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We can find everywhere that $\phi :\mathbb A^1_K \to V(y^2-x^3)$ sending $t$ to $(t^2,t^3)$ defines a bijective map but not a morphism of affine varieties. To me it's not clear why it's even surjective. Here
Parametrization of the cuspidal cubic we say that we can parametrize the curve by considering a line passing through 0, and it's the argument I saw the most. I'm not used to parametrizations and I'm not sure to understand why this works, it looks arbitrary to me. How can we be sure that if we have $(u,v)\in V(y^2-x^3)$, then it's necessarily of this form ?

Also just to be sure, the reason of $K[x^2,x^3]\subsetneq K[x]$ (and so it's not an isomorphism of affine varieties) is because of the degree of elements in $K[x^2,x^3]$, which is either 0 or bigger than 2 is it correct ?

Thank you

raisinsec
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"Each rational line $y=kx$ (where $k\neq0$) meets the variety (apart from the origin) at exactly one rational point. This shows that $\phi$ is an injective function.

Conversely, each rational point on the variety (leaving aside the origin) lies on a rational line $y=kx$ (where $k\neq0$). This shows that $\phi$ is a surjective function."

I'll let you formalise these statements yourself.

$\\$

$\phi$, as you defined it in the question, is indeed a morphism of affine varieties because $t^2,t^3\in K[t]$.

But for $K$ algebraically closed, $\phi$ is not an isomorphism of affine varieties, because indeed $t\notin K[t^2,t^3]$.

Chris Sanders
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  • Indeed it's clear now why $\phi$ is bijective. Yes it's a morphism of affine varieties, but why is it an isomorphism for $K=\mathbb F_2$ for example ? Does $P(x,y)=2x+y$ work since $P(t^2,t^3)=t$ ? – raisinsec Mar 20 '22 at 11:48
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    yes, your example does work for $\mathbb{F}_2$ – Chris Sanders Mar 20 '22 at 11:58
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    The first half of this answer is good and correct, but $\phi$ is never an isomorphism no matter what field you pick. If it were, this would imply that $k[x,y]_{(x,y)}/(y^2-x^3)$ is regular, which it is not. Alternatively, it would imply $k[t^2,t^3]$ integrally closed, which it clearly is not. The key error is that the formula you've presented for $\phi^{-1}$ is not a regular function because the second formula is not defined on an open set. – KReiser Mar 20 '22 at 18:02
  • sorry, in the case $K=\mathbb{F}_2$, is it not enough for $\phi^{-1}(x,y)$ to agree with a polynomial $p(x,y)$ on the points $(0,0)$ and $(1,1)$? – Chris Sanders Mar 21 '22 at 01:05
  • I thought that for a finite field $K$, since each point on the affine space $K^n$ is a variety, every subset of $K^n$ is open. @KReiser :) – Chris Sanders Mar 21 '22 at 01:12
  • @ChrisSanders Thinking about varieties over non-algebraically-closed fields that way is really just not the right thing to do. It totally breaks the equivalence between affine varieties and their coordinate algebras. (Your case-by-case definition in your post doesn't match that last link you're attempting to cite - what's your experience level with algebraic geometry?) – KReiser Mar 21 '22 at 01:24
  • @KReiser admittedly not much experience, so I'll edit my answer then assuming $K$ is algebraically closed – Chris Sanders Mar 21 '22 at 01:27