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According to this question, we have: $B:= \mathbb{C}[x^2,x^3][T] / \langle T^2-x^2, x^2T-x^3, x^3T-x^4 \rangle = \mathbb{C}[x]$

Consider $A:= \mathbb{C}[x^2,x^3][T]/\langle x^2T-x^3 \rangle$.

Clearly, $A \to B=\mathbb{C}[x]$ is onto.

If I am not wrong:

(i) $A/I=B$, where $I=\frac{\langle T^2-x^2, x^2T-x^3, x^3T-x^4 \rangle}{\langle x^2T-x^3 \rangle}$.

(ii) $B$ is an integral domain, so $I$ is a prime ideal.

Question (1) Is it possible to somehow characterize that $A$, not with quotients? Is such $A$ contained in $\mathbb{C}(x)$ -- probably not? Is such $A$ contained in some algebraic ring extension of $\mathbb{C}[x]$?

Question (2) Same question with $A:= \mathbb{C}[x^2,x^3][T]/\langle x^2T-x^3 \rangle$ replaced by $A:= \mathbb{C}[x^2,x^3][T]/\langle T^2-x^2 \rangle$.

Thank you very much!

user26857
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user237522
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  • Since $A$ has dimension one and not equal to $B$, it is not a domain, so it can not be contained in any field. – Mohan Aug 31 '20 at 16:17
  • @Mohan, thank you very much for your comment. Nice idea. Could you please give an example of a (non-trivial) zero divisor in $A$? – user237522 Aug 31 '20 at 17:48
  • $x^2(x^3T-x^4)=x^3(x^2T-x^3)=0$, so both $x^2, x^3T-x^4$ are zero divisors. – Mohan Aug 31 '20 at 19:14
  • Thank you! Please, what about zero-divisors in the 'second' algebra $\mathbb{C}[x^2,x^3][T]/\langle T^2-x^2 \rangle$? – user237522 Aug 31 '20 at 21:07
  • $(x^2T-x^3)(x^2T+x^3)=x^4T^2-x^6=x^4(T^2-x^2)$ – Mohan Aug 31 '20 at 22:49
  • Very nice, thank you! You have shown that $T^2-x^2$ is not prime in $\mathbb{C}[x^2,x^3][T]$, so $\mathbb{C}[x^2,x^3][T]/\langle T^2-x^2 \rangle$ is not a domain. (Also, $x^2T-x^3$ is not prime in $\mathbb{C}[x^2,x^3][T]$, so $\mathbb{C}[x^2,x^3][T]/\langle x^2T-x^3 \rangle$ is not a domain). – user237522 Sep 01 '20 at 12:48
  • What if $A$ would have been a domain? Is there a way to characterize $A$ more specifically? I do not have a one dimensional example, but a two-dimensional one: $\mathbb{C}[y,xy][T]/\langle yT-xy \rangle$. This algebra contains $x$, so it contains $\mathbb{C}[x,y]$, but I guess it is bigger than $\mathbb{C}[x,y]$? (it seems that this depends on whether $\mathbb{C}[x,xy]$ is a UFD or not). – user237522 Sep 01 '20 at 13:19
  • Is it true that $\mathbb{C}[y,xy]/\langle yT-xy \rangle$ is isomorphic to $k[x,y]$? – user237522 Sep 01 '20 at 19:35
  • Yes. $C[y,xy]\cong C[y,u]$, so your ring is just $C[y,u,T]/(yT-u)=C[y,T]$. – Mohan Sep 01 '20 at 19:59
  • Thank you. Should this work ($A$ is isomorphic to $B$) whenever $A$ is a UFD? BTW, I am trying to ask a similar question here: https://math.stackexchange.com/questions/3810982/k-subsetneq-r-subseteq-kx-y-a-ufd-implies-that-ry-xy-a-ufd – user237522 Sep 01 '20 at 20:04
  • Where: $A=D/\langle vT-u \rangle$, $D$ is an integral domain having field of fractions $Q(D)$, $u,v \in D=k$, $D[u/v]$ is isomorphic to $B$. In this situation $A$ is isomorphic to $B$. – user237522 Sep 01 '20 at 20:13

1 Answers1

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Since $\mathbb{C}[x^2,x^3]\simeq\mathbb C[U,V]/\langle U^2-V^3 \rangle$ ($U$ corresponds to $x^3$ and $V$ to $x^2$) we have

$$\mathbb{C}[x^2,x^3][T]/\langle T^2-x^2 \rangle\simeq\mathbb C[U,V,T]/\langle U^2-V^3,T^2-V \rangle\simeq\mathbb C[U,T]/\langle U^2-T^6\rangle.$$

user26857
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