LEC ALG3 16

Towers of algebraic extensions

Definition 322.1(Definition).

We say ${\text{dim}}_F(K)$ is the degree of the field extension $K/F$ and denote it by $[K:F]$.
If $[K:F]$ is finite, we say $K/F$ is a finite (field) extension.

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Proposition 322.2(Multiplicativity).

Given a tower of fields $L/K/F$,

$$[L:F]=[L:K][K:F].$$

Proof.

Let $a=[L:K]$ and $b=[K:F]$. Let ${\alpha }_1,\dots ,{\alpha }_a\in L$ be a $K$-basis for $L$. Let ${\beta }_1,\dots ,{\beta }_b\in K$ be a $F$-basis for $K$. For $l\in L$, write:

$$\begin{array}{rlr}l& =\sum _{i=1}^a{k}_i{\alpha }_i& k_i\in K\\ & =\sum _{i=1}^a\left(\sum _{j=1}^b{f}_{i,j}{\beta }_j\right){\alpha }_i& f_{i,j}\in F\end{array}$$

Thus, $\mathcal{B}=\{{\beta }_j{\alpha }_i:1⩽j⩽b,1⩽i⩽a\}$ spans $L$. $\mathcal{B}$ is also linearly independent, since

$$\begin{array}{rlr}& \sum _{i=1}^a\left(\sum _{j=1}^b{f}_{i,j}{\beta }_j\right){\alpha }_i=0& \\ & ⟹\sum _{j=1}^b{f}_{i,j}{\beta }_j=0& \forall i\\ & ⟹f_{i,j}=0& \forall i,j\end{array}$$

So, ${\dim }_F(L)=|\mathcal{B}|=ab$.□

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Remark 322.3(Remark).

Given a tower $L/K/F$, if $[L:F]$ is finite, then $[L:K]$ and $[K:F]$ are finite too - the former is immediate, and the latter follows from the fact that subspaces of finite dimensional vector spaces are finite dimensional. Thus, $L/F$ is finite iff $L/K$ is finite and $K/F$ is finite.

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Proposition 322.4(Proposition).

If $[K:F]$ is finite, then $K$ is algebraic over $F$.

Proof.

Suppose $[K:F]=n$. Choose $\alpha \in K$. Then the elements $1,\alpha ,\dots ,{\alpha }^n$ are linearly dependent over $F$. A relation of linear dependence now gives the desired polynomial in $F[x]$ that $\alpha$ must satisfy.□

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In general, it is false that an algebraic extension is finite.

Lemma 322.5(Lemma).

Let $F,K$ be fields. Let $\{F_i{\}}_{i\in \mathbb{N}}$ be fields such that $F_i\subseteq F_{i+1}$ and $F\subseteq F_i\subseteq K$ for every $i$. Then ${\bigcup }_{i\in \mathbb{N}}{F}_i$ is a subfield of $K$.

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Remark 322.6(Remark).

If $K/F$ is an algebraic extension, is $K/F$ finite? No. Consider

$$\mathbb{Q}\subseteq \mathbb{Q}[\sqrt[2]{2}]\subseteq \mathbb{Q}[\sqrt[4]{2}]\subseteq \mathbb{Q}[\sqrt[8]{2}]\subseteq \cdots \subseteq \mathbb{Q}[\sqrt[2^n]{2}]\subseteq \cdots \subseteq \mathbb{C}.$$

$[\mathbb{Q}[\sqrt[2^i]{2}]:\mathbb{Q}]=2^i$. ${\bigcup }_{i\in \mathbb{N}}\mathbb{Q}[\sqrt[2^i]{2}]$ is a field by Lem 5, and is clearly an algebraic extension of $\mathbb{Q}$. However, $\left[{\bigcup }_{i\in \mathbb{N}}\mathbb{Q}[\sqrt[2^i]{2}]:\mathbb{Q}\right]$ is clearly not finite.

By Prp 119.13, If $K/F$ and $\alpha \in K$ is algebraic, then $F(\alpha )/F$ is finite. We can state a more general result:

Definition 322.7(Definition).

Let $K/F$ and ${\alpha }_1,\dots ,{\alpha }_r\in K$ be algebraic over $F$. Then $F({\alpha }_1,\dots ,{\alpha }_n)$ is the smallest subfield of $K$ containing ${\alpha }_1,\dots ,{\alpha }_r$ and $F$.

Proposition 322.8(Proposition).

Let $F$ be a field, and let ${\alpha }_1,\dots ,{\alpha }_n$ be elements of some extension field such that each ${\alpha }_i$ is algebraic over $F$. Then the extension $F({\alpha }_1,\dots ,{\alpha }_n)/F$ is finite and algebraic.

That is, a finitely generated algebraic extension is finite.

Proof.

Each extension $F({\alpha }_1,\dots ,{\alpha }_{i+1})/F({\alpha }_1,\dots ,{\alpha }_i)$ is generated by one element and algebraic, hence finite. Therefore, we get a tower $F({\alpha }_1,\dots ,{\alpha }_n)/\dots /F({\alpha }_1)/F$ of finite extensions. It follows from Prp 2 that $F({\alpha }_1,\dots ,{\alpha }_n)/F$ is finite. The extension is algebraic by Prp 4.□

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Finitely generated extensions

Definition 322.9(Definition).

We say that $E$ is finitely generated over $F$ if there is a finite family of elements ${\alpha }_1,\dots ,{\alpha }_n$ of $E$ such that $E=F({\alpha }_1,\dots ,{\alpha }_n)$.

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Proposition 322.10(Proposition).

A finite extension of fields is a finitely generated extension. The converse is not true in general.

Proof.

If $\{{\alpha }_1,\dots ,{\alpha }_n\}$ is a basis of $E$ as a vector space of $F$, then of course $E=F({\alpha }_1,\dots ,{\alpha }_n)$.

For the converse: If $F$ is any field, then the rational function field $F(t)$ is not a finite extension - the elements $\{t^n:n\in \mathbb{Z}\}$ are independent over $F$.□

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