galois.Poly.is_square_free

galois.Poly.is_square_free() → bool

Determines whether the polynomial $f (x)$ over $GF (q)$ is square-free.

Why is this a method and not a property?

This is a method to indicate it is a computationally expensive task.

Returns:¶: True if the polynomial is square-free.

Notes

A square-free polynomial $f (x)$ has no irreducible factors with multiplicity greater than one. Therefore, its canonical factorization is

f (x) = \prod_{i = 1}^{k} g_{i} (x)^{e_{i}} = \prod_{i = 1}^{k} g_{i} (x) .

Examples

Generate irreducible polynomials over $GF (3)$ .

In [1]: GF = galois.GF(3)

In [2]: f1 = galois.irreducible_poly(3, 3); f1
Out[2]: Poly(x^3 + 2x + 1, GF(3))

In [3]: f2 = galois.irreducible_poly(3, 4); f2
Out[3]: Poly(x^4 + x + 2, GF(3))

Determine if composite polynomials are square-free over $GF (3)$ .

In [4]: (f1 * f2).is_square_free()
Out[4]: True

In [5]: (f1**2 * f2).is_square_free()
Out[5]: False