Why does $\Bbb{F_2}[X]/(X^2)$ have characteristic $0$?

84 Views Asked by At

I am trying to understand dual numbers. From Wikipedia:

In abstract algebra terms, the dual numbers can be described as the quotient of the polynomial ring $R[X]$ by the ideal generated by the polynomial $X^2$,

$$R[X]/(X^2)$$

The image of $X$ in the quotient is the unit $ε$. With this description, it is clear that the dual numbers form a commutative ring with characteristic $0$.

So I tried to make a simple example with $L=\Bbb{F_2}[X]/(X^2)$, which is $\{0,1,X,X+1\}$ (or is it $\{0,1\}$, since from $X^2=0 \in L$ it would follow that $X=0$?).

Now it says that the characteristic of this ring (I think it is even a field in both cases) is $0$, but $1+1=0$ in both cases. Did I misunderstand something?