of $$f$$ adjoined to $$K$$ as above for it to be a splitting field. Is this the case, or is it enough to adjoin only one root of $$f$$ to $$K$$ (to get $$K(α)$$) then perhaps apply the Primitive Element Theorem to show that it is the same as $$L = K(α_1, α_2, . . . , α_n)$$?

I'm studying the following proof:

Suppose $$L$$ is a finite extension of a field $$K$$. If $$L$$ is Galois over $$K$$, then $$L$$ is a splitting field over $$K$$ of some separable irreducible polynomial $$f \in K[X]$$.

Proof:

If $$L$$ is Galois over $$K$$, then $$L$$ is separable over $$K$$, so $$L = K(α)$$ for some $$α ∈ L$$ by the Primitive Element Theorem. Let $$f ∈ K[X]$$ be the minimal polynomial of $$α$$ over $$K$$. Then $$f$$ is separable (since $$α$$ is separable over $$K$$) and irreducible. Furthermore $$f$$ splits completely over $$L$$ (by Proposition 7.5 or 7.18). Since $$L$$ is generated over $$K$$ by (one of) the roots of $$f$$, we conclude that $$L$$ is a splitting field for $$f$$ over $$K$$.

I'm fine with all except the bit in bold. What is the justification for having a single root adjoined to $$K$$ to make it a splitting field?

Proposition 7.5. Let $$L$$ be a finite extension of $$K$$. Then the following are equivalent:

(i) $$L$$ is normal over $$K$$;

(ii) for all $$α ∈ L$$, the minimal polynomial $$m_{α,K}$$ splits completely over $$L$$;

(iii) $$L$$ is a splitting field of some polynomial $$f ∈ K[X]$$.

Proposition 7.18. Suppose that $$L$$ is a finite extension of $$K$$. Then the following are equivalent:

(i) $$L$$ is Galois over $$K$$;

(ii) for all $$α ∈ L$$, the minimal polynomial $$m_{α,K}$$ has deg $$m_{α,K}$$ distinct roots in $$K$$;

(iii) $$\#Aut_K(L) = [L : K]$$.

• Sometimes when you add one of the roots, the remaining roots get added automatically to the field generated, but sometimes more roots need to be added. For example: $x^2-2\in\mathbb{Q}[x]$ doesn't have rational roots. If we add the root $\sqrt{2}$ to $\mathbb{Q}$, then $-\sqrt{2}$ already belongs to the field generated $\mathbb{Q}(\sqrt{2})$. But in the case of $x^3-2\in\mathbb{Q}[x]$, if we add $\sqrt{2}$ we get a field $\mathbb{Q}(\sqrt{2})$ that only contains real numbers. Therefore, its other roots are still not there. – user551819 Apr 16 '18 at 14:55

• $L=K(\alpha)$, where $\alpha$ is one of the zeros of $f(x)$, so $L$ is contained in a splitting field $K_f=K(\alpha_1,\alpha_2,\ldots,\alpha_n),\alpha=\alpha_1,$ of $f$ over $K$.
• On the other hand, you know that i) $L/K$ is normal, ii) $f(x)=m_{\alpha,K}(x)$ is irreducible over $K$, iii) $f$ has at least one zero, namely $\alpha$ in $L$. So Proposition 7.5. says that $L$ contains all the zeros $\alpha_i, i=1,2,\ldots,n$, of $f$. Therefore $K_f\subseteq L$.
• We have $K_f\subseteq L\subseteq K_f$, so $L=K_f$.
No. Say $f\in\Bbb Q[x]$, $f(x)=(x^2-2)(x^2+1)$, and note that $\Bbb Q(\sqrt 2)\subset\Bbb R$.