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Are the subrings $\mathbb{C}\lbrack t^2,t^3\rbrack$ and $\mathbb{C}\lbrack t^3+3t^2,t^2+2t\rbrack$ of $\mathbb{C}\lbrack t\rbrack$ isomorphic? I do not think that they are, but I was not able to prove it.

My attempt was focused on comparing their integral closures, but I was not quite able to derive the integral closure of $\mathbb{C}\lbrack t^3+3t^2,t^2+2t\rbrack$.

TheoPatr
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  • Is $f\mapsto f+f'$ (essentially) the only isomorphism between them as complex vector spaces? Because I think a ring isomorphism must preserve the vector space structure, and that's clearly not a ring homomorphism. – Arthur Mar 07 '18 at 09:47
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    The integral closure of the second ring is pretty clearly $\mathbb C[t]$, since $t$ is integral over it (satisfies $x^2+2x-(t^2+2t)$). If you plot the second one you'll see that the lone singular point has two branches, whereas in the first one it only has one -- you'd need to do a change of coordinates to bring the singular point of the first one to the origin, but then one sees that the two rings are not even isomorphic mod the square of the maximal ideal at the origin -- one has nilpotents and the other does not. – John Brevik Mar 07 '18 at 09:55
  • I see, my attempt was bound to fail. Isn't there a simpler of showing that they are not isomorphic? I'm not supposed to us any algebraic geometry to solve this question, or, at least, not this much. – TheoPatr Mar 08 '18 at 05:54
  • The first ring is isomorphic to $\mathbb C[X,Y]/(X^2-Y^3)$ while the second is isomorphic to $\mathbb C[X,Y]/(X^2-4X-Y^3+3Y^2)$ which at its turn is isomorphic to $\mathbb C[X,Y]/(X^2-Y^3-3Y^2)$ (by a change of variables $X\mapsto X+2$ and $Y\mapsto Y+2$). Note that both rings have only one maximal ideal where their localization isn't regular, that is, $(x,y)$. This means that $(x,y)$ must correspond to $(x,y)$ is an isomorphism, and then the completions are also isomorphic. But the first completion is an integral domain while the second is not. – user26857 Mar 09 '18 at 19:46

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