The Stacks project

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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 $F$).

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 $E$.

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 10.34.11.

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_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.

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 $\mathbf{Q} \subset K$ 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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