# Field extensions

Recall that since a field does not have a proper ideal, a field morphism is either injective or the zero morphism (the kernel of the morphism is an ideal so it must be or the whole field). Recall also that for any field there is a unique ring homomorphism and we define the characteristic of K, to be the integer such that generates the principal ideal

Note that since is an integral domain (fields are integral domains) must be a prime ideal hence either (if is injective) or is a prime number.

Now, recall the construction of the quotient field of a ring, which is, in a universal way, the smallest field containing the ring. Since the quotient field of is and is a field, a field of characteristic 0 (resp. of characteristic ) contains a copy of (resp. a copy of ). Moreover, these are their respective smallest subfield, called their *prime* subfield.

Since a field morphism is injective, for every morphism we can identify as a subfield of it is then said that is a field extension of noted or So every field is a field extension of its prime subfield, this is why theory of fields is essentially about the study of field extensions.

Note that for a field extension, we have so fields of different characteristic do not interact with each others.

If is a field extension, then is a -algebra hence a vector space over A field extension is finite of degree if the dimension of over as a vector space is The degree of a field extension is noted or simply

We have already seen an example of a finite extension in last post where I talked about adjoining a root of a polynomial in a ring. There we saw that for a ring, the ring is a minimal ring with respect to rings containing and a root for In fact, if is a field, then is a principal ideal domain (PID) so if is irreducible, the ideal will be maximal and will be a field extension of minimal with respect to field extensions containing a root for Although, as we have seen in the theorem of last post, this extension is minimal but not universal: if is another field extension of having a root for we certainly have morphisms

but the morphism from to may not be unique.

Let be a field extension and Then the smallest subfield of containing both and is is denoted A field extension is called a *simple extension* if there is an element such that

For example, if for an irreducible monic polynomial of degree d, the extension is simple of degree d. Indeed, if is the coset of in any subfield of containing and must contain all polynomials in over since we must have

In fact, we will see that this is characteristic of finite simple extensions:

Theorem:Let be a simple extension and consider the evaluation morphism defined by Then

- is injective iff is infinite. In this case, the field of rational functions over
- is not injective iff is finite. In this case there is a unique monic irreducible polynomial of degree such that

**Proof:** Recall that the quotient field of a ring is the smallest field containing that ring. So if is injective, then it extends uniquely to a field homomorphism from the quotient field of namely to the field But then the image of contains and so since is minimal with this property, we must have hence the isomorphism between and Moreover, since is injective, the image of the powers of are linearly independent so the extension is infinite.

On the other hand, if is not injective, then as seen in my last post, is prime and generated by a unique monic irreducible non-constant polynomial. The first isomorphism theorem then gives us a homomorphism and since the image of is a subfield containing both and it is the whole As seen in my last post, we then have

Finally, since either is injective or it isn’t, the converse of both points of the theorem is proved.

QED

In the first case of this theorem, is said to be *transcendantal *over * *and in the second it is said to be *algebraic*.

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

M. Artin, Algebra, 2nd edition

E. Artin, Galois Theory.

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