$F[x]/\langle x^2\rangle$ is not an integral domain

Question

The ring $F[x]/\langle x^2\rangle$ for an infinite field $F$ is an infinite commutative ring with identity which isn't a domain.

I'm still stuck in understanding why is it not a integral domain.i.e. which element it contains are non-zero zero divisors..Please help ...

Answer 1

Hint: For $p\in F[x]$ denote $\bar{p}=p+\langle x^{2}\rangle$. What is $\bar{x}\cdot\bar{x}\,?$

Answer 2

Hint:

$$\left(x+\left\langle x^2\right\rangle\right)^2=\overline 0$$

Answer 3

Similar to $\mathbb{Z}/\langle 7^2 \rangle$.