Let $(F_2,+,\cdot)$ be the finite field with $2$ elements. We define the quotient ring $$R=F_{2}[X]/{\langle X^3\rangle}$$
Personal question: does $X^3$ have any roots in $R$? I would say no :/
How many elements does R have?
$R$ has $2^3=8$ elements since $F_2$ has two elements and $3$ is the degree of the polynomial $X^3$.
I am really confused about how to answer to the following questions, any suggestions? I would love to receive hints, I am not asking for the solution. Thanks.
- Find a zerodivisor in the ring $(R, +, ·)$.
- Find the multiplicative inverse of the element $X + 1 + \langle X^3\rangle$
- How many elements does the set of units in R contain?
Everything in the ideal generated by $X$ is a root of $f(x)=x^3$. These elements are: $0, X, X^2, X+X^2$.
Well, $X^3=0$ already, and neither $X$ nor $X^2$ are zero, so we already have.
Handy to know: if $X^n=0$, then $X+1$ is a unit with inverse $1-X+X^2-X^3+...+(-1)^{n-1}X^{n-1}$. So, $(X+1)(1-X+X^2)=1+x^3=1$. In this way, $1+N$ is invertible for any nilpotent element $N$.
The ring is an Artinian local ring, so everything is either a unit or a zero divisor. We've already established that everything in $X+\langle X^3\rangle$ is nilpotent, and that $1+$ all of these elements is invertible. How many elements does that account for?