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Field Extensions and the Complex Numbers: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-13Open item page →

$Q (\sqrt{2}) ≅ Q [x] / (x^{2} - 2)$ with basis $1, \sqrt{2}$

Example

Over the rational field $Q$ (The rationals form a field), let $a = \sqrt{2}$ , whose existence and positive choice are supplied by the completeness of $R$ (The Cauchy-sequence reals have the least-upper-bound property) and Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ . Then $Q (a) ≅ Q [x] / (x^{2} - 2),$ and every element is uniquely $r + s a$ with $r, s \in Q$ .

Facts & Assumptions

Given: The element $a = \sqrt{2}$ over $Q$ .

[F1]

Eisenstein's criterion makes a primitive integer polynomial irreducible over $Q$ when some prime divides every nonleading coefficient, does not divide the leading coefficient, and its square does not divide the constant coefficient (Eisenstein criterion over the integers).

[F2]

A simple algebraic extension is its minimal-polynomial quotient and has the associated power basis (A simple algebraic extension is its minimal-polynomial quotient and has power basis $1, a, \dots, a^{n - 1}$ and degree $n$ ).

[F3]

The real numbers are a complete ordered field, so $\sqrt{2}$ exists and satisfies $(\sqrt{2})^{2} = 2$ (The Cauchy-sequence reals have the least-upper-bound property, Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

[F4]

An element is algebraic when a nonzero polynomial vanishes at it (Algebraic and transcendental elements and algebraic extensions); for an algebraic element with minimal polynomial $m_{a}$ , $f (a) = 0$ exactly when $m_{a}$ divides $f$ (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).

Verification

technique · direct

1.1

The polynomial $x^{2} - 2$ satisfies [F1] at the prime $2$ , so it is monic and irreducible over $Q$ ; since it vanishes at $a$ , [F4] makes it the minimal polynomial of $a$ .

F1F3F4algebra

2.1

Apply [F2] to obtain the quotient isomorphism, the basis $1, a$ , and the unique form $r + s a$ .

F2step 1.1

3.1

For instance, $(1 + a)^{- 1} = a - 1$ , because $(1 + a) (a - 1) = a^{2} - 1 = 1$ .

step 2.1algebra∎

ExampleConstruction: Literature-sourcedVerification: Literature-sourcedprecheck passaudited 2026-08-13Open item page →

The four-element field $(Z / 2) [x] / (x^{2} + x + 1)$

Example

Let $F = Z / 2$ , with its quotient arithmetic (For every $n \in N$ , the congruence-class ring $Z / n$ is the quotient ring $Z / n Z$ ). This is a field (For every prime $p$ , the two operations on $Z / p$ make it a field). Then $K = F [x] / (x^{2} + x + 1) = {0, 1, a, 1 + a}, a = x + (x^{2} + x + 1),$ is a field of four elements.

Facts & Assumptions

Given: The field $F = Z / 2$ and the polynomial $x^{2} + x + 1$ .

[F1]

A quadratic over a field is irreducible exactly when it has no root in that field (A polynomial of degree two or three over a field is irreducible exactly when it has no root in the field).

[F2]

The quotient by a monic irreducible degree- $n$ polynomial is a field and has unique representatives of degree below $n$ ( $F [x] / (p)$ for monic irreducible $p$ is a field extension containing the root $x + (p)$ with unique reduced representatives).

Verification

technique · direct

1.1

The values of $x^{2} + x + 1$ at $0$ and $1$ are both $1$ in $F$ , so [F1] makes it irreducible.

F1algebra

2.1

By [F2], $K$ is a field and its unique linear representatives are exactly $0, 1, x, 1 + x$ , giving the displayed four classes.

F2step 1.1

3.1

The defining relation is $a^{2} + a + 1 = 0$ , hence $a^{2} = a + 1$ in characteristic two. Consequently $a (a + 1) = 1$ and $(a + 1)^{2} = a$ , which determines the products of the nonzero elements.

step 2.1algebra∎

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-13Open item page →

$Q [x] / (x^{2} - 1)$ is not a field: reducibility creates nonzero zero divisors

Statement refuted

The quotient by any nonconstant polynomial over a field is a field.

Counterexample

In the rational field $Q$ (The rationals form a field), take $R = Q [x] / (x^{2} - 1) .$ The two classes $x - 1 + (x^{2} - 1)$ and $x + 1 + (x^{2} - 1)$ are nonzero, but their product is zero.

Facts & Assumptions

Given: The quotient $R = Q [x] / (x^{2} - 1)$ .

[F1]

Division by a nonzero polynomial gives a unique remainder of smaller degree (Division algorithm for polynomials over a field).

