0
$\begingroup$

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 $\Bbb R[x,y]/J \cong \Bbb C$.

$\endgroup$
2
  • $\begingroup$ you tried something? $\endgroup$
    – user87543
    Jun 28, 2014 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. $\endgroup$ Jun 28, 2014 at 16:50

2 Answers 2

3
$\begingroup$

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 $y$.)

$\endgroup$
2
$\begingroup$

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 $\mathbb{R}[i]\cong \mathbb{C}$

$\endgroup$
2
  • $\begingroup$ Thank you for the hint, but why $\mathbb{R}[x,y]/(x-2)$ is in some sense same as $\mathbb{R}[y]$? $\endgroup$ Jun 28, 2014 at 16:45
  • $\begingroup$ that is a hint.. you should think for a while i guess... $\endgroup$
    – user87543
    Jun 28, 2014 at 16:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.