Why is $\mathbb{Q}(t,\sqrt{t^3-t})$ not a purely transcendental extension of $\mathbb{Q}$? This question is taken from Dummit and Foote (14.9 #6). Any help will be appreciated:

Show that if $t$ is transcendental over $\mathbb{Q}$, then $\mathbb{Q}(t,\sqrt{t^3-t})$ is not a purely transcendental extension of $\mathbb{Q}$.

Here's what I've got so far:
Abbreviate $\sqrt{t^3-t}$ as $u$.
I've shown that the transcendence degree is 1, so the problem boils down to showing that $\mathbb{Q}(t,u) \supset \mathbb{Q}(f(t,u)/g(t,u))$ strictly, for all polynomials $f,g$ in two variables.
Suppose for contradiction that $\mathbb{Q}(t,u) = \mathbb{Q}(f(t,u)/g(t,u))$. Look at this field as $\mathbb{Q}(t)[x]/(x^2-(t^3-t))$, with $\bar{x}=u$.
Then since $t$ and $u$ are generated by $f/g$, we have that for some polynomials $a,b,c,d$ in 1 variable, $a(\frac{f(t,x)}{g(t,x)})/b(\frac{f(t,x)}{g(t,x)})-t \in (x^2-(t^3-t))$, and $c(\frac{f(t,x)}{g(t,x)})/d(\frac{f(t,x)}{g(t,x)}) - x \in (x^2-(t^3-t))$.
I then tried playing with degrees, but I haven't found a contradiction.
 A: For convenience change the base field to $k=\mathbb{C}$
(if the extension were purely transcendental before
then it would be purely transcendental after).
As the extension has transcendence degree $1$, then
if the extension were purely transcendental it would equal $k(x)$
for some $x$. Then there are nonconstant rational functions
$f(x)$ and $g(x)$ such that
$$g(x)^2=f(x)^3-f(x).$$
It's easy to see we can write $f(x)=u(x)/w(x)^2$
and $g(x)=v(x)/w(x)^3$ in lowest terms
where $u$, $v$ and $w$ are polynomials.
We find that
$$\phi(x)=\frac{g'(x)}{3f(x)^2-1}=\frac{f'(x)}{2g(x)}.$$
In fact $\phi(x)$ is a polynomial. Otherwise the denominator
of $\phi$ has a factor $x-a$. Expressing $\phi$ in terms of $u$, $v$
and $w$ (details?) we see that this implies that
$$2g(a)=3f(a)^2-1=0$$
as well as
$$g(a)^2=f(a)^3-f(a).$$
This is impossible. As $f$ and $g$ are nonconstant, $\phi$ is a nonzero
polynomial.
Now replace $f(x)$ and $g