Let $F = Z_5[x]/(x^3-x^2-1) = Z_5(u)$ where $u = [x]$ Give a basis for $F$ over $Z_5$

Question

How can you have $Z_5([x])$ when x is not any particular element?

Answer 1

Consider the polynomial $f=x^3-x^2-1\in \Bbb{F}_5[x]$. Checking each element in $\Bbb F_5$, we see $f$ has no roots, and is thus irreducible. To adjoin an element $u$ to $\Bbb{F}_5$, such that $\Bbb{F}_5(u)$ has a root for $x^3-x^2-1$ in $\Bbb{F}_5(u)[x]$ we simply take $$\Bbb{F}_5[x]/(x^3-x^2-1),$$ where the coset $[x]$ is the root we desired. We then denote $[x]:=u$. Now $\Bbb{F}_5(u)$ is a field extension of $\Bbb F_5$, and we wish to know the degree of this extension (dimension of $\Bbb F_5(u)$ as an $\Bbb F_5$-vector space). Consider an element $b= a_0+a_1u+a_2u^2+a_3u^3$, you can use the relation coming from the quotient (meaning $u^3=u^2+1$) to express $b$ as $$b=(a_0+a_3)+a_1u+(a_2+a_3)u^2,$$ meaning any such element can be written in terms of the $\Bbb F_5$-basis $\{1,u,u^2\}$. Can you see how to finish the argument (taking an arbitrary element of $\Bbb F_5(u)$)?

In particular $[x]$ is a particular element of the quotient ring!