The Stacks project

9.6 Field extensions

In general, though, we are interested not so much in fields by themselves but in field extensions. This is perhaps analogous to studying not rings but algebras over a fixed ring. The nice thing for fields is that the notion of a “field over another field” just recovers the notion of a field extension, by the next result.

Lemma 9.6.1. If $F$ is a field and $R$ is a nonzero ring, then any ring homomorphism $\varphi : F \to R$ is injective.

Proof. Indeed, let $a \in \mathop{\mathrm{Ker}}(\varphi )$ be a nonzero element. Then we have $\varphi (1) = \varphi (a^{-1} a) = \varphi (a^{-1}) \varphi (a) = 0$. Thus $1 = \varphi (1) = 0$ and $R$ is the zero ring. $\square$

Definition 9.6.2. If $F$ is a field contained in a field $E$, then $E$ is said to be a field extension of $F$. We shall write $E/F$ to indicate that $E$ is an extension of $F$.

So if $F, F'$ are fields, and $F \to F'$ is any ring-homomorphism, we see by Lemma 9.6.1 that it is injective, and $F'$ can be regarded as an extension of $F$, by a slight abuse of language. Alternatively, a field extension of $F$ is just an $F$-algebra that happens to be a field. This is completely different than the situation for general rings, since a ring homomorphism is not necessarily injective.

Let $k$ be a field. There is a category of field extensions of $k$. An object of this category is an extension $E/k$, that is a (necessarily injective) morphism of fields

\[ k \to E, \]

while a morphism between extensions $E/k$ and $E'/k$ is a $k$-algebra morphism $E \to E'$; alternatively, it is a commutative diagram

\[ \xymatrix{ E \ar[rr] & & E' \\ & k \ar[ru] \ar[lu] & } \]

The set of morphisms from $E \to E'$ in the category of extensions of $k$ will be denoted by $\mathop{\mathrm{Mor}}\nolimits _ k(E, E')$.

Definition 9.6.3. A tower of fields $E_ n/E_{n - 1}/\ldots /E_0$ consists of a sequence of extensions of fields $E_ n/E_{n - 1}$, $E_{n - 1}/E_{n - 2}$, $\ldots $, $E_1/E_0$.

Let us give a few examples of field extensions.

Example 9.6.4. Let $k$ be a field, and $P \in k[x]$ an irreducible polynomial. We have seen that $k[x]/(P)$ is a field (Example 9.3.3). Since it is also a $k$-algebra in the obvious way, it is an extension of $k$.

Example 9.6.5. If $X$ is a Riemann surface, then the field of meromorphic functions $\mathbf{C}(X)$ (Example 9.3.6) is an extension field of $\mathbf{C}$, because any element of $\mathbf{C}$ induces a meromorphic — indeed, holomorphic — constant function on $X$.

Let $F/k$ be a field extension. Let $S \subset F$ be any subset. Then there is a smallest subextension of $F$ (that is, a subfield of $F$ containing $k$) that contains $S$. To see this, consider the family of subfields of $F $ containing $S$ and $k$, and take their intersection; one checks that this is a field. By a standard argument one shows, in fact, that this is the set of elements of $F$ that can be obtained via a finite number of elementary algebraic operations (addition, multiplication, subtraction, and division) involving elements of $k$ and $S$.

Definition 9.6.6. Let $k$ be a field. If $F/k$ is an extension of fields and $S \subset F$, we write $k(S)$ for the smallest subfield of $F$ containing $k$ and $S$. We will say that $S$ generates the field extension $k(S)/k$. If $S = \{ \alpha \} $ is a singleton, then we write $k(\alpha )$ instead of $k(\{ \alpha \} )$. We say $F/k$ is a finitely generated field extension if there exists a finite subset $S \subset F$ with $F = k(S)$.

For instance, $\mathbf{C}$ is generated by $i$ over $\mathbf{R}$.

Exercise 9.6.7. Show that $\mathbf{C}$ does not have a countable set of generators over $\mathbf{Q}$.

Let us now classify extensions generated by one element.

Lemma 9.6.8 (Classification of simple extensions). If a field extension $F/k$ is generated by one element, then it is $k$-isomorphic either to the rational function field $k(t)/k$ or to one of the extensions $k[t]/(P)$ for $P \in k[t]$ irreducible.

We will see that many of the most important cases of field extensions are generated by one element, so this is actually useful.

Proof. Let $\alpha \in F$ be such that $F = k(\alpha )$; by assumption, such an $\alpha $ exists. There is a morphism of rings

\[ k[t] \to F \]

sending the indeterminate $t$ to $\alpha $. The image is a domain, so the kernel is a prime ideal. Thus, it is either $(0)$ or $(P)$ for $P \in k[t]$ irreducible.

If the kernel is $(P)$ for $P \in k[t]$ irreducible, then the map factors through $k[t]/(P)$, and induces a morphism of fields $k[t]/(P) \to F$. Since the image contains $\alpha $, we see easily that the map is surjective, hence an isomorphism. In this case, $k[t]/(P) \simeq F$.

If the kernel is trivial, then we have an injection $k[t] \to F$. One may thus define a morphism of the quotient field $k(t)$ into $F$; given a quotient $R(t)/Q(t)$ with $R(t), Q(t) \in k[t]$, we map this to $R(\alpha )/Q(\alpha )$. The hypothesis that $k[t] \to F$ is injective implies that $Q(\alpha ) \neq 0$ unless $Q$ is the zero polynomial. The quotient field of $k[t]$ is the rational function field $k(t)$, so we get a morphism $k(t) \to F$ whose image contains $\alpha $. It is thus surjective, hence an isomorphism. $\square$

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