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Consider the integral domain $R:=\mathbb{C}[x,y]/(y^2-x^3)$. Over its field of fraction let $t:=y/x\in R_{frac}$. I want to show $\mathbb{C}[t] \cong R[t]$. How can I show this? and what is geometric meaning over this? I think they are isomorphic because $\mathbb{C}[t]$ is a polynomial over the complex line which maps to a function defined on $V(y^2-x^3)\subset \mathbb{C}^2$. Is this right interpretation? Thank you.

Jiya
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    Your titular question and the question in the body of the post are somewhat different. It souns like you're asking specifically why $\mathbb{C}[t]\cong R[t]$ which essentially is asking, secretly, why the normalization of the cuspidal cubic is $\mathbb{A}^1$. Is this what you're interested in, or are you actually interested in the general geometric intrution for integral extensions? If it's the former, draw it to see that you get a 'horned sphere' and then its normalization, its desingularization, is clearly a sphere. For the latter they're just finite maps in all reasonable situations. – Alex Youcis Oct 05 '20 at 23:59
  • @AlexYoucis I'm sorry I just thought the isomorphism has relation with that t=y/x is integral over R : t^2-x =0. – Jiya Oct 06 '20 at 00:49
  • Welcome to MSE. I agree with Alex Youcis that the title could match the post better. The title is an important part of asking a question here - see Meta for discussion of what a good title is. It's not unreasonable to interpret your title as asking for a geometric characterization of all integral ring extensions but that's not what you're asking about. I suggest editing your title to make it easier to tell what your question is from just reading the title. – KReiser Oct 06 '20 at 01:17
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    Related – Viktor Vaughn Oct 06 '20 at 02:32

1 Answers1

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We can start building the isomorphism $R[t] \to \mathbb{C}[t]$ as follows:

$$\mathbb{C}[x, y] \to \mathbb{C}[t],\quad x\mapsto t^2, y \mapsto t^3 $$

This sends $y^2-x^3$ to $t^6-t^6=0$, and so we have a map $R \to \mathbb{C}[t]$ and this map is injective becuase the kernel is exactly $(y^2-x^3)$. (here's why: writing elements of $R$ as having a representative $p(x) + q(x)y$ (since we can replace $y^2$ by $x^3$) we see that $p(x) + q(x)y$ maps to $p(t^2) + q(t^2)t^3$ and if this is $0$ then by looking at the parity (even/oddness) of powers of $t$ we see that $p(x)=0$ and $q(x)=0$. Then adjoining $y/x$ we get $R(y/x) \to \mathbb{C}[t]$ makes the map surjective, but we have to check we didn't lose injectivity. Writing any element of $R(y/x)$ as represented by $p(x) + q(x)x (y/x)$ the same argument as before works.

The geometric meaning of this is blowing up the cuspidal cubic at the origin (with ideal $(x,y)$) gives you something which on one chart is isomorphic to $\mathbb{A}^1$ (and on the other chart where you adjoin $x/y$ is also isom to $\mathbb{A}^1$, together they glue to a $\mathbb{P}^1$.

usr0192
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