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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?

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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 $\overline{p(x,y)} = 0$.

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thanks, care to explain the part "write $h(t)=h_{0}(t^{2})+th_{1}(t^{2})$? –  user10 Apr 11 '12 at 19:42
    
Sure: that was just meant as a compact way of saying "break $h(t)$ into its even-degree and odd-degree terms". For example, if $h(t) = t^5 + 2t^3 - t^2 - 7t + 4 = (-t^2 + 4) + t(t^4 + 2t^2 - 7)$, then $h_0(t) = -t + 4$ and $h_1(t) = t^2 + 2 - 7$. The point is to set things up to use $f^*(x+1) = t^2$ and $f^*(y) = t f^*(x)$. –  Michael Joyce Apr 11 '12 at 19:49
    
oh, nice, cheers. –  user10 Apr 11 '12 at 20:44
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