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More on finite simple extensions.

May 5, 2012

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.

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