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I'm reading through Shafarevich volume 1 and am a bit confused about some of his examples. So I get how if we use the line $y=tx$ then we can show certain curves are birational to a line. One of his later examples takes a function $f$ with terms of degree $n$, $n-1$, and $n-2$, i.e.,$f=u_{n-2}+u_{n-1}+u_n$, where the $u_i$ are homogeneous of degree $i$. We again set $y=tx$, divide out the $x^{n-2}$ at which point we have $f=a(t)x^2+b(t)x+c(t)$. He then uses $s=2ax+b$ to complete the square at which point he declares that $f$ is birational to $s^2=p(t) = b^2-4ac$ (hyperelliptic curve).

What's confusing is that in the earlier example of $y^2=x^2+x^3$ he did the same thing, found $x =t^2-1$ (analogous to the $s$ above) which isn't the equation of a line (which the curve is birational to). How is it that our function $s^2(t)$ above was the birational curve but this function $x(t)$ isn't?

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Maybe the following could be of help. Let us compare the following two curves:

$$y=x^3+x,\;\;\; \mathrm{and} \;\;\;y^2=x^3+x.$$ The first curve is rational, because clearly for each value of $x$ there is a unique value of $y$, so there is a one to one map from a line to the curve, $x\to (x, x^3+x)$. On the other hand in the second example there is no such an obvious map, because if you know the $x$ coordinate of a point on the curve you can not say what is the value of $y$ -- the quadratic equation has two solution. And indeed, the curve in the second example is an elliptic curve, it is not rational.

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  • $\begingroup$ I get what's going on. What's bothering me is how we know to parameterize in the first example to directly produce the equation of the birat curve while in the second we have to notice that we've obtained a map from a single parameter to our curve and say that the birat curve must be Z(t=0). $\endgroup$
    – Mike C.
    Commented Feb 15, 2013 at 3:56

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