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Let $\varphi:\Bbb C[x,y]\to\Bbb C[t]$ be homomorphism that sends $x\to t$ and $y\to t^2$. This is surjective map with kernel $(y-x^2)$. The correspondence theorem relates ideals $I$ of $\Bbb C[x,y]$ that contain $y-x^2$ to ideals $J$ of $\Bbb C[t]$ by $J=\varphi(I)$ and $I=\varphi^{-1}(J)$. Since $J$ will be principal ideal $(p(x))$, each ideal $I$ of $\Bbb C[x,y]$ containing $(y-x^2)$ will be $(y-x^2,p(x))$.

Now let's consider $\Phi:\Bbb R[x,y]\to\Bbb R[t]$ be the homomorphism that is the identity on real numbers and sends $x\to t^2, y\to t^3$. We see that the kernel is $(y^2-x^3)$. Can we have some similar info abut kernels of $R[x,y]$ containing $(y^2-x^3)$? (In the aforementioned example, we had the luxury of $x\to t$.)

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    $\begingroup$ The real problem is that $\Phi$ is not onto all of $\mathbb R[t].$ So the ideals containing $(y^2-x^3)$ correspond to ideals of a subring of $\mathbb R[t]$ consisting of polynomials with no linear term. $\endgroup$ – Thomas Andrews Apr 30 at 15:56
  • $\begingroup$ @ThomasAndrews, Thank you very much $\endgroup$ – Silent Apr 30 at 17:41
  • $\begingroup$ Can you please let me know why not surjective? @ThomasAndrews $\endgroup$ – Silent Apr 30 at 17:49
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    $\begingroup$ The image of $\Phi: \mathbb C[x,y]\to \mathbb C[t]$ does not include $t.$ More generally, if $p(t)=a_0+a_1t+a_2t^2+\cdots + a_nt^n$ then $p(t)\in\operatorname{Im}(\Phi)$ if and only if $a_1=0.$ @Silent $\endgroup$ – Thomas Andrews Apr 30 at 18:07

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