0
$\begingroup$

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!

Thanks!

$\endgroup$
2
$\begingroup$

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 $f(g(t)=f([x])=t$.

$g(f([x]))=g(t)=[x]$,

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

$g(f([z]))=g(1)=1=[z]$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.