# yes/no : Is $\mathbb{Z}[x,y]/((x+1, y+1) \cong \mathbb{Z}$?

Is $$\mathbb{Z}[x,y]/((x+1, y+1) \cong \mathbb{Z}$$ ?

# My attempt : i think yes

consider the map $$\phi : \mathbb{Z}[x,y] \to \mathbb{Z}$$ defined by $$\phi(f(x,y)) = f(-1,-1)$$. $$\phi$$ is a ring homomorphism with $$\ker(\phi) = \{ f(x,y) \in \mathbb{Z}[x,y] : f(-1,-1) = 0 \}$$. We will show that the kernel is the principal ideal $$(x+1,y+1)$$. This will imply, from the first isomorphism theorem, that $$\operatorname{im}(\phi) \cong \mathbb{Z}[x,y]/((x+1, y+1)$$, which gives an explicit description of the quotient.

Is its true ?

• You are definitely on the right track. You also need to show $\varphi$ is surjective, so that $\operatorname{im}\varphi = \mathbb Z$. By the way, the ideal $(x+1,y+1)$ is not principal - it is not generated by one element. – lisyarus Aug 28 '19 at 10:53
• okss @lisyarus.. – jasmine Aug 28 '19 at 10:55

The factor ring $${\Bbb Z}[x,y]/\langle x+1,y+1\rangle$$ has the monomial standard basis $$\{1\}$$ and so is isomorphic to $$\Bbb Z$$ as a $$\Bbb Z$$-module, but also a ring.