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Let $J=(x-2,x-y^2-3)$ ideal in the polynomial ring $\Bbb R[x,y]$.
Please help me prove that $J$ is a maximal ideal which isn't principal, and that $\Bbb R[x,y]/J \cong \Bbb C$.

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  • $\begingroup$ you tried something? $\endgroup$ – user87543 Jun 28 '14 at 15:53
  • $\begingroup$ I tried to find an extension for $\Bbb R[x] \rightarrow \Bbb R$ ($f(x)=f(2))$ to the homomorphism $\Bbb R[x,y] \rightarrow \Bbb R[y]$ but I couldn't find how. $\endgroup$ – Ran Kashtan Jun 28 '14 at 16:50
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Note that $$ J=(x-2,x-y^2-3) = (x-2, y^{2} + 1). $$ In fact $y^{2} + 1 = x - 2 - (x - y^{2} -3)$, and $x - y^{2} - 3 = x - 2 - (y^{2} + 1)$.

Now consider the homomorphism $$ \mathbb{R}[x, y] \to \mathbb{C}, \qquad x \mapsto 2, y \mapsto i. $$

As to $J$ being non-principal, what are the elements $z \in \mathbb{R}[x, y]$ which divide both $x-2$ and $y^{2} + 1$? (Just look at the degrees in $x$ and $y$.)

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Hint :

See $\mathbb{R}[x,y]/(x-2,x-y^2-3)$ as $\mathbb{R}[x,y]/(x-2)$ and then as resultant mod$(x-y^2-3)$

It should be clear that :

  • $\mathbb{R}[x,y]/(x-2)$ is in some sense same as $\mathbb{R}[y]$

Spend some time and see that $\mathbb{R}[x,y]/(x-2,x-y^2-3)$ is in some sense same as $\mathbb{R}[y]/(2-y^2-3)$ which is same as $\mathbb{R}[i]\cong \mathbb{C}$

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  • $\begingroup$ Thank you for the hint, but why $\mathbb{R}[x,y]/(x-2)$ is in some sense same as $\mathbb{R}[y]$? $\endgroup$ – Ran Kashtan Jun 28 '14 at 16:45
  • $\begingroup$ that is a hint.. you should think for a while i guess... $\endgroup$ – user87543 Jun 28 '14 at 16:45

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