advanced26 min studyLesson 152 in Mathematics

Minimal polynomials: the first relation determines the dimension

Prove irreducibility of a minimal relation, identify a power basis, and reduce arithmetic in a simple extension.

On this page

What you’ll understand

  • Explain why a minimal polynomial is unique, monic, and irreducible.
  • Derive a power basis from polynomial division.
  • Compute multiplication and inverses modulo a minimal polynomial.

Many equations can vanish at one number. The number √2 satisfies X² − 2, X⁴ − 4, and every multiple of either polynomial, but these relations contain different amounts of redundancy. The minimal polynomial removes that redundancy: it is the monic relation of least positive degree over the chosen base field. Its degree is not merely the length of a convenient equation. It equals the number of independent base-field coordinates needed for arithmetic after adjoining the number.

Use the least relation, not just any relation

Let α be algebraic over F. Choose a nonzero polynomial of least degree vanishing at α, and divide by its leading coefficient to make it monic. Its degree is positive because a nonzero constant cannot vanish. If this polynomial factored into two lower positive-degree polynomials over F, evaluating at α would make their product zero in a field. One factor would vanish, contradicting minimality. Thus the minimal polynomial is irreducible over F.

Polynomial division proves both uniqueness and a useful divisibility test. Divide any polynomial g vanishing at α by the minimal polynomial m. The remainder has smaller degree and also vanishes at α, so it must be zero. Therefore m divides every relation. Two monic minimal relations divide each other and consequently agree. Notice the phrase over F: X² − 2 is minimal for √2 over Q, whereas X − √2 is minimal once √2 is already allowed as a coefficient.

g(α)=0⟺mα,F(X)∣g(X)(g∈F[X])g(\alpha)=0\quad\Longleftrightarrow\quad m_{\alpha,F}(X)\mid g(X)\qquad(g\in F[X])
For an algebraic element, every polynomial relation over F is a multiple of its monic minimal polynomial.↔ Scroll wide equations horizontally.

A quotient ring becomes a coordinate system

If m has degree d, dividing by m reduces every polynomial expression to degree below d. Evaluation identifies F[α] with the quotient F[X]/(m). Since m is irreducible, this quotient is a field, so F[α] already equals F(α). Explicitly, for a polynomial h with h(α) nonzero, the greatest common divisor of h and m is one. A polynomial Bézout identity then produces a polynomial representing the inverse of h(α).

The expressions 1, α, up to α raised to d − 1 span the extension. They are also independent: any nontrivial linear dependence would be a lower-degree polynomial relation. Hence they form a basis over F. The extension degree [F(α):F] is defined as this vector-space dimension. It counts independent coordinates, not the number of field elements and not automatically the degree of the first polynomial someone happens to write down.

F(α)≅F[X]/(mα,F),[F(α):F]=deg⁡mα,F=dF(\alpha)\cong F[X]/(m_{\alpha,F}),\qquad [F(\alpha):F]=\deg m_{\alpha,F}=d
An algebraic simple extension has a power basis consisting of 1 through α to the power d − 1.↔ Scroll wide equations horizontally.

Reduce products while preserving an irreducibility certificate

For the real cube root α of 2, the polynomial X³ − 2 is irreducible over Q by Eisenstein at the prime 2. Thus every element has a unique expression a + bα + cα² with rational coefficients. Multiplication is ordinary expansion followed by replacing α³ with 2. Uniqueness depends on irreducibility, not merely on the fact that the cubic vanishes at α. A reducible relation would give a spanning list that could still contain unnecessary coordinates.

Minimal-polynomial computations therefore have two obligations: show that the proposed polynomial vanishes and show it is irreducible over the stated field. Numerical approximation verifies neither obligation exactly. A decimal close to a root could instead be a nearby rational number, whose minimal polynomial is linear. When the base field changes, repeat the irreducibility argument; a polynomial irreducible over Q can factor after adjoining another root or adding new coefficients.

Find an inverse in a cubic extension

Let α be the real cube root of 2. Compute the inverse of 1 + α in Q(α), using the power basis 1, α, α².

  1. Eisenstein at 2 makes X³ − 2 irreducible over Q, so the extension has degree three and α³ = 2 is the reduction rule. Also 1 + α is nonzero because α is positive.

  2. Use the polynomial identity (1 + α)(1 − α + α²) = 1 + α³. The middle terms cancel before any approximation is used.

  3. Replace α³ by 2 to obtain product 3, then divide the second factor by 3. Multiplication by 1 + α now gives exactly one.

The result

The inverse is (1 − α + α²)/3; its three rational coefficients are unique.

Explore the model

Interactive lab

Interactive lab · function

A plotted relation is not an irreducibility proof

The lab plots y = ax² and shows its tangent at x. Predict how doubling a changes the height at a fixed nonzero x, then test it. This illustrates evaluating and manipulating a polynomial family. The display has no field selector or polynomial factorization control, so it cannot determine minimal polynomials or the extension degree.

A plotted relation is not an irreducibility proofThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forEvaluation supplies a relation; an algebraic argument is needed to establish minimality.

Check your understanding

An algebraic element α has minimal polynomial of degree four over F. What follows?

Try it yourself

Find the minimal polynomial of √7 over Q and express 1/(√7 − 2) in its power basis.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

A rational root of X² − 7 would be a rational square root of 7. Then multiply the denominator by its conjugate.

Compare with the solution

X² − 7 vanishes and is irreducible over Q because 7 is not a rational square. The degree is two and the basis is 1, √7. Multiplying (√7 − 2) by (√7 + 2) gives 3, so the requested inverse is (√7 + 2)/3. Its denominator is a nonzero rational number, and multiplication checks the answer exactly.

Take it with you

  • The minimal polynomial divides every relation over the same base field.
  • Its irreducibility makes the polynomial quotient into a field.
  • Extension degree equals the size of the power basis, not a field's cardinality.

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