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

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  • $\begingroup$ 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 $\sqrt[3]{2}$ we get a field $\mathbb{Q}(\sqrt[3]{2})$ that only contains real numbers. Therefore, its other roots are still not there. $\endgroup$ – user551819 Apr 16 '18 at 14:55
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Trying to add details to the part in bold.

  • $L=K(\alpha)$, 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 $L=K_f$.
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No. Say $f\in\Bbb Q[x]$, $f(x)=(x^2-2)(x^2+1)$, and note that $\Bbb Q(\sqrt 2)\subset\Bbb R$.

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