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I have the following problem:

Let $R:=\mathbb{R}[X,Y]/(Y-X^2,Y+X)$

i) Show that $R\cong_\text{Ring} \mathbb{R}\times \mathbb{R}$

ii) Conclude that $|\operatorname{Spec}R|=2$. How do these two ideals look like?

So for the first one I got the hint to eliminate one variable at a time and see what happens. Doing that I got:

$1.) \mathbb{R}[X]/(-X^2,X)\cong \mathbb{R}[X]/(-X^2)\times\mathbb{R}[X]/(X)\cong\mathbb{R}[X](-X^2)\times \mathbb{R}$

$2.) \mathbb{R}[Y]/(Y,Y)=\mathbb{R}[Y]/(Y)\cong\mathbb{R}$

using the chinese remainder theorem in $1.)$. But since $\mathbb{R}[X]/(-X^2,X)\times\mathbb{R}[Y]/(Y)\ncong\mathbb{R}[X,Y]/(Y-X^2,Y+X)$ and the fact that there is still $\mathbb{R}[X]/(-X^2)$ left in the first equation, I don't see how this leads me to the desired isomorphism. For the second one I don't see how an isomorphism can lead me to the number of ideals, because I'm pretty sure $A\cong B$ doesn't imply $\operatorname{Spec}A \cong \operatorname{Spec}B$ for rings $A,B$ (I've seen it in a book).

Can someone help me with this problem? Thanks in advance.

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    $\begingroup$ I dont see how $A\cong B$ doesn't imply $Spec A\cong Spec B$ can possibly be, unless the first isomorphism is less than a ring isomorphism. $\endgroup$ – Rene Schipperus May 10 '18 at 18:31
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    $\begingroup$ The Chinese Remainder Theorem $R/(I\cap J)\cong R/I\oplus R/J$ only applies when $I+J= R$. Here, you ahve $I=(X)$ and $J=(X^2)$, so you do not have $I+J=R$. $\endgroup$ – Mike Earnest May 10 '18 at 18:53
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The hint to 'eliminate one variable at a time' suggests to replace $Y$ by $-X$ as modulo $\langle Y - X^2, Y + X\rangle$ they are equal (or $Y$ by $X^2$ or $X$ by $-Y$). So

$$R = {\mathbb R}[X,Y]/\langle Y-X^2,Y+X\rangle \cong {\mathbb R}[X]/\langle -X - X^2\rangle = {\mathbb R}[X]/\langle X(X+1)\rangle.$$

Now you can apply the Chinese Remainder Theorem to get

$$R \cong {\mathbb R}[X]/\langle X\rangle \times {\mathbb R}[X]/\langle X+1\rangle \cong {\mathbb R} \times {\mathbb R}.$$

The prime ideals of the final ring ${\mathbb R} \times {\mathbb R}$ are ${\mathbb R} \times \{0\}$ and $\{0\} \times {\mathbb R}$, so $| \text{Spec}(R) | = 2$. (Note that it is true that if $A \cong B$ as rings, then $\text{Spec}(A) \cong \text{Spec}(B)$).

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  • $\begingroup$ Thanks! The hint makes a lot more sense now. $\endgroup$ – Yasuduck May 21 '18 at 21:52
  • $\begingroup$ How did you use the Chinese remainder theorem? Do you show that X and X+1 are comaximal? (Are they comaximal because X+1 - X = 1?) $\endgroup$ – Sally G May 25 '19 at 6:25

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