# 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 ??

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 $$\Bbb Q\times\Bbb Q$$?
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, $$\Bbb Q[x]/\left\langle (x+1)^2\right\rangle \not\cong \Bbb Q\times\Bbb Q$$. .
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.