On page 76 of Reid's book Undergraduate Algebraic Geometry, he says that

Over an infinite field $k$, the polynomial map $\phi: k\rightarrow C: (y^2=x^3) \subset k^2$ given by $\phi(t)=(t^2,t^3)$ is not an isomorphism.

This is because the induced homomorphism $\phi^*: k[C]\rightarrow k[t]$ maps $k[C]$ to $k[t^2,t^3]$, where $t^2,t^3$ cannot generate $k[t]$, so $k[C]$ is not isomorphic to $k[t]$.

I checked that $\phi$ is bijective by the following. Let $\psi: C\rightarrow k$ be defined by $\psi(x,y)=\frac{y}{x}$. Then $$\phi\circ\psi(x,y)=\phi\left(\frac{y}{x}\right)=\left(\left(\frac{y}{x}\right)^2,\left(\frac{y}{x}\right)^3\right)=(x,y)\\ \psi\circ\phi(t)=\psi(t^2,t^3)=t$$ My doubts:

I think the argument should work over a finite field too. I cannot see why $t^2,t^3$ can generate $k[t]$ even in a finite field. But since the author says "over an infinite field" at the beginning of this example, I cannot be sure whether finite field is a different case.

Thank you for your help.

  • $\begingroup$ What do you mean by isomorphism ? Did you check what happens when $k=\mathbb F_2$ ? $\endgroup$
    – Bebop
    Jan 30 '16 at 10:34
  • $\begingroup$ @Bebop: Thank you for your reply! I checked the bijectivity and wouldn't it be the same in the case where $k$ is infinite? By the book's definition, the isomorphism of $k$ and $C$ is equivalent as the ring isomorphism of $k[C]$ and $k[t]$. $\endgroup$
    – KittyL
    Jan 30 '16 at 10:36
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    $\begingroup$ Can someone explain why the downvote so I can improve my question? $\endgroup$
    – KittyL
    Jan 30 '16 at 10:51
  • $\begingroup$ @Bebop: I misunderstood the definition of polynomial isomorphism. They have to be polynomials. So in $\mathbb{F}_2$, can we use $\phi(x,y)=x$, since $t^2=t$? $\endgroup$
    – KittyL
    Jan 30 '16 at 11:32

The reason he assumes the field is infinite is that over a finite field, it is not true that the coordinate ring of $k$ is $k[t]$. For the coordinate ring is defined as $k[t]$ modulo the ideal of polynomials that vanish on every point of $k$, and if $k$ is finite this ideal is nontrivial! (If $k$ has $q$ elements, then the ideal is generated by the polynomial $t^q-t$). So your argument that $\phi$ is not an isomorphism breaks down. In fact, over a finite field $\phi$ is an isomorphism, since every map between algebraic sets over a finite field is given by a polynomial (in particular, $\phi^{-1}$ is). You can prove this by a variant of Lagrange interpolation.

(For this and other related reasons, Reid's definition of "coordinate ring" is not really the "right" definition when $k$ is not algebraically closed. However, giving the right definition requires a more sophisticated notion of "spaces" that you are defining coordinate rings of than merely algebraic sets.)

  • $\begingroup$ Thank you! I never thought about it from this direction. I have been always assuming that the coordinate ring of $k$ is $k[t]$. $\endgroup$
    – KittyL
    Jan 30 '16 at 10:56

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