# More on finite simple extensions.

Recall that in last post, we saw that a finite simple extension is such that for a monic irreducible nonconstant polynomial. This polynomial is called the *irreducible polynomial* for over 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 Indeed, there are as many distinct isomorphisms as there are roots of in

Now, recall that the construction of the polynomial ring over a ring is functorial : every ring morphism lifts uniquely to a ring morphism agreeing with on 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 to itself doesn’t lift uniquely to a morphism from to fixing even if and have the same irreducible polynomial over Indeed, we will certainly have isomorphisms but there will be as many distinct isomorphisms fixing as there are roots of in We thus need to specify the image of if we want unicity. More generally, here is the situation:

Theorem:Let and be two simple field extensions. Let and be the respective irreducible polynomials of and respectively over and Let be an isomorphism such that (See the above diagram for polynomials). Then there exists a unique isomorphism such that and

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

**Proof:** Consider the map

defined as where is the canonical projection. Since is injective and we must have On the other hand, since is a maximal ideal, we have hence the isomorphism

Moreover, in the theorem of last post we saw that finite simple extensions were characterized as quotients of the polynomial ring over by the ideal generated by the irreducible polynomial of over More explicitly, for the irreducible polynomial for over we have an isomorphism identifying the coset of with Composing this all together, we have an isomorphism

such that as desired.

For the unicity of note that every element of is a finite linear combination of powers of over so is completely and uniquely determined by its action on together with its action on Thus, since and the isomorphism is unique.

QED

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

**Definition:** Let be a field extension. A -automorphism, or an automorphism of -extension, is a field automorphism such that The group of -automorphism of will be denoted by

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

Corollary:Let be a simple finite extension and let be the minimal polynomial of over (where Then is the number of distinct roots of in In particularwith equality iff factors over in a product of distinct polynomials of degree 1.

**Proof:**** **Let Since is completely determined by But it must send to a root of because

Hence there are at most the number of roots of elements in But by the theorem, every choice of a root of determines a unique lift of the identity hence the equality. Since the number of roots of is smaller or equal to the proof is complete.

QED

**Reference: **P. Aluffi, Algebra : Chapter 0.