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.

QED

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.