[F2]

$F [x] / (p)$ is a field if and only if the nonconstant polynomial $p$ is irreducible (For a nonconstant $p$ in $F [x]$ , the ideal $(p)$ is maximal and $F [x] / (p)$ is a field exactly when $p$ is irreducible).

Verification

technique · counterexample

1.1

Neither $x - 1$ nor $x + 1$ lies in $(x^{2} - 1)$ : each already has degree below $2$ , so uniqueness of the remainder in [F1] would otherwise make it zero.

2.1

Yet $(x - 1) (x + 1) = x^{2} - 1$ , so the product of their nonzero classes is zero.

step 1.1algebra

3.1

Thus $R$ has nonzero zero divisors and is not a field, agreeing with [F2] because $x^{2} - 1$ is reducible.

F2step 2.1∎

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-13Open item page →

The minimal polynomial of $\sqrt{2 + \sqrt{3}}$ over $Q$ is $x^{4} - 4 x^{2} + 1$

Example

Let $a = \sqrt{2 + \sqrt{3}}$ , with the nonnegative real square roots supplied by the completeness of $R$ (The Cauchy-sequence reals have the least-upper-bound property) and Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ . Its minimal polynomial over $Q$ is $p (x) = x^{4} - 4 x^{2} + 1,$ so $[Q (a) : Q] = 4$ .

Facts & Assumptions

Given: The real number $a = \sqrt{2 + \sqrt{3}}$ .

[F1]

Eisenstein's criterion proves irreducibility over $Q$ under its prime divisibility hypotheses (Eisenstein criterion over the integers).

[F2]

The polynomial-ring universal property gives substitution homomorphisms such as $f (x) \mapsto f (x + 1)$ (Universal property of $R [x]$ : a coefficient homomorphism and the image of $x$ determine a unique ring homomorphism).

[F3]

A root of a nonzero polynomial is algebraic (Algebraic and transcendental elements and algebraic extensions), and the minimal polynomial of an algebraic element is the monic irreducible polynomial generating its evaluation kernel (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).

[F4]

Its degree equals the degree of the associated simple extension (A simple algebraic extension is its minimal-polynomial quotient and has power basis $1, a, \dots, a^{n - 1}$ and degree $n$ ).

[F5]

The real numbers are a complete ordered field, so the displayed nonnegative square roots exist and square to their radicands (The Cauchy-sequence reals have the least-upper-bound property, Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

Verification

technique · direct

1.1

From $a^{2} = 2 + \sqrt{3}$ one obtains $(a^{2} - 2)^{2} = 3$ , hence $p (a) = a^{4} - 4 a^{2} + 1 = 0$ .

F5algebra

1.2

Substitution from [F2] gives $p (x + 1) = x^{4} + 4 x^{3} + 2 x^{2} - 4 x - 2$ , which satisfies [F1] at $2$ and is therefore irreducible.

F1F2algebra

2.1

Substitution by $x - 1$ is inverse to substitution by $x + 1$ , so a factorization of $p$ would produce one of $p (x + 1)$ . Thus $p$ is irreducible.

F2step 1.2algebra

3.1

Since $p$ is monic, irreducible, and annihilates $a$ , [F3] identifies it as the minimal polynomial; [F4] gives degree $4$ .

F3F4step 1.1step 2.1∎

ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-13Open item page →

The square roots of $i$ are $\pm (1 + i) / \sqrt{2}$

Example

The two square roots of $i$ are $\frac{1 + i}{\sqrt{2}} and - \frac{1 + i}{\sqrt{2}} .$

Facts & Assumptions

Given: The imaginary unit $i \in C$ .

[F1]

Complex multiplication satisfies $(a + b i)^{2} = (a^{2} - b^{2}) + 2 a b i$ ( $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ).

[F2]

The real numbers are a complete ordered field, so the nonnegative real square root satisfies $(\sqrt{2})^{2} = 2$ (The Cauchy-sequence reals have the least-upper-bound property, Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

[F3]

Every complex number has a square root (Every complex number has a square root, by an explicit Cartesian formula).

Verification

technique · direct

1.1

By [F1] and [F2], $((1 + i) / \sqrt{2})^{2} = i$ ; its negative has the same square.

F1F2algebra

1.2

Conversely, if $(a + b i)^{2} = i$ , [F1] gives $a^{2} = b^{2}$ and $2 a b = 1$ . Consequently $a, b \neq 0$ , they have the same sign, and $a = b$ .

F1algebra

2.1

Then $2 a^{2} = 1$ , so [F2] gives $a = b = 1 / \sqrt{2}$ or $a = b = - 1 / \sqrt{2}$ . These are exactly the two values in step 1.1.

