I am trying to convince myself of an isomorphism between:

$$k[x,y,z]/(x^2-yz,z-1) \rightarrow k[t]$$

In trying to show that these rings are isomorphic, I have constructed a map sending: $x \rightarrow t, y \rightarrow t^2, z \rightarrow 1$. Now this map is clearly surjective, and I'm pretty certain that the kernel of this map is indeed (the ideal): $(x^2-yz,z-1)$, however I'm not entirely certain how I would prove that it is...Is this the best way to show that these rings are isomorphic?

Any help would be greatly appreciated!



The morphism $x\rightarrow t, y\rightarrow t^2$, $z\rightarrow 1$ induces a morphism

You have $f:k[x,y,z)/(x^2-yz,z-1)\rightarrow k[t]$ defined by $f([x])=t, f([y])=t^2, f([z])=1$ where $[x]$ is the class of $x$

Consider $g:k[t]\rightarrow k[x,y,z]/(x^2-yz,z-1)$ defined by $g(t)=[x]$, you have $f(g(t)=f([x])=t$.


$g(f([y]))=g(t^2)=[x^2]=[y]$ since $[x^2]=[y][z]$ and $[z]=1$,


  • $\begingroup$ Ahh right, can I just clarify for my understanding, so if we can show that a surjective map has a well defined inverse, then this shows that these rings are isomorphic? $\endgroup$ – Kendrick Easley May 29 '16 at 12:55
  • $\begingroup$ $f$ and $g$ verify $f\circ g=Id$ and $g\circ f=Id$ thus they are isomorphisms. $\endgroup$ – Tsemo Aristide May 29 '16 at 12:56
  • $\begingroup$ Ahh right! Thanks so much! $\endgroup$ – Kendrick Easley May 29 '16 at 12:57

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