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Let $k$ be a field, it's claimed in algebraic geometry textbook that $k[v^2, v^3]\cong k[t, u]/(t^2-u^3)$ via $v^2\mapsto u, v^3\mapsto t$. But I can't show it's well-defined, since an integer can have many ways to write as $2i+3j$. Any idea?

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  • $\begingroup$ Yes, for instance $20=2\times1+3\times 6$ and $20=2\times7+3\times 2$. So $v^{20}$ should go to $ut^6$ and $u^7t^2$? Luckily that's not a problem, as $ut^6-u^7t^2$ is a multiple of $t^2-u^3$. $\endgroup$ Jun 26, 2020 at 14:41
  • $\begingroup$ OK, I think I know how to do it. $\endgroup$
    – Lao-tzu
    Jun 26, 2020 at 15:03

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Instead you should try to show that $\phi : k[t,u] \longrightarrow k[v^2,v^3]$ has kernel generated by $(t^2-u^3).$ This map will be surjective, and well defined.The first isomorphism theorem will give you the desired result.

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    $\begingroup$ Indeed I did this, but similar problem arise. It's not easy to show the kernel is of that form. $\endgroup$
    – Lao-tzu
    Jun 26, 2020 at 14:40
  • $\begingroup$ Write out some of your work on the kernel in the question block then, and let's see if we can get to the root of the problem. $\endgroup$ Jun 26, 2020 at 14:40
  • $\begingroup$ I can only show the coefficients of the beginning terms vanish. I want to do induction on 2i+3j on t^iu^j, but seems massy. $\endgroup$
    – Lao-tzu
    Jun 26, 2020 at 14:42
  • $\begingroup$ You don't need to worry about using induction. What you should take $\phi$ as described, and ask yourself if you had a polynomial $\phi(p(t,u))=0,$ that is $p(v^2,v^3)=0,$ then what could you say about the kernel of this map? Angina Seng gave you the way to see the answer in their comment on your original post. $\endgroup$ Jun 26, 2020 at 14:47
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    $\begingroup$ Why do you always say "what could you say..."? That's where I was stuck. If you have an answer, please write it inside above. $\endgroup$
    – Lao-tzu
    Jun 26, 2020 at 14:55

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