Definition 9.7.1. Let F/E be an extension of fields. The dimension of F considered as an E-vector space is called the degree of the extension and is denoted [F : E]. If [F : E] < \infty then F is said to be a finite extension of E.
9.7 Finite extensions
If F/E is a field extension, then evidently F is also a vector space over E (the scalar action is just multiplication in F).
Example 9.7.2. The field \mathbf{C} is a two dimensional vector space over \mathbf{R} with basis 1, i. Thus \mathbf{C} is a finite extension of \mathbf{R} of degree 2.
Lemma 9.7.3. Let K/E/F be a tower of algebraic field extensions. If K is finite over F, then K is finite over E.
Proof. Direct from the definition. \square
Let us now consider the degree in the most important special example, that given by Lemma 9.6.8, in the next two examples.
Example 9.7.4 (Degree of a rational function field). If k is any field, then the rational function field k(t) is not a finite extension. For example the elements \left\{ t^ n, n \in \mathbf{Z}\right\} are linearly independent over k.
In fact, if k is uncountable, then k(t) is uncountably dimensional as a k-vector space. To show this, we claim that the family of elements \{ 1/(t- \alpha ), \alpha \in k\} \subset k(t) is linearly independent over k. A nontrivial relation between them would lead to a contradiction: for instance, if one works over \mathbf{C}, then this follows because \frac{1}{t-\alpha }, when considered as a meromorphic function on \mathbf{C}, has a pole at \alpha and nowhere else. Consequently any sum \sum c_ i \frac{1}{t - \alpha _ i} for the c_ i \in k^*, and \alpha _ i \in k distinct, would have poles at each of the \alpha _ i. In particular, it could not be zero.
Amusingly, this leads to a quick proof of the Hilbert Nullstellensatz over the complex numbers. For a slightly more general result, see Algebra, Theorem 10.35.11.
Lemma 9.7.5. A finite extension of fields is a finitely generated field extension. The converse is not true.
Proof. Let F/E be a finite extension of fields. Let \alpha _1, \ldots , \alpha _ n be a basis of F as a vector space over E. Then F = E(\alpha _1, \ldots , \alpha _ n) hence F/E is a finitely generated field extension. The converse is not true as follows from Example 9.7.4. \square
Example 9.7.6 (Degree of a simple algebraic extension). Consider a monogenic field extension E/k of the form discussed in Example 9.6.4. In other words, E = k[t]/(P) for P \in k[t] an irreducible polynomial. Then the degree [E : k] is just the degree d = \deg (P) of the polynomial P. Indeed, say
9.7.6.1
\begin{equation} \label{fields-equation-P} P = a_ d t^ d + a_{d - 1} t^{d - 1} + \ldots + a_0. \end{equation}
with a_ d \not= 0. Then the images of 1, t, \ldots , t^{d - 1} in k[t]/(P) are linearly independent over k, because any relation involving them would have degree strictly smaller than that of P, and P is the element of smallest degree in the ideal (P).
Conversely, the set S = \{ 1, t, \ldots , t^{d - 1}\} (or more properly their images) spans k[t]/(P) as a vector space. Indeed, we have by (9.7.6.1) that a_ d t^ d lies in the span of S. Since a_ d is invertible, we see that t^ d is in the span of S. Similarly, the relation t P(t) = 0 shows that the image of t^{d + 1} lies in the span of \{ 1, t, \ldots , t^ d\} — by what was just shown, thus in the span of S. Working upward inductively, we find that the image of t^ n for n \geq d lies in the span of S.
This confirms the observation that [\mathbf{C}: \mathbf{R}] = 2, for instance. More generally, if k is a field, and \alpha \in k is not a square, then the irreducible polynomial x^2 - \alpha \in k[x] allows one to construct an extension k[x]/(x^2 - \alpha ) of degree two. We shall write this as k(\sqrt{\alpha }). Such extensions will be called quadratic, for obvious reasons.
The basic fact about the degree is that it is multiplicative in towers.
Lemma 9.7.7 (Multiplicativity). Suppose given a tower of fields F/E/k. Then
[F:k] = [F:E][E:k]
Proof. Let \alpha _1, \ldots , \alpha _ n \in F be an E-basis for F. Let \beta _1, \ldots , \beta _ m \in E be a k-basis for E. Then the claim is that the set of products \{ \alpha _ i \beta _ j, 1 \leq i \leq n, 1 \leq j \leq m\} is a k-basis for F. Indeed, let us check first that they span F over k.
By assumption, the \{ \alpha _ i\} span F over E. So if f \in F, there are a_ i \in E with
f = \sum \nolimits _ i a_ i \alpha _ i,
and, for each i, we can write a_ i = \sum b_{ij} \beta _ j for some b_{ij} \in k. Putting these together, we find
f = \sum \nolimits _{i,j} b_{ij} \alpha _ i \beta _ j,
proving that the \{ \alpha _ i \beta _ j\} span F over k.
Suppose now that there existed a nontrivial relation
\sum \nolimits _{i,j} c_{ij} \alpha _ i \beta _ j = 0
for the c_{ij} \in k. In that case, we would have
\sum \nolimits _ i \alpha _ i \left( \sum \nolimits _ j c_{ij} \beta _ j \right) = 0,
and the inner terms lie in E as the \beta _ j do. Now E-linear independence of the \{ \alpha _ i\} shows that the inner sums are all zero. Then k-linear independence of the \{ \beta _ j\} shows that the c_{ij} all vanish. \square
We sidetrack to a slightly tangential definition.
Definition 9.7.8. A field K is said to be a number field if it has characteristic 0 and the extension K/\mathbf{Q} is finite.
Number fields are the basic objects in algebraic number theory. We shall see later that, for the analog of the integers \mathbf{Z} in a number field, something kind of like unique factorization still holds (though strict unique factorization generally does not!).
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