details in an answer about $\Bbb Q[x]/\left<(x+1)^2\right> \cong \Bbb Q\times\Bbb Q$

Question

I have some doubt in this answer:

Hint: In $\Bbb Q[x]/\left<(x+1)^2\right>$ there exists an element $a \neq 0$ for which $a^2 = 0$.

Can anybody explain /elaborate more about this answer?

Answer 1

He's remarking that in the factor ring, we have a non-trivial nilpotent element:

$$\overline w:=(x+1)+ \langle (x + 1)^2 \rangle\;,\;\;\overline w\neq 0\;,\;\;\text{but}\;\;\overline w^2=0$$

He then asks: is there such a (non-trivial nilpotent) element in $\;\Bbb Q\times \Bbb Q\;$ ? The answer is no, there is none...and thus both rings cannot be isomorphic.

Answer 2

I was remarking that in $\mathbb{Q}[x]/\langle (x+1)^2 \rangle$ there exists a nilpotent element; that is $a:=\pi(x+1)$ where $\pi$ is the projection. The element $a$ is not zero, since it is not in $\langle (x+1)^2 \rangle$. However, its square is. Thus it is nilpotent of degree $2$.

In $\mathbb{Q} \times \mathbb{Q}$ there does not exist such an element. Pick some element $(x,y) \in \mathbb{Q} \times \mathbb{Q}$, which is non-zero, meaning $x \neq 0$ or $y \neq 0$. Then square it, and see that the result is non-zero again.

Hence the two rings cannot be isomorphic.