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On p.28 (here) Reid proves that the curve $y^2=x(x-1)(x-\lambda)$ for $\lambda\neq 0,1$ has no rational paramaterisation. At one point (on p.29), he has the equation $$r^2=ap(p-q)(p-\lambda q)$$ where $p$ and $q$ are polynomials over a field $k$, and $a$, $\lambda$ are constants in the field. He then says that "by considering factorisation into primes, there exist nonzero constants $b,c,d\in k$ such that $$bp,c(p-q),d(p-\lambda q)$$ are all squares in $k[t]$.

Edit: $p$ and $q$ are known to be coprime.

I feel like this should be obvious, but I don't follow his argument. Why do we know that those factors are squares (up to units)?

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p, (p-q), (p-lambda*q) are pairwise coprime. Shared irreducible factors between r^2 and p have to appear an even number of times in p. –  T.. Nov 18 '10 at 7:14
    
For an alternative argument, see math.stackexchange.com/questions/5278/… . This can be adapted to work for the curve considered by Reid. –  Robin Chapman Nov 18 '10 at 7:37
    
Thanks for both answers. I'm not quite at the stage where I'm comfortable translating between algebra and geometry yet, but hopefully I'll get there one day. –  Lost In Math Nov 19 '10 at 7:07
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up vote 5 down vote accepted

Because $p$ and $q$ are coprime, then $p$ and $p-q$ are coprime: any irreducible factor of both would divide $p-(p-q)=q$, so it would divide both $p$ and $q$. Likewise, since $\lambda\neq 0$, $p$ and $p-\lambda q$ are coprime (same reason). And since $\lambda\neq 1$, then $p-\lambda q$ and $p-q$ are coprime; any shared factor between them is a factor of $\lambda(p-q)-(p-\lambda q) = (\lambda-1)p$ (hence of $p$) and of $(p-q)-(p-\lambda q) = (\lambda-1)q$ (hence of $q$), but $p$ and $q$ have no common factors other than units.

So $p$, $(p-q)$, and $(p-\lambda q)$ are pairwise coprime as well.

At this point, the argument is the same as in any UFD (e.g., the integers): you have $r^2 = xyz$, with $x$, $y$ and $z$ relatively prime; looking at each irreducible factor of $r$, and at each irreducible factor of $x$, of $y$, and of $z$, you get that each irreducible factor of $x$, of $y$, and of $z$ must appear to an even exponent, so each of $x$, $y$, and $z$ are squares.

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Thank you very much for the answer! –  Lost In Math Nov 19 '10 at 7:03
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