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This is taken from the book Algebra by Thomas W. Hungerford ;

Theorem. Let $K$ be an extension of $F$. The following are equivalent:

  1. $K$ is algebraic and Galois over $F$.
  2. $K$ is separable over $F$ and $K$ is a splitting field over $F$ of a set $S$ of polynomials in $F[x]$.
  3. $K$ is the splitting field over $F$ of a set $T$ of separable polynomials in $F[x]$.

Proof. (1)$\implies$(2),(3) Let $u\in K$ and let $f(x)\in F[x]$ be the monic irreducible polynomial of $u$. Let $u=u_1,\ldots,u_r$ be the distinct roots of $f$ in $K$; then $r\leq n=\deg(f)$. If $\tau\in\mathrm{Aut}_F(K)$, then $\tau$ permutes the $u_i$. So the coefficients of the polynomial $g(x) = (x-u_1)(x-u_2)\cdots(x-u_r)$ are fixed by all $\tau\in\mathrm{Aut}_F(K)$, and therefore $g(x)\in F[x]$ (since the extension is Galois, so the fixed field of $\mathrm{Aut}_F(K)$ is $F$). Since $u$ is a root of $g$, then $f(x)|g(x)$. Therefore, $n=\deg(f)\leq \deg(g) = r \leq n$, so $\deg(g)=n$. Thus, $f$ has $n$ distinct roots in $K$, so $u$ is separable over $F$. Now let $\{u_i\}_{i\in I}$ be a basis for $K$ over $F$; for each $i\in I$ let $f_i\in F[x]$ be the monic irreducible of $u_i$. Then $K$ is the splitting field over $F$ of $S=\{f_i\}_{i\in I}$, and each $f_i$ is separable. This establishes (2) and (3).

What I don't understand is why $K$ is the splitting field of $S=\{f_i\}_{i\in I}$. In order to $K$ to be the splitting field of the said set, $K$ needs to be equal to $F(X)$ (according to the definition of the book) where $X = \{v | v$ is a root of $f_i$ and $i \in I\} $

It can be easily seen that $F(X) \subseteq K$. How do I show the other side?

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1 Answer 1

up vote 2 down vote accepted

As $F \subseteq F(X)$, it is sufficient to see, that all $u_i$ are in $F(X)$. But $u_i$ is a root of $f_i$, thus $u_i \in F(X)$. $$K = F((u_i)_{i\in I}) \subseteq F(X)$$

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Thank you, for the answer. Remarkably elegant. Thank you, again! –  Guest_000 Dec 28 '13 at 16:52

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