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I have come across the following two definitions of a normal field extension.

$\textbf{Definition 1:}$ An algebraic field extension $L/K$ is said to be normal if $L$ is the splitting field of a family of polynomials in $K[X]$

$\textbf{Defintion 2:}$ Let $\overline{k}$ be an algebraic closure of $k$ containing $L$. Any field extension $L$ of $k$ is normal if all $k$-algebra homomorphisms of $L$ into a fixed algebraic closure, $\overline{k}$ of $k$ have the same image.

The first definitions says that if $L$ is normal, then $L$ contains all the roots to some set of polynomials in $K[x]$, i.e. $L$ is a splitting field of $K$ with respect to some set of polynomials.

The second definition says that $L$ is normal if it's image under ANY $k$-algebra homomorphism,$\rho$, from $L$ to $\overline{k}$, have the same image. That is, $\rho_i(L)=\rho_j(L)$, $\forall i,j$.

How are these definitions related? I don't see a connection between the roots of a family of polynomials and the image of a $k$-algebra homomorphism.

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    $\begingroup$ You don't get to choose the polynomials (unless you also get to choose the field $L$). $\endgroup$ – Eoin Jul 18 '15 at 18:15
  • $\begingroup$ Good point. Will edit. $\endgroup$ – Yuugi Jul 18 '15 at 18:17
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Alright, I think I got the $\implies$ direction for finite and infinite extensions. The $\impliedby$ direction is done only for finite extensions. I'm sure it could be worked out where $k[\alpha_1,\alpha_2,...]$ is an infinite extension to define a map for each of the minimal polynomials of $\alpha_1,\alpha_2,...$ and have the family of polynomials defining the extension to be the set of all of these minimal polynomials. But I'm sure that uses the axiom of choice and I'll let this be as is.


Theorem: Definition 1 and Definition 2 are equivalent. (With the caveat stated above).

Proof. Let $L$ be a normal field extension of $k$ satisfying definition 1, and denote by the family of polynomials for which $L$ is the splitting field $\mathscr{F}$. Let $k[\alpha_i]$ denote the field by adjoining the roots, $\alpha_i$ of the polynomials of $\mathscr{F}$. Since $L$ contains the roots of the polynomials of $\mathscr{F}$ we have $L\supset k[\alpha_i]$ and, since $L$ is the splitting field of the polynomials of $\mathscr{F}$, we have $L\subset k[\alpha_i]$. This shows the equality $L=k[\alpha_i]$.

Let $\bar{k}$ be an algebraic closure of $L=k[\alpha_i]\supset k$. Let $\phi:L\rightarrow \bar{k}$ be a $k$-algebra homomorphism. Since this is a $k$-algebra homomorphism we have $0=\phi(0)=\phi(f(\alpha_i))=f(\phi(\alpha_i))$ so that each $\alpha_i$ must be mapped to a conjugate root of its minimal polynomial. But $L$ is generated over $k$ by $\alpha_i$ and its conjugates so that $\phi(L)=L$.

Conversely, suppose $L$ is a normal field extension of finite degree over $k$ satisfying definition 2. By the primitive element theorem there is an $\alpha$ such that $L=k[\alpha]$. Let $p$ be the minimal polynomial of $\alpha$ over $k$. Since every $k$-homomorphism maps into $L$, the map defined by $\phi(k)=k$ and $\phi(\alpha)=\beta$ where $\beta$ is a root of $p$ is a $k$-homomorphism and therefore its image is contained in $L$. But this means $L$ is the splitting field of $p$, proving the other direction.

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