Is $\Bbb Q[x]/\left\langle (x+1)^2\right\rangle
\cong \Bbb Q\times\Bbb Q$ ?
My attempt : i thinks yes , $\frac{\Bbb Q[x]}{\left\langle(x+1)^2\right\rangle }
= \frac{\Bbb Q[x]}{\left\langle (x+1)\right\rangle } \times
\frac{\Bbb Q[x]}{\left\langle(x+1)\right\rangle }
= \Bbb Q\times\Bbb Q$
Here $\frac{\Bbb Q[x]}{\left\langle(x+1)\right\rangle } \cong \mathbb{Q}$
consider the map $\phi : \mathbb{Q}[x] \to \mathbb{Q}$ defined by $\phi(f(x)) = f(-1)$. $\phi$ is a ring homomorphism with $\ker(\phi) = \{ f(x) \in \mathbb{Q}[x] : f(-1) = 0 \}$. We will show that the kernel is the principal ideal $(x+1)$. This will imply, from the first isomorphism theorem, that $\operatorname{im}(\phi) \cong \mathbb{Q}[x]/((x+1)$, which gives an explicit description of the quotient.
Is its true ??
Best Answer
Hint: In $\Bbb Q[x]/\left<(x+1)^2\right>$ there exists an element $a \neq 0$ for which $a^2 = 0$.
Does such an element exist in $\Bbb Q\times\Bbb Q$?