TITLE: Product polynomial in $\mathbb{F}_7$
QUESTION [6 upvotes]: I need to compute the product polynomial $$(x^3+3x^2+3x+1)(x^4+4x^3+6x^2+4x+1)$$
when the coefficients are regarded as elements of the field $\mathbb{F}_7$.
I just want someone to explain to me what does it mean when a cofficient (let us take 3 for example) is in $\mathbb{F}_7$ ? I know that $\mathbb{F}_7= \mathbb{Z}/7\mathbb{Z}$

REPLY [3 votes]: Well, you can multiply two polynomials in the usual way first, then you look at the coefficients and reduce them to modulo 7. For example:
$(3x^2+x+1)(x^2+3x+6) = 3x^2(x^2+3x+6)+x(x^2+3x+6)+1.(x^2+3x+6)=(3x^4+9x^3+18x^2)+(x^3+3x^2+6x)+(x^2+3x+6)=3x^4+10x^3+22x^2+9x+6$ which is the same as $3x^4+3x^3+x^2+2x-1$ in $\mathbb{Z}_7$ because $10 \equiv 3 \pmod{7}$, $22 \equiv 1 \pmod{7}$, $9 \equiv 2 \pmod{7}$ and $6 \equiv -1 \pmod{7}$.