Field extensions and finite field examples

The earlier quadratic and eight-element constructions are examples of field extensions. We now name that structure precisely, calculate extension degrees, and explain how a polynomial relation controls both arithmetic and vector-space size. Finite fields and integer residue rings must be distinguished even when their numbers of elements agree.

1 Extensions and degree

A field extension $K/F$ means that $F$ is identified with a subfield of $K$. Scalar multiplication by elements of $F$ makes $K$ an $F$-vector space. Its dimension, finite or infinite, is the degree $[K:F]$.

For $\alpha\in K$, $F(\alpha)$ is the smallest subfield containing $F$ and $\alpha$. If $\alpha$ satisfies a nonzero polynomial over $F$, it is algebraic over $F$. Its monic polynomial of least positive degree vanishing at $\alpha$ is its minimal polynomial $p$.

Theorem. Let $d=\deg p$. The minimal polynomial is irreducible, the evaluation kernel is $(p)$, and $F(\alpha)\cong F[x]/(p)$. Consequently $[F(\alpha):F]=\deg p$ and the powers $1,\alpha,\ldots,\alpha^{d-1}$ form a basis.

Proof. If $p=uv$ with both factors of smaller positive degree, then $u(\alpha)v(\alpha)=0$ in a field. One factor vanishes, contradicting minimality. For any $f$ vanishing at $\alpha$, divide $f=qp+r$ with smaller-degree remainder. Then $r(\alpha)=0$, so minimality forces $r=0$. Thus the kernel is $(p)$ and the image $F[\alpha]$ is isomorphic to the irreducible quotient field. Being already a field, the image equals $F(\alpha)$. Unique degree-below-$d$ remainders prove the basis and dimension. $\square$

For $\alpha=\sqrt2$ over $\mathbb Q$, $p=x^2-2$, so every element is uniquely $a+b\sqrt2$ and the degree is 2. Over $\mathbb R$, the same element lies in the base field already, so the minimal polynomial is $x-\sqrt2$ and the degree is 1. Minimal polynomials depend on the base field.

2 Building an extension in two stages

Tower law. If $F\subseteq K\subseteq L$ with both $[K:F]=m$ and $[L:K]=n$ finite, then $[L:F]=mn$.

Proof. Let $u_1,\ldots,u_m$ be an $F$-basis of $K$ and $v_1,\ldots,v_n$ a $K$-basis of $L$. Expanding coefficients shows the $mn$ products $u_iv_j$ span $L$ over $F$. If $\sum_{i,j}c_{ij}u_iv_j=0$ with $c_{ij}\in F$, independence of the $v_j$ over $K$ gives $\sum_i c_{ij}u_i=0$ for each $j$. Independence of the $u_i$ then gives every $c_{ij}=0$. Thus the products form a basis. $\square$

Worked biquadratic example. Let $L=\mathbb Q(\sqrt2,\sqrt3)$. The second square root does not lie in $\mathbb Q(\sqrt2)$. If $\sqrt3=a+b\sqrt2$ with rational $a,b$, squaring and uniqueness of rational and $\sqrt2$ parts gives $2ab=0$ and $a^2+2b^2=3$. The case $b=0$ would make $\sqrt3$ rational; the case $a=0$ would make $b^2=3/2$, impossible for rational $b$ by prime-factor parity. Hence the second quadratic remains irreducible over the first extension, and $[L:\mathbb Q]=4$. A basis is $1,\sqrt2,\sqrt3,\sqrt6$.

For $\alpha=\sqrt2+\sqrt3$, we have $1/\alpha=\sqrt3-\sqrt2$. Thus $\sqrt3=(\alpha+\alpha^{-1})/2$ and $\sqrt2=(\alpha-\alpha^{-1})/2$, so $\mathbb Q(\alpha)=L$. Squaring $\alpha^2=5+2\sqrt6$ gives

$$\alpha^4-10\alpha^2+1=0.$$

Because its extension degree is 4, this monic degree-4 polynomial is the minimal polynomial and is irreducible. Merely checking rational roots would not justify that quartic conclusion.

3 The four-element field

Over $\mathbb F_2$, $x^2+x+1$ has no root and is therefore irreducible. Let $K=\mathbb F_2[x]/(x^2+x+1)$ and $\beta=[x]$. The four elements are $0,1,\beta,1+\beta$, with $\beta^2=\beta+1$.

Addition is coefficientwise modulo 2, so every element is its own additive inverse. Multiplication reduces powers using the relation. In particular $\beta(\beta+1)=1$, so the two nontrivial elements invert each other, and $\beta^3=1$. The nonzero multiplicative group has order 3, while the additive group is $\mathbb F_2^2$, not cyclic of order 4.

