Calculator
Example data table
| Field | Element | Inverse | Reason |
|---|---|---|---|
| GF(11) | 7 | 8 | 7×8 = 56 ≡ 1 (mod 11) |
| GF(13) | 10 | 4 | 10×4 = 40 ≡ 1 (mod 13) |
| GF(5^3) | a(x)=1+x^2 | Computed by Euclid | Inverse depends on chosen modulus m(x) |
Formula used
In a prime field GF(p), the multiplicative inverse of a is the number a⁻¹ satisfying a·a⁻¹ ≡ 1 (mod p). The calculator finds a⁻¹ using the Extended Euclidean Algorithm, which returns integers x,y such that ax + py = gcd(a,p). When gcd(a,p)=1, x mod p is the inverse.
In an extension field defined by a modulus polynomial m(x), the inverse of a(x) is a polynomial b(x) such that a(x)·b(x) ≡ 1 (mod m(x)). This is computed by running Extended Euclid on polynomials over GF(p).
How to use this calculator
- Pick a mode: prime field or extension field.
- Enter a prime p. The tool checks primality.
- For GF(p), enter any integer a; it is reduced mod p.
- For GF(p^m), enter coefficients for a(x) and m(x), low-to-high.
- Enable “Show Euclid steps” to view the full derivation.
- Use CSV/PDF downloads to save the computed inverse and steps.
Article
Field selection and validity
A finite field needs a prime modulus p or an irreducible modulus polynomial m(x) over GF(p). This calculator verifies p using trial division up to √p. For GF(11), every nonzero element has an inverse; for example, 7⁻¹=8 because 7×8≡1 (mod 11). For GF(13), 10⁻¹=4 since 10×4=40 and 40≡1 (mod 13).
Extended Euclid performance
The inverse is computed with the Extended Euclidean Algorithm. Integer Euclid typically finishes in O(log p) divisions, so even p near one billion is handled quickly. If you enable steps, you can audit each quotient q and remainder r to see exactly where the inverse arises. For a=7, p=11, the tool returns 8 and verifies it immediately.
Polynomial representation and degree
In GF(p^m), elements are polynomials a(x)=a0+a1x+… with coefficients reduced mod p. Enter coefficients low-to-high, such as 1,0,1 to represent 1+x². The modulus m(x) sets the degree m; a useful teaching example is degree 3 where each element fits in three coefficients. With p=5 and m=3, the field has 5³=125 elements, and every operation is performed modulo m(x).
Invertibility checks in GF(p^m)
An element is invertible exactly when gcd(a(x), m(x))=1. When m(x) is irreducible, every nonzero element passes this test, matching field theory. If you accidentally enter a reducible modulus, some nonzero inputs may fail and the tool will report that no inverse exists. The step log shows polynomial divisions and remainders, helping you diagnose a wrong modulus choice.
Exportable audit trail
The CSV export records the field, the normalized element, the inverse, and optional step lines for reproducibility. The PDF export is a compact, one-page report suited for assignments and verification logs. Use exports to compare multiple runs, share results, or attach a proof of correctness to your work. If you compute inverses for several a values, the CSV becomes a quick dataset for checking group properties.
Graph-based verification
The Plotly chart visualizes correctness. For GF(p), it plots k versus (a·k mod p) and highlights the unique k that maps to 1, which is the inverse. For GF(p^m), it shows coefficient bars for a(x) and a(x)⁻¹, making it easy to spot patterns and confirm reductions. A flat check line at 1 indicates success, while missing highlights signal an invalid or zero element.
FAQs
1) Why must p be prime in GF(p)?
A prime modulus guarantees every nonzero residue has a multiplicative inverse. If p is composite, some nonzero values share factors with p, so inverses fail and the structure is not a field.
2) What does “coefficients low-to-high” mean?
It means you enter a0,a1,a2,… where a0 is the constant term, a1 multiplies x, and a2 multiplies x². For example, 1,0,1 represents 1 + x².
3) Do I need an irreducible modulus polynomial?
Yes for a true extension field. An irreducible m(x) ensures every nonzero polynomial class has an inverse. With a reducible modulus, you may create zero divisors and some inverses will not exist.
4) Why does 0 never have an inverse?
Because 0·b = 0 for every b, it can never equal 1. Fields define inverses only for nonzero elements, so the calculator blocks 0 immediately in both prime and polynomial modes.
5) How does the Plotly graph confirm correctness?
For GF(p), the point where (a·k mod p) equals 1 identifies the inverse k. For GF(p^m), the bar chart shows coefficients of a(x) and its inverse, supporting visual sanity checks.
6) What if the tool says “no inverse exists” in polynomial mode?
It usually indicates gcd(a(x), m(x)) ≠ 1, often caused by a reducible modulus or an element sharing a factor with it. Try an irreducible modulus and ensure a(x) is not zero modulo m(x).