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Is there any nice way to understand the following quotient ring: $$R=\frac{\mathbb{R}[x,y]}{(y+x^2)\cap(y-x^2-1)}$$ I think that $y+x^2$ and $y-x^2-1$ are irreducible but I'm not too familiar with polynomial rings. In particular, I am curious to see whether or not $R\cong \mathbb{R}[s]\times \mathbb{R}[t]$.

user26857
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J. Moeller
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  • Well, $y+x^2$ is a degree 1 polynomial (wrt $y$) over the integral domain $\mathbb{R}[x]$... – Daniel Schepler Feb 16 '18 at 22:57
  • I'm not sure where you are leading with the comment, @DanielSchepler. Would you mind saying a bit more? – J. Moeller Feb 16 '18 at 23:19
  • Daniel is indicating a way to verify your thoughts about irreducibility. – Gerry Myerson Feb 16 '18 at 23:41
  • So does it follow that $(y+x^2)\cap(y-x^2-1)=(y+x^2)(y-x^2-1)$? I think it does. However, I don't believe that $k[x,y]=(y+x^2)+(y-x^2-1)$. So, if the result is true that $R\cong \mathbb{R}[s]\times \mathbb{R}[t]$, it will not follow from CRT in the usual way. – J. Moeller Feb 16 '18 at 23:56
  • You think it does or you know it does? – Mariano Suárez-Álvarez Feb 18 '18 at 20:32
  • @Prototank $R$ has only trivial idempotents, and therefore can't be isomorphic to a direct product. – user26857 Feb 19 '18 at 07:37
  • @user26857 How do we know that $R$ has only trivial idempotents? – J. Moeller Feb 19 '18 at 13:57
  • @Prototank It seems you tried and failed. Set $p_1=y+x^2$ and $p_2=y-x^2-1$. If $f$ is idempotent modulo $(p_1)\cap(p_2)=(p_1p_2)$, then $p_1p_2\mid f^2-f$. (No idea from where you came up with $f^2+1$.) Since $p_i$ are irreducible (hence prime) we have $p_i\mid f$ or $p_i\mid f-1$. If $p_1\mid f$ and $p_2\mid f$ you are done. Similarly for $p_1\mid f-1$ and $p_2\mid f-1$. Now suppose $p_1\mid f$ and $p_2\mid f-1$. We get $f=p_1u$ and $f-1=p_2v$, and then $p_1u-p_2v=1$. Now try to find a contradiction. – user26857 Feb 19 '18 at 19:30
  • Your arguments have lead to "show that $(p_1)+(p_2)=(1)$ is impossible", which I agree would prove the result but more directly by CRT. – J. Moeller Feb 19 '18 at 20:16
  • @Prototank I have no idea how CRT can be involved in this. As far as I can remember CRT is about comaximal ideals, and says nothing about the case when the ideals are not comaximal. (All you can say is that you can't use CRT for $R$.) – user26857 Feb 20 '18 at 08:53
  • For a counterexample to the converse of CRT see https://math.stackexchange.com/questions/895428/can-r-times-r-be-isomorphic-to-r-as-rings – user26857 Feb 20 '18 at 09:10

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$R$ can't be isomorphic to a direct product of rings since it has only trivial idempotents.

Set $p_1=y+x^2$ and $p_2=y−x^2−1$. If $f$ is idempotent modulo $(p_1)∩(p_2)=(p_1p_2)$, then $p_1p_2∣f^2−f$. Since $p_i$ are irreducible (hence prime) we have $p_i∣f$ or $p_i∣f−1$. If $p_1∣f$ and $p_2∣f$ then $f=0$ in $R$. Similarly for $p_1∣f−1$ and $p_2∣f−1$. Now suppose $p_1∣f$ and $p_2∣f−1$. We get $f=p_1u$, $f−1=p_2v$, and therefore $p_1u−p_2v=1$, a contradiction. (Similarly for $p_1\mid f-1$ and $p_2\mid f$.)

user26857
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