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.
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.
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 $\mathop{Mor}\nolimits _ k(E, E')$.
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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