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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 ?

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    $\begingroup$ 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. $\endgroup$ – lisyarus Aug 28 '19 at 10:53
  • $\begingroup$ okss @lisyarus.. $\endgroup$ – jasmine Aug 28 '19 at 10:55
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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.

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