This field is not $\mathbb Z/4\mathbb Z$, whose nonzero element $[2]$ has square zero. A field of four elements has characteristic 2; four elements do not force characteristic 4.

4 An eight-element example and a general constraint

The polynomial $x^3+x+1$ has no root in $\mathbb F_2$, hence is irreducible cubic. Its quotient gives a field with eight elements and basis $1,\gamma,\gamma^2$, where $\gamma^3=\gamma+1$. For example, $\gamma^{-1}=\gamma^2+1$ and $\gamma^4=\gamma^2+\gamma$.

Every finite field has prime characteristic. If its characteristic were composite $ab$ with $1<a,b$ below it, the nonzero elements $a\cdot1$ and $b\cdot1$ would multiply to zero. Thus the characteristic is a prime $p$, and the field contains $\mathbb F_p$. As a finite vector space over that subfield it has finite dimension $d$ and exactly $p^d$ elements. This proves a necessary size form, not the existence or uniqueness theorem for every prime power; the explicit quotients above establish the stated examples.

The rational obstruction in the two-stage example

The impossibility claims in the biquadratic calculation can be proved with the same prime, 3. If $\sqrt3=m/n$ in lowest terms with $n>0$, then $m^2=3n^2$. Squares modulo 3 are 0 and 1, so $3\mid m$; writing $m=3k$ gives $n^2=3k^2$, hence $3\mid n$, a contradiction. If a reduced rational $b=m/n$ satisfied $b^2=3/2$, then $2m^2=3n^2$ again gives $3\mid m$; substituting gives $n^2=6k^2$, hence $3\mid n$. Both branches in the exclusion of $\sqrt3\in\mathbb Q(\sqrt2)$ are therefore justified. The minimal-polynomial conclusion for $\sqrt2+\sqrt3$ uses the already proved degree four, not a failed rational-root search.

Why finite fields have prime-power size

Here is also the full finite-field size argument. In a finite field the sequence $0,1,2\cdot1,\ldots$ repeats, so there is a least positive $m$ with $m\cdot1=0$. Since $1\ne0$, $m>1$. If $m=ab$ with $1<a,b<m$, then $(a\cdot1)(b\cdot1)=0$ while both factors are nonzero by minimality, contradicting the field property. Thus $m=p$ is prime. The map $\mathbb Z\to K$, $a\mapsto a\cdot1$, therefore has kernel $p\mathbb Z$ and embeds $\mathbb F_p$ as the prime subfield. Choose a maximal linearly independent list over $\mathbb F_p$ by successively adding a vector outside the current span. The list must stop because $K$ is finite; maximality makes it a basis. If it has length $d$, uniqueness of basis coordinates gives exactly $p^d$ elements, one for each $d$-tuple of coefficients. This proves the necessary cardinality statement. Existence or uniqueness of fields for all prime powers and cyclicity of every finite field's multiplicative group are not proved by this counting argument.

5 Graded practice

E1 Foundation. Find the minimal polynomial and degree of $i$ over $\mathbb R$, then over $\mathbb C$.

E2 Quadratic arithmetic. In the four-element field, compute $(1+\beta)^2$ and the inverse of $1+\beta$.

E3 Extension degree. If $[K:F]=3$ and $[L:K]=2$, find $[L:F]$. Explain why simply adding the two degrees is wrong.

E4 Finite fields. Can a field have six elements? Which characteristic must a field with 27 elements have?

6 Full solutions

S1. Over $\mathbb R$, the minimal polynomial is $x^2+1$ and the degree is 2. Over $\mathbb C$, $i$ is already in the base field, so the minimal polynomial is $x-i$ and degree is 1.

S2. In characteristic 2, $(1+\beta)^2=1+\beta^2=1+(1+\beta)=\beta$. Since $\beta(1+\beta)=1$, its inverse is $\beta$.

S3. The tower law gives 6. The proof constructs products of basis vectors, giving $3\cdot2$ independent spanning vectors; a union of the two bases is not generally a basis over the bottom field.

S4. No, because six is not a prime power. For size 27, the contained prime field has size $p$ and $p^d=27$, forcing $p=3$ and $d=3$. This identifies characteristic and degree, not a new construction without an irreducible polynomial.

Summary

A field extension is a vector space over a specified base field. Minimal polynomials convert algebraic elements into quotient fields and determine their degree. Tower bases multiply degrees, while finite fields necessarily have prime-power size. Explicit four- and eight-element examples distinguish field arithmetic from integer arithmetic modulo a composite number.