Recall that in last post, we saw that a finite simple extension $K(\alpha)/K$ is such that $K(\alpha) \simeq K[x]/(f)$ for $f \in K[x]$ a monic irreducible nonconstant polynomial. This polynomial is called the irreducible polynomial for $\alpha$ over $K.$ However, by the theorem in my post about adjunction of roots in a ring, we know that in general there is not a unique isomorphism $K[x]/(f) \to K(\alpha).$ Indeed, there are as many distinct isomorphisms as there are roots of $f$ in $K(\alpha).$

Now, recall that the construction of the polynomial ring over a ring is functorial : every ring morphism $\varphi : R \to S$ lifts uniquely to a ring morphism $\tilde{\varphi} : R[x] \to S[x]$ agreeing with $\varphi$ on $R.$ ie. such that this diagram commutes :

We would like a similar situation with finite simple field extensions but as remarked, even the identity morphism from a field $K$ to itself doesn’t lift uniquely to a morphism from $K(\alpha_1)$ to $K(\alpha_2)$ fixing $K,$ even if $\alpha_1$ and $\alpha_2$ have the same irreducible polynomial $f(x)$ over $K.$ Indeed, we will certainly have isomorphisms $K(\alpha_1) \simeq K[x]/(f) \simeq K(\alpha_2)$ but there will be as many distinct isomorphisms $\varphi : K(\alpha_1) \to K(\alpha_2)$ fixing $K$ as there are roots of $f(x)$ in $K(\alpha_2).$ We thus need to specify the image of $\alpha_1$ if we want unicity. More generally, here is the situation:

Theorem: Let $K_1(\alpha_1)/K_1$ and $K_2(\alpha_2)/K_2$ be two simple field extensions. Let $f_1(x) \in K_1[x]$ and $f_2(x) \in K_2[x]$ be the respective irreducible polynomials of $\alpha_1$ and $\alpha_2$ respectively over $K_1$ and $K_2.$ Let $\varphi : K_1 \to K_2$ be an isomorphism such that $\tilde{\varphi}(f_1(x)) = f_2(x)$ (See the above diagram for polynomials). Then there exists a unique isomorphism $\phi : K_1(\alpha_1) \to K_2(\alpha_2)$ such that $\phi | _{K_1} = \varphi$ and $\phi(\alpha_1) = \alpha_2.$

Here is a diagram to help picture what has just been said:

Proof: Consider the map

$\psi : K_1[x] \to K_2[x]/(f_2(x))$

defined as $\psi = \pi \circ \tilde{\varphi},$ where $\pi : K_2[x] \to K_2[x]/(f_2(x))$ is the canonical projection. Since $\tilde{\varphi}$ is injective and $\tilde{\varphi}(f_1(x)) = f_2(x),$ we must have $\ker(\psi) = \tilde{\varphi}^{-1}(\ker(\pi)) \supset (f_1(x)).$ On the other hand, since $(f_1(x))$ is a maximal ideal, we have $\ker(\psi) = (f_1(x))$ hence the isomorphism

$K_1[x]/(f_1) \simeq K_2[x]/(f_2).$

Moreover, in the theorem of last post we saw that finite simple extensions $K(\alpha)$ were characterized as quotients of the polynomial ring over $K$ by the ideal generated by the irreducible polynomial of $\alpha$ over $K.$ More explicitly, for $f(x)$ the irreducible polynomial for $\alpha$ over $K,$ we have an isomorphism $K[x]/(f) \simeq K(\alpha)$ identifying the coset of $x$ with $\alpha.$ Composing this all together, we have an isomorphism

$\phi : K_1(\alpha_1) \simeq K_1[x]/(f) \simeq K_2[x](f) \simeq K_2(\alpha_2)$

such that $\phi(\alpha_1) = \alpha_2,$ as desired.

For the unicity of $\phi,$ note that every element of $K_1(\alpha_1)$ is a finite linear combination of powers of $\alpha_1$ over $K_1$ so $\phi$ is completely and uniquely determined by its action on $K_1$ together with its action on $\alpha_1.$ Thus, since $\phi(K_1) = \varphi(K_1)$ and $\phi(\alpha_1) = \alpha_2,$ the isomorphism $\phi$ is unique.

QED

More often than not, we will be concerned with morphisms of field extensions that extend the identity:

Definition: Let $F/K$ be a field extension. A $K$-automorphism, or an automorphism of $K$-extension, is a field automorphism $\varphi : F \to F$ such that $\varphi|_K = \text{Id}_K.$ The group of $K$-automorphism of $F$ will be denoted by $\text{Aut}_K(F).$

A corollary of the theorem hints at the interaction between group theory and field theory:

Corollary: Let $F/K$ be a simple finite extension and let $f(x)$ be the minimal polynomial of $\alpha$ over $K$ (where $F \simeq K(\alpha)).$ Then $|\text{Aut}_K(F)|$ is the number of distinct roots of $f(x)$ in $F.$ In particular

$|\text{Aut}_K(F)| \leq [F : K]$

with equality iff $f(x)$ factors over $K$ in a product of distinct polynomials of degree 1.

Proof: Let $\sigma \in \text{Aut}_K(F).$ Since $\sigma|_K = \text{Id}_K,$ $\sigma$ is completely determined by $\sigma(\alpha).$ But it must send $\alpha$ to a root of $f(x)$ because

$f(\sigma(\alpha)) = \sigma ( f(\alpha)) = 0.$

Hence there are at most the number of roots of $f(x)$ elements in $\text{Aut}_K(F).$ But by the theorem, every choice of a root of $f(x)$ determines a unique lift of the identity $\sigma \in \text{Aut}_K(F)$ hence the equality. Since the number of roots of $f(x)$ is smaller or equal to $\text{deg } f(x) = [K : F],$ the proof is complete.

QED

Reference: P. Aluffi, Algebra : Chapter 0.