# Determining ideals by correspondence theorem for $(y^2-x^3)$

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 $$x\to t$$.)

• 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. – Thomas Andrews Apr 30 at 15:56
• @ThomasAndrews, Thank you very much – Silent Apr 30 at 17:41
• Can you please let me know why not surjective? @ThomasAndrews – Silent Apr 30 at 17:49
• 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 – Thomas Andrews Apr 30 at 18:07