F2step 1.2algebra

3.1

Thus the existence guaranteed abstractly by [F3] is realized by exactly the two displayed roots.

F3step 1.1step 2.1∎

False statementConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-13Open item page →

FALSE: the complex field admits an order making it an ordered field

Statement refuted

There is a total order on $C$ compatible with its field operations.

Facts & Assumptions

Given: The proposed ordered-field structure on $C$ .

[F1]

In an ordered field every nonzero square is positive (Squares of nonzero elements are positive).

[F2]

$i^{2} = - 1$ (The complex numbers as $R [x] / (x^{2} + 1)$ , with the real embedding and imaginary unit $i$ ), and every complex number is uniquely $a + b i$ ( $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ), so $i = 0 + 1 i$ is nonzero.

Refutation

technique · contradiction

1.1

Suppose such an ordered-field structure existed. Since $i \neq 0$ , [F1] and [F2] would give $- 1 = i^{2} > 0$ .

assume-contraF1F2

2.1

Also $1 = 1^{2} > 0$ by [F1], so compatibility with addition gives $0 = 1 + (- 1) > 0$ , impossible in a strict order.

F1step 1.1algebra

3.1

Therefore no order compatible with the complex field operations exists.

step 2.1discharge-contradiction∎

CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-13Open item page →

An annihilating polynomial need not be minimal: $\sqrt{2}$ is a root of both $x^{2} - 2$ and $x^{4} - 4$

Statement refuted

Every nonzero polynomial that vanishes at an algebraic element is that element's minimal polynomial.

Counterexample

Let $a = \sqrt{2}$ , whose nonnegative real value satisfies $a^{2} = 2$ because $R$ is complete (The Cauchy-sequence reals have the least-upper-bound property, Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ). Then both $x^{2} - 2$ and $x^{4} - 4$ vanish at $a$ , but only the first is minimal.

Facts & Assumptions

Given: The algebraic number $a = \sqrt{2}$ over $Q$ .

[F1]

Eisenstein's criterion proves irreducibility over $Q$ under its prime divisibility hypotheses (Eisenstein criterion over the integers).

[F2]

The minimal polynomial is the unique monic irreducible generator of the evaluation kernel, and it divides every annihilating polynomial (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).

[F3]

Verification

technique · counterexample

1.1

Evaluation gives $a^{2} - 2 = 0$ and $a^{4} - 4 = 0$ .

F3algebra

2.1

The polynomial $x^{2} - 2$ satisfies [F1] at $2$ , so [F2] identifies it as the minimal polynomial of $a$ over $Q$ .

F1F2step 1.1

3.1

The other annihilator factors as $x^{4} - 4 = (x^{2} - 2) (x^{2} + 2)$ and has larger degree, so it is a proper multiple of the minimal polynomial.

step 2.1algebra

4.1

Thus a polynomial may annihilate an algebraic element without being its minimal polynomial.

step 1.1step 3.1∎

Sources

Standard references

Recommended treatments; not extraction sources.

Field Extensions and the Complex Numbers: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Q(2)≅Q[x]/(x2−2)Q(2​)≅Q[x]/(x2−2) with basis 1,21,2​

Example

Facts & Assumptions

Verification

The four-element field (Z/2)[x]/(x2+x+1)(Z/2)[x]/(x2+x+1)

Example

Facts & Assumptions

Verification

Q[x]/(x2−1)Q[x]/(x2−1) is not a field: reducibility creates nonzero zero divisors

Statement refuted

Counterexample

Facts & Assumptions

Verification

The minimal polynomial of 2+32+3​​ over QQ is x4−4x2+1x4−4x2+1

Example

Facts & Assumptions

Verification

The square roots of ii are ±(1+i)/2±(1+i)/2​

Example

Facts & Assumptions

Verification

FALSE: the complex field admits an order making it an ordered field

Statement refuted

Facts & Assumptions

Refutation

An annihilating polynomial need not be minimal: 22​ is a root of both x2−2x2−2 and x4−4x4−4

Statement refuted

Counterexample

Facts & Assumptions

Verification

Sources

Standard references

$Q (\sqrt{2}) ≅ Q [x] / (x^{2} - 2)$ with basis $1, \sqrt{2}$

The four-element field $(Z / 2) [x] / (x^{2} + x + 1)$

$Q [x] / (x^{2} - 1)$ is not a field: reducibility creates nonzero zero divisors

The minimal polynomial of $\sqrt{2 + \sqrt{3}}$ over $Q$ is $x^{4} - 4 x^{2} + 1$

The square roots of $i$ are $\pm (1 + i) / \sqrt{2}$

An annihilating polynomial need not be minimal: $\sqrt{2}$ is a root of both $x^{2} - 2$ and $x^{4} - 4$