Skip to content

Field extensions

May 3, 2012

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 \{ 0 \} or the whole field). Recall also that for any field K, there is a unique ring homomorphism \iota :\textbf{Z} \to K, and we define the characteristic of K, \text{char } K, to be the integer p \in \textbf{N} such that p generates the principal ideal \ker \iota.

Note that since \iota(\text{Z}) is an integral domain (fields are integral domains) \ker \iota must be a prime ideal hence either \text{char } K = 0 (if \iota is injective) or \text{char } K 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 \textbf{Z} is \textbf{Q} and \textbf{Z}/p\textbf{Z} is a field, a field K_0 of characteristic 0 (resp. K_p of characteristic p) contains a copy of \textbf{Q} (resp. a copy of \textbf{Z}/p\textbf{Z}). Moreover, these are their respective smallest subfield, called their prime subfield.

Since a field morphism is injective, for every morphism K \to F, we can identify K as a subfield of F, it is then said that F is a field extension of K, noted K \subseteq F or F/K. 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 F/K a field extension, we have \text{char }K = \text{char } F so fields of different characteristic do not interact with each others.

If F/K is a field extension, then F is a K-algebra hence a vector space over K. A field extension F/K is finite of degree n if the dimension of F over K as a vector space is \dim F = n \in \textbf{N}. The degree of a field extension F/K is noted [F : K] or simply (F/K).

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 R a ring, the ring R[x]/(f) is a minimal ring with respect to rings containing R and a root for f(x). In fact, if R = K is a field, then K[x] is a principal ideal domain (PID) so if f(x) is irreducible, the ideal (f) will be maximal and K[x]/(f) will be a field extension of K, minimal with respect to field extensions containing a root for f(x). Although, as we have seen in the theorem of last post, this extension is minimal but not universal: if F is another field extension of K having a root for f(x), we certainly have morphisms

K \hookrightarrow K[x]/(f) \hookrightarrow F

but the morphism from K[x]/(f) to F may not be unique.

Let F/K be a field extension and \alpha \in F. Then the smallest subfield of F containing both K and \alpha is is denoted K(\alpha). A field extension F/K is called a simple extension if there is an element \alpha \in F such that F = K(\alpha).

For example, if F = K[x]/(f), for f(x) an irreducible monic polynomial of degree d, the extension F/K is simple of degree d. Indeed, if \alpha is the coset of x in F, any subfield of F containing \alpha and K must contain all polynomials in \alpha over K, since K[x]/(f) \simeq K[\alpha], we must have K(\alpha) = F.

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

Theorem: Let K(\alpha)/K be a simple extension and consider the evaluation morphism \text{ev}_{\alpha} : K[x] \to K(\alpha) defined by f(x) \mapsto f(\alpha). Then

  1. \text{ev}_{\alpha} is injective iff K(\alpha)/K is infinite. In this case, K(\alpha) \simeq K(x), the field of rational functions over K.
  2. \text{ev}_{\alpha} is not injective iff K(\alpha)/K is finite. In this case there is a unique monic irreducible polynomial f(x) \in K[x] of degree [K(\alpha) : K] such that K(\alpha) \simeq K[x] / (f).

Proof: Recall that the quotient field of a ring is the smallest field containing that ring. So if \text{ev}_{\alpha} is injective, then it extends uniquely to a field homomorphism \phi from the quotient field of K[x], namely K(x), to the field K(\alpha). But then the image of \phi contains K and \alpha so since K(\alpha) is minimal with this property, we must have \phi(K(x)) = K(\alpha), hence the isomorphism between K(x) and K(\alpha). Moreover, since \text{ev}_{\alpha} is injective, the image of the powers of x; (1, \alpha, \alpha^2, \ldots ) are linearly independent so the extension K(\alpha)/K is infinite.

On the other hand, if \text{ev}_{\alpha} is not injective, then as seen in my last post, \ker \text{ev}_{\alpha} is prime and generated by a unique monic irreducible non-constant polynomial. The first isomorphism theorem then gives us a homomorphism \phi : K[x] / (f) \to K(\alpha) and since the image of \phi is a subfield containing both K and \alpha, it is the whole K(\alpha). As seen in my last post, we then have [K(\alpha) : K] = \deg f.

Finally, since either \text{ev}_{\alpha} is injective or it isn’t, the converse of both points of the theorem is proved.


In the first case of this theorem, \alpha is said to be transcendantal over K 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.

One Comment

Trackbacks & Pingbacks

  1. More on finite simple extensions. « arbourj's blog

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: