Marra

Brazil

Oct
26
comment Computing these multiplicities
I meant $y\rightarrow x^3$, sorry for that. I'll start another question after I work a little more on this one, thanks :)
Oct
26
comment Computing these multiplicities
It actually could, the result would be the same...
Oct
26
comment Computing these multiplicities
I just don't see why you are choosing this particular ring isomorphism (did you mean surjective ring morphism?). Is it because of the first ideal $(y-x^2)$? Then could it be, and would it work, if the morphism were $x\rightarrow x^3$?
Oct
26
comment Computing these multiplicities
That's probably the reason why I'll never get it.
Oct
26
comment Computing these multiplicities
Why is $(\mathbb C[x,y]/(y-x^2,y-x^3))_{(x,y)}=( \mathbb C[x]/(x^2-x^3))_{(x)}$? I'm guessing that the ideals $(y-x ^2,y-x^3)$ and $(y,x^2-x^3)$ are equal?
Oct
26
accepted Computing these multiplicities
Oct
26
revised Computing these multiplicities
added 75 characters in body
Oct
26
comment Computing these multiplicities
Damn, that's right, I was thinking about something else. Lemme correct that.
Oct
26
asked Computing these multiplicities
Oct
24
asked Why aren't those Cartier Divisors equivalent?
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awarded Nice Answer
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