Build a field from an irreducible polynomial

Try

Over F₂, the polynomial x²+x+1 takes the value one at both zero and one. A quadratic with no root is irreducible, so its quotient is a field. Let α denote the class of x. The relation is α²=α+1, since subtraction equals addition in characteristic two.

Every element has a unique form a+bα with a,b in F₂. There are four possibilities: zero, one, α, and α+1. This field is not the integer residue ring Z/4Z.

Predict

Predict α³ by multiplying α²=α+1 by α and reducing again. Then replace the modulus by x²+1=(x+1)² over F₂. The new quotient still has four coordinate pairs. Does matching cardinality guarantee a field? Predict the square of the nonzero class x+1 in that second quotient.

Experiment

Use the exact quotient explorer to compare the irreducible and reducible moduli. Change one operand, predict the coefficient expansion, then inspect reduction modulo two and by the selected quadratic. The full multiplication table is finite evidence for these specific rings.

Live experiment: a quotient ring is not automatically a field

Work over F₂, whose coefficients are 0 and 1 with addition and multiplication modulo 2. Let α be the class of the polynomial variable t in a quadratic quotient. Every class has a unique representative a₀+a₁α, so both choices below have four elements. Predict whether every nonzero element has an inverse.

Quotient modulus t²+t+1 (irreducible)t²+1 (reducible)First element a 01αα+1Second element b 01αα+1Reset experiment

1. Multiply representatives: (α)(α+1) expands to α²+α before reducing coefficients.
2. Reduce coefficients modulo 2: α²+α. A coefficient 2 becomes 0; this is characteristic 2 arithmetic.
3. Reduce the quadratic term: α²=α+1 because α²+α+1=0. The product is 1.

The sum a+b is 1, adding the two coefficients separately modulo 2. The inverse of a=α is α+1. Multiplying by that element produces the multiplicative identity 1.

Complete multiplication table: row times column. Selected product (α)·(α+1)=1.

×	0	1	α	α+1
0	0	0	0	0
1	0	1	α	α+1
α	0	α	α+1	1
α+1	0	α+1	1	α

This quotient is a field with four elements. Its units are 1, α, α+1. Its nonzero zero divisors are none.

The polynomial t²+t+1 has no root in F₂: substitution at 0 and 1 both gives 1. A quadratic over a field is reducible exactly when it has a root, so this polynomial is irreducible and its ideal is maximal. That theorem proves the quotient is a field; the table checks this particular example. Its nonzero multiplicative group is cyclic of order 3, generated by α.

Squaring and squaring again

z	z²	z⁴
0	0	0
1	1	1
α	α+1	α
α+1	α	α+1

Squaring swaps α and α+1 and fixes 0,1. It is the Frobenius automorphism; applying it twice returns every element, so z⁴=z throughout F₄. The general finite-field theorem requires a field, not merely a finite ring of prime-power size.

Choose a=b=α+1 and switch the modulus. Explain each changed product through polynomial reduction. Then return to the irreducible modulus and find all inverses without opening a separate solution. These tables exhaust two exact four-element examples; general quotient, irreducibility and field theorems still require proof. Settings are temporary.

For the field case, the key reductions are:

Product	Reduction
α²	α+1
α(α+1)	α²+α=1
(α+1)²	α²+1=α

Thus α and α+1 are inverses, and both have multiplicative order three. In the reducible quotient, the nonzero class x+1 squares to zero, so it cannot have an inverse. The two quotients share a vector-space size but not a field structure.

Understand

Every finite field has prime characteristic. Finiteness makes some positive multiple of one vanish. If the least such integer n factored as ab with both factors smaller than n, then (a·1)(b·1)=0 with neither factor zero, impossible in a field. Thus n=p is prime, and the multiples of one form a copy of F_p.

The field is a finite-dimensional vector space over this prime subfield. If the dimension is n, a basis gives p choices for each of n coordinates, hence exactly pⁿ elements. This proves the prime-power necessity. Existence for every prime power and uniqueness up to isomorphism will be proved using splitting fields, rather than assumed from the count alone.

An irreducible polynomial m of degree n over F_p gives a field quotient F_p[x]/(m) with pⁿ elements. Unique remainders of degree below n supply the coordinates. For example x²+1 has no root over F₃, because its values are one, two, and two. Its quotient has nine elements, with β²=−1=2. The inverse of 1+β is 2+β, since their product reduces to one modulo three.

The field with pⁿ elements is generally not Z/pⁿZ when n exceeds one. The latter has characteristic pⁿ and nonzero zero divisors, while the field has characteristic p. The notation F_(pⁿ) refers to a field of that size, not integer arithmetic modulo the cardinality.

Apply

Verify every nonzero inverse in the four-element field, then identify a zero-divisor witness in the reducible quotient. Explain why one failed inverse disproves the field claim, while the quotient construction still gives a valid ring.

For the nine-element example, reduce a product in the basis one and β without borrowing the characteristic-two sign rules. Coefficient arithmetic depends on the prime subfield, and a rule such as minus equals plus must not be transferred to odd characteristic.

Transfer: a new coordinate is not automatically a field map

In F₉=F₃[β] with β²=2, the pair 1,β is a basis over F₃. So is 1,γ where γ=1+β: if a+bγ=0, then (a+b)+bβ=0, forcing b=a=0. A coordinate change is an invertible linear operation. It does not follow that replacing β by γ preserves multiplication.

Compute γ²=(1+β)²=1+2β+2=2β=2γ+1 in characteristic three. Thus γ satisfies t²+t+2=0, not t²+1=0. The polynomial t²+t+2 takes the values 2,1,2 at 0,1,2, so it is irreducible over F₃. Consequently there is an isomorphism

$$\mathbb F_3[t]/(t^2+t+2)\longrightarrow\mathbb F_9,\qquad [t]\longmapsto1+\beta.$$

Why this rule works. The defining polynomial vanishes at 1+β, so evaluation descends through the ideal. The image contains β=γ−1, so it reaches the whole field. Equivalently, on coordinates the map is (a,b)↦(a+b,b), an invertible linear map. Preserving the defining relation is the extra multiplication check; bijectivity of coordinates alone is insufficient.

In contrast, trying to define an endomorphism of the original presentation by β↦1+β fails: it would have to send β²=2 to 2, yet its proposed image has square 2β. A valid automorphism is β↦−β, because (−β)²=2. It sends a+bβ to a−bβ and applying it twice returns the original element.

Predict, then inspect the certificate

Changed case A valid generator imageA tempting invalid image

In F₉=F₃[β] with β²=2, does β↦−β define an F₃-automorphism?

Your prediction Choose a predictionYesNo Check reasoning Reset prediction

Two authored cases are practice, not a proof of the general theorem. Explain which hypothesis changed. Responses are temporary and do not change saved progress.

Changed task. Express (1+γ)(2+γ) first using γ²=2γ+1, then translate both factors to the β basis and multiply independently. Explain why the answers agree without treating the two symbols as equal.

Feedback

In the γ basis, the product is 2+3γ+γ²=2+(2γ+1)=2γ. In the β basis the factors are 2+β and β, whose product is 2β+2=2(1+β)=2γ. The coordinate conversion is consistent with a homomorphism between two different quotient presentations, not the invalid substitution inside the same presentation. All coefficient arithmetic is modulo three.

Continue

Frobenius next raises elements to the characteristic prime. In a finite field it is an automorphism whose iterates organize subfields and later give the complete Galois group over the prime field.