2
$\begingroup$

This is problem $4$ from "Basic Algebraic Geometry I" page $4$.

Let $X$ be the curve defined by the equation $y^{2}=x^{2}+x^{3}$ and $f:\mathbb{A}^{1} \rightarrow X$ defined by $f(t)=(t^{2}-1,t(t^{2}-1))$. Prove $f^{*}$ maps the coordinate ring $k[X]$ isomorphically to the subring of the polynomial ring $k[t]$ consisting of polynomials $g(t)$ such that $g(1)=g(-1)$. (Assume that chark $\neq 2$).

Let $W$ be the above subring. I can see that the image of $f^{*}$ is contained in $W$, I don't see the reverse inclusion, why is this?, also why $f^{*}$ is injective?

1 Answers 1

1

Let $g(t)$ be a polynomial such that $g(1) = g(-1)$. Then $g(t) - g(1) = (t^2 - 1)h(t)$. Write $h(t) = h_0(t^2) + t h_1(t^2)$. It follows that $g(t) = f^*(g(1) + x h_0(x+1) + y h_1(x+1))$. Thus, $f^*$ is surjective onto $W$.

To see that $f^*$ is injective, let $p(x,y) = p_0(x) + y p_1(x)$ be the unique polynomial in $k[x][y]$ of $y$-degree $\leq 1$ that represents $\overline{p(x,y)} \in k[X]$ and suppose that $f^*(\overline{p(x,y)}) = 0$. Then $p_0(t^2 - 1) + t(t^2-1) p_1(t^2 - 1) = 0$. Since the first summand contains only even-degree terms and the second summand contains only odd-degree terms, it follows that $p_0(x) = p_1(x) = 0$ (as polynomials in one variable), so $\overline{p(x,y)} = 0$.

  • 0
    oh, nice, cheers.2012-04-11