# Finite polynomial rings and degree of polynomial

Suppose we have a ring of polynomials over a field $\mathbb{Z}_5 = {0,1,2,3,4}$. Both addition and multiplication are subject to modular arithmetic.

If we want to calculate $P\odot Q$ with $P=-x ^3$ and $Q=3x^2$, do I need to make sure that both coefficients and exponents shouldn't be grater than 4 (i.e. the result would be 2 because $x^5$ is congruent to $x^0$ and $-3 \mod 5 = 2$), or is it ok if I say that the result is $2x^5$ (i.e exponents don't need to be from $\mathbb{Z}_5$).

• Note that $2^5$ is not congruent to $2^0$ modulo $5$. If you treat exponents as "regular" numbers, you will be in a lot of trouble. – Lord Soth Nov 6 '13 at 0:11
• Does "circle-dot" between $P$ and $Q$ denote multiplication of polynomials? – hardmath Nov 6 '13 at 0:12
• You can do as you wish unless required otherwise. In your example, $$-x^3\cdot 3x62=-3x^5=2x^5\pmod 5$$ and there's no a priori way to decide which one is better. And, of course, do not touch the variable's exponents: those must be natural numbers and zero by definition and we work with them under the usual exponents' laws. – DonAntonio Nov 6 '13 at 0:16
• The "circle-dot" is multiplication modulo 5. I am wondering because, I've read something about irreducible polynomials and thought it might apply here. – random guy Nov 6 '13 at 0:31

Be careful, only the coefficients have operations subject to modular arithmetic, $x^nx^m = x^{n+m}$, no mod 5 reduction! The indeterminate $x$ is free, meaning it satisfies no algebraic relation (for example $x^5 \ne x^0$, as pointed out in the comments). What can happen is that $x^n = x^m$ for $n\neq m$ when you evaluate $x$ at all elements of $\Bbb Z_5$, for example, every $a \in \Bbb Z_5$ satisfies $a^5 = a$, in this case, when viewed as functions $\Bbb Z_5 \to \Bbb Z_5$, the polynomials $x^5$ and $x$ are equal, but as polynomials they are different. This is important, for example, once in the future we will extend the ring $\Bbb Z_5$ to a bigger ring where the relation $a^5 = a$ will not hold always.