core30 min studyLesson 150 in Mathematics

Polynomial interpolation over fields

Construct a polynomial from prescribed values, prove uniqueness, and connect interpolation to a quotient-ring decomposition.

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What you’ll understand

  • Build Lagrange basis polynomials from distinct field nodes.
  • Prove the degree-bounded interpolation theorem by a root count.
  • Separate exact algebraic existence from numerical stability and predictive accuracy.

Suppose you prescribe a value at each of several distinct field elements. A polynomial can be built that meets every prescription, and a degree restriction makes it unique. This interpolation problem joins evaluation, root bounds, and the Chinese remainder theorem in one construction. Over finite fields it is exact modular algebra; over real or complex fields its symbolic existence must still be distinguished from the numerical behavior of a particular implementation and from any claim that the data reveal a physical law.

Build a basis that selects one node at a time

Choose n distinct nodes in a field and label their desired values. For a particular node, multiply x minus every other node, producing a polynomial that vanishes at all the other locations. Divide by the value of that product at the selected node so that the selected value becomes one. Every denominator factor is the difference of two distinct field elements, hence is nonzero and invertible. This constructs a Lagrange basis polynomial of degree n minus one.

Multiply each basis polynomial by its prescribed value and add them. At a selected node, all but one summand vanish and the remaining summand gives exactly the desired value. This proves existence without solving a coefficient system. The construction works over any field with enough distinct elements for the nodes, including a finite field. Over a general ring, distinct nodes alone are insufficient because their differences may be nonunits.

Lj(x)=∏k≠jx−akaj−ak,P(x)=∑j=1nyjLj(x),Lj(ai)=δijL_j(x)=\prod_{k\neq j}\frac{x-a_k}{a_j-a_k},\qquad P(x)=\sum_{j=1}^{n}y_jL_j(x),\qquad L_j(a_i)=\delta_{ij}
The n nodes are distinct elements of a field. Each node difference is invertible, and δ is one when its indices agree and zero otherwise.↔ Scroll wide equations horizontally.

Uniqueness comes from the degree bound

If two polynomials of degree below n meet the same n prescriptions, their difference has degree below n and vanishes at all n nodes. A nonzero polynomial over a field cannot have more roots than its degree, so the difference is zero. The degree restriction is essential. Adding any multiple of the product of x minus all nodes gives another polynomial with exactly the same prescribed values, generally at higher degree.

This is also a quotient-ring statement. The simultaneous evaluation map has kernel generated by that node product, since the distinct linear ideals are pairwise comaximal. Its target is the product of n copies of the field, and the CRT makes it surjective. Polynomial division then selects a unique representative of degree below n for each quotient class. Lagrange interpolation provides an explicit inverse from the value tuple to that representative.

F[x]/(∏j=1n(x−aj))≅Fn,[P]⟼(P(a1),…,P(an))F[x]\Big/\left(\prod_{j=1}^{n}(x-a_j)\right)\cong F^n,\qquad [P]\longmapsto(P(a_1),\ldots,P(a_n))
The product Fⁿ has componentwise addition and multiplication. Distinct field nodes make the linear ideals comaximal and the displayed map an isomorphism.↔ Scroll wide equations horizontally.

Exact fitting is not a guarantee of a good model

Over the reals, the theorem guarantees exact interpolation for exact inputs, but it does not guarantee numerical stability. Closely spaced nodes can make basis denominators small, amplifying roundoff or measurement noise in some representations. A different basis or node choice can change computational behavior without changing the unique exact polynomial. These numerical questions require error analysis beyond the existence proof, and should not be dismissed simply because a symbolic formula is available.

Likewise, matching observations does not prove that the interpolating polynomial describes the mechanism between nodes or outside the sampled interval. In a finite field, different formal polynomials can even define the same function at every field element: the node product over the whole field vanishes everywhere. Keep three statements separate: a unique low-degree interpolant exists under the hypotheses, an algorithm computes an approximation to it, and a chosen model predicts useful new observations. Only the first follows from this algebraic theorem.

Interpolate exactly modulo five

Find the degree-below-three polynomial over F5\mathbb{F}_5 with values one, two, and zero at nodes zero, one, and two.

  1. Write P=ax2+bx+cP=ax^2+bx+c. The value at zero gives c = 1. At one, a+b+1=2a+b+1=2, so a+b=1a+b=1 in the field.

  2. At two, 4a+2b+1=04a+2b+1=0, giving 4a+2b=44a+2b=4 modulo five. Subtract twice the previous equation to obtain 2a=22a=2. The inverse of two is three modulo five, so a = 1 and then b = 0.

  3. The polynomial is x2+1x^2+1. Its evaluations are one, two, and five, the last equal to zero modulo five. Any other degree-below-three candidate would differ by a degree-at-most-two polynomial with three distinct roots, so uniqueness follows.

The result

The unique required representative is P=x2+1∈F5[x]P=x^2+1\in\mathbb{F}_5[x].

Explore the model

Interactive lab

Interactive lab · function

Read interpolation data from a known quadratic

Choose a in the real lab's y=ax2y=ax^2 family and record outputs at x equal to negative one, zero, and one. Predict the unique polynomial of degree below three fitting these exact values before changing a. This samples a known real quadratic; the lab does not construct arbitrary interpolants or perform finite-field arithmetic.

Read interpolation data from a known quadraticThis 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 forThree distinct exact field samples determine a polynomial of degree below three, even though higher-degree polynomials can share them.

Check your understanding

Why do n distinct field samples determine at most one polynomial of degree below n?

Try it yourself

Over the rationals, find the degree-below-two interpolant with P(0) = 3 and P(2) = 7. Then give a different polynomial with those same two values.

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

Give me a hint

First solve for a line. Then add a polynomial vanishing at both nodes.

Compare with the solution

The line is P=3+2xP=3+2x: its constant is three and its slope is (7−3)/(2−0)=2(7-3)/(2-0)=2. The polynomial Q=3+2x+x(x−2)Q=3+2x+x(x-2) has the same prescribed values because the added term vanishes at both nodes. It simplifies to x2+3x^2+3 and has degree two, so it does not contradict uniqueness among polynomials of degree below two. This example isolates the exact role of the degree restriction.

Take it with you

  • Lagrange basis polynomials turn each data value into an independent node selector.
  • Uniqueness follows from the field root bound together with a strict degree restriction.
  • Exact interpolation, stable computation, and reliable prediction are separate claims.

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