Sympy symbol and the cls parameter

Question

I have seen f = sympy.symbols('f', cls=Function) but not any documentation. Python does not like x = sympy.symbols('x', cls=FF(8)), it complains about

raise CoercionFailed("expected an integer, got %s" % a) CoercionFailed: expected an integer, got x

Whan is the purpose of the cls parameters and what must I do so that cls=FF(8) is meaning full?

With x = sympy.symbols('x', cls=FF(8)) I want x to be a symbol in the field FF(8), i.e x^(2^8-1) must give me 1.

Answer 1

There are a few issues here:

• The FF object does not allow Symbols. It only works for exact numerical entries, like FF(3)(2).

• Therefore, the cls parameter of symbols will not work. That just changes what object is used to create the symbol, so it must take a string as an input (the default is Symbol).

• SymPy does not currently support Symbols over finite fields. The best bet you can get is to use the Poly object with the modulus flag.

• FF currently only supports finite fields of prime cardinality. FF(8) has actually created the ring Z_8, not the finite field with 8 elements.

• You probably know this, but ^ does not do exponentiation in SymPy/Python. Use **.