Yes/no : Is $\Bbb Q[x]/\left<(x+1)^2\right> \cong \Bbb Q\times\Bbb Q$.? 
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 ??
 A: 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 $\Bbb Q\times\Bbb Q$?
A: If $m$ is a maximal ideal of $\Bbb Q[x]/\left\langle (x+1)^2\right\rangle $, then $m=M/\left\langle (x+1)^2\right\rangle $, where $M$ is a maximal ideal of  $\Bbb Q[x]$ contaning $(x+1)^2$. So $M $ contans $x+1$, and since  $\left\langle x+1\right\rangle $ is a maximal ideal of $\Bbb Q[x]$, we have $m=\left\langle x+1\right\rangle /\left\langle (x+1)^2\right\rangle$. Thus, the ring $\Bbb Q[x]/\left\langle (x+1)^2\right\rangle $ is local, but $\Bbb Q\times\Bbb Q$ has two maximal ideals $\Bbb Q\times 0$ and $0\times\Bbb Q$. Hence, $\Bbb Q[x]/\left\langle (x+1)^2\right\rangle 
\not\cong \Bbb Q\times\Bbb Q$. .
A: Hint: The decomposition ${\Bbb Q}[x,y]/\langle f,g\rangle $ into the direct product ${\Bbb Q}[x,y]/\langle f\rangle \times {\Bbb Q}[x,y]/\langle g\rangle $ certainly works if $f,q$ are relatively prime. But this is not the case in your problem.
