Field extensions and nine element arithmetic

A polynomial quotient explains both: it adds a root while retaining exact arithmetic. We examine one quadratic real extension, one cubic extension, and a nine-element field whose addition and multiplication differ from integer arithmetic modulo nine.

1 Extensions as vector spaces

An extension $K/F$ is a field containing an identified copy of the base field $F$. Multiplying by base-field scalars makes $K$ a vector space over $F$; its dimension is the degree $[K:F]$. The smallest subfield containing $F$ and an element $\alpha$ is $F(\alpha)$.

If $\alpha$ satisfies a nonzero polynomial over $F$, choose the monic vanishing polynomial $p$ of least positive degree. It is irreducible: if it factored into two smaller positive-degree polynomials, their evaluated product would be zero in a field, forcing one smaller factor to vanish. Polynomial division proves every polynomial vanishing at $\alpha$ is a multiple of $p$, because a smaller-degree remainder would contradict minimality.

Thus evaluation has kernel $(p)$ and identifies $F[\alpha]$ with $F[x]/(p)$. The irreducible quotient is a field by Bézout inverses, so $F[\alpha]=F(\alpha)$. If $p$ has degree $d$, unique degree-below-$d$ remainders give the basis $1,\alpha,\ldots,\alpha^{d-1}$ and degree $d$.

2 Two characteristic-zero examples

For $\alpha=\sqrt5$ over $\mathbb Q$, $p=x^2-5$ has no rational root and is irreducible. Every element is uniquely $a+b\sqrt5$. Its inverse when nonzero is

$$(a+b\sqrt5)^{-1}=\frac{a-b\sqrt5}{a^2-5b^2}.$$

The rational denominator is nonzero: if $b\ne0$ and it vanished, $a/b=\pm\sqrt5$ would be rational, contrary to the prime-divisibility proof of irrationality; if $b=0$, nonzero input gives $a\ne0$. For example $(2+\sqrt5)^{-1}=\sqrt5-2$.

For $\beta=\sqrt[3]2$, the polynomial $x^3-2$ has no rational root by the rational-root test, hence is irreducible cubic. Therefore $\mathbb Q(\beta)$ has basis $1,\beta,\beta^2$ and degree 3. Reducing powers uses $\beta^3=2$. In particular,

$$\beta^{-1}=\frac{\beta^2}{2},\qquad (1+\beta)^{-1}=\frac{\beta^2-\beta+1}{3},$$

because $(1+\beta)(\beta^2-\beta+1)=1+\beta^3=3$. A field extension gives actual inverses for all nonzero elements, not merely a vector space containing a new symbol.

3 Constructing a field with nine elements

In $\mathbb F_3$, the only squares are 0 and 1. Hence $x^2+1$ has no root and is irreducible. Let

$$K=\mathbb F_3[x]/(x^2+1),\qquad\gamma=[x],\qquad\gamma^2=-1=2.$$

Every element is uniquely $a+b\gamma$ with $a,b\in\{0,1,2\}$, giving nine elements. Operations reduce coefficients modulo 3 and powers using $\gamma^2=2$.

For nonzero $a+b\gamma$,

$$(a+b\gamma)(a-b\gamma)=a^2+b^2.$$

The denominator is nonzero in $\mathbb F_3$: one nonzero coordinate gives square sum 1, while two nonzero coordinates give 2. Thus the conjugate-over-norm inverse formula works here as well. For $1+\gamma$, denominator is 2, whose inverse modulo 3 is 2, so $(1+\gamma)^{-1}=2+\gamma$.

This field is not $\mathbb Z/9\mathbb Z$. The latter has the nonzero zero divisor $[3]$ with square zero. Our field has characteristic 3 and is a two-dimensional vector space over $\mathbb F_3$; its additive group is not cyclic of order 9.

4 Multiplicative and additive structure

The element $\gamma$ has order 4, since $\gamma^2=2$ and $\gamma^4=1$. It does not generate all eight nonzero elements. But $g=1+\gamma$ does: $g^2=2\gamma$, $g^4=2$, and $g^8=1$. Its order divides 8 by Lagrange, but does not divide 4, so it is 8. Thus this example's nonzero multiplicative group is cyclic, generated by $g$.

A general finite field has prime characteristic, since a composite characteristic would create zero divisors from smaller integer multiples of 1. It contains a prime subfield $\mathbb F_p$, and as a finite-dimensional vector space over it has $p^d$ elements. This necessary size form explains why 9 is possible but 6 is not. It does not alone prove existence and uniqueness for every prime power, which is a separate theorem beyond these examples.

Algebraicity, basis independence, and finite-field counts

The least-degree polynomial argument assumes that $\alpha$ is algebraic; a transcendental element has no such polynomial. If $p$ has degree $d$, the powers $1,\alpha,\ldots,\alpha^{d-1}$ span by division, and they are linearly independent because a nonzero relation would be a vanishing polynomial of degree below $d$. This verifies both parts of the basis assertion. The constant-field inclusion is injective because a nonzero constant cannot be divisible by nonconstant $p$.

The irreducibility tests in characteristic zero can be justified directly. If $(a/b)^2=5$ in lowest terms, with $b>0$, then $a^2=5b^2$. Squares modulo 5 are $0,1,4$, so $5\mid a^2$ implies $5\mid a$. Substitution then gives $5\mid b$, contradicting coprimality. If $(a/b)^3=2$ in lowest terms, $a^3=2b^3$ forces $b=1$: a prime dividing $b$ would also divide $a$. An integer root must divide 2 by the equation $a^3=2$, but none of $1,-1,2,-2$ satisfies it. The degree-two/three criterion proved in the preceding chapter now supplies irreducibility, not a rule for arbitrary degrees.

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. In $\mathbb Q(\sqrt5)$, calculate $(1+\sqrt5)(3-\sqrt5)$ and the inverse of $1+\sqrt5$.

E2 Cubic reduction. In $\mathbb Q(\beta)$ with $\beta^3=2$, reduce $\beta^5+\beta^3+1$ into the stated basis.

E3 Finite multiplication. In the nine-element field, compute $(2+\gamma)(1+2\gamma)$.

E4 Inverse and order. Verify the inverse of $1+\gamma$ and explain why its multiplicative order is 8 rather than 4.

6 Full solutions

S1. The product is $3-\sqrt5+3\sqrt5-5=-2+2\sqrt5$. The inverse is $(1-\sqrt5)/(1-5)=(\sqrt5-1)/4$, checked by multiplication.

S2. $\beta^5=2\beta^2$ and $\beta^3=2$, so the result is $3+2\beta^2$. Unique basis coordinates are $(3,0,2)$ over $\mathbb Q$.

S3. Multiply to get $2+5\gamma+2\gamma^2=2+2\gamma+4=2\gamma$ after reducing coefficients modulo 3. The real-number-looking intermediate integers represent classes in the prime field.

S4. $(1+\gamma)(2+\gamma)=2+3\gamma+\gamma^2=2+0+2=1$ modulo 3. For the order, $g^4=2\ne1$ and $g^8=1$. The divisors of 8 below 8 all divide 4, so none can be its order. Its order is exactly 8.

Summary

Field extensions combine polynomial relations with vector-space structure. Irreducible degrees determine basis sizes, while Bézout or conjugate identities calculate inverses. A nine-element field has characteristic 3 and distinct additive and multiplicative structures; it is not the composite-modulus integer ring with the same number of elements.