Finite extension fields

Overview

Finite extension

If $E$ is an extension field of $F$, then $E$ is a finite extension of $F$ if $E$ is a finite-dimensional $F$-vector space, and the degree of $E$ over $F$ is the positive integer $[E:F]={\dim }_{F}E,$ where $[E:F]=1$ if and only if $E=F$.

If $E$ is a finite extension of $F$, then $E$ is also an algebraic extension of $F$, i.e., every element $\alpha \in E$ is algebraic over $F$. More precisely, every finite extension $E$ of a field $F$ is of the form

$$E=F({\alpha }_1,\dots ,{\alpha }_n)$$

for some ${\alpha }_i\in E$ which are all the roots of some polynomial $f\in F[x]$ (not necessarily irreducible).

Relevant theorems:

• (Theorem) The number of extensions of a field homomorphism is at most the degree of the field extension

Related notes:

• Minimal polynomial of an element in an extension field
• Finite extensions of the rationals
• Automorphisms and polynomial roots

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Generating finite fields

Every finite extension is generated by polynomial roots

Every finite extension $E$ of a field $F$ is of the form

$$E=F({\alpha }_1,\dots ,{\alpha }_n),$$

where ${\alpha }_i$ are the roots of some polynomial $f\in F[x]$, not necessarily irreducible.

Proof from Modern Algebra II. We know that a finite extension has the form

$$E=F({\beta }_1,\dots ,{\beta }_k)$$

where ${\beta }_i$ are algebraic over $F$. The general idea is to construct $f$ by as the product of each minimal polynomial $\text{irr}({\beta }_i,F)$: for each $i$, let $f_i=\text{irr}({\beta }_i,F)$ and let ${\alpha }_1,\dots ,{\alpha }_n\in E$ be the elements which are the root of some $f_i$. Since each $f_i$ has only finitely many roots, this is a finite set. Define

$$f=f_1\cdots f_k=\prod _{i=1}^k\text{irr}({\beta }_i,F).$$

Then an element $\alpha \in E$ is a root of $f$ if and only if $\alpha$ is a root of some $f_i$, so that $\{{\alpha }_1,\dots ,{\alpha }_n\}$ is precisely the set of roots of $f$ in $E$. Further, each $f_i$ has ${\beta }_i$ as a root, so ${\beta }_i\in \{{\alpha }_1,\dots ,{\alpha }_n\}$ for all $i$. Thus

$$E=F({\beta }_1,\dots ,{\beta }_k)\le F({\alpha }_1,\dots ,{\alpha }_n)\le E,$$

which gives the claimed equality. $\square$

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Computing degrees of extensions

Tower law for degrees of finite extensions

If $E$ is a finite extension field of $F$ and $V$ is an $E$-vector space:

• (i) $V$ is a finite-dimensional $F$-vector space if and only if $V$ is a finite-dimensional $E$-vector space.
• (ii) In the case above, we have ${\dim }_{F}V=[E:F]{\dim }_{E}V.$

In particular, if $F\le E\le K$ is a sequence of fields, then $K$ is a finite extension of $F$ iff $K$ is a finite extension of $E$, and in this case we have $[K:F]=[K:E][E:F].$

Corollary

If $F\le E\le K$ and $K$ is a finite extension of $F$, then $[K:E]$ and $[E:F]$ both divide $[E:F]$.

Corollary

If $K$ is a finite extension field of $E$ and $E$ is a finite extension field of $F$ with bases ${\beta }_1,\dots ,{\beta }_n,{\alpha }_1,\dots ,{\alpha }_m,$ respectively, then $\{{\alpha }_i{\beta }_j:1\le i\le m,1\le j\le n\}$ is an $F$-basis of $K$.

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Connections to other topics

• Galois groups: If $E$ is a finite extension of a field $F$ and $\sigma :E\to E$ is any ring homomorphism that fixes $F$, i.e., $\sigma (a)=a$ for all $a\in F$, then $\sigma$ is a surjective, hence an automorphism, hence an element of the Galois group $\text{Gal}(E/F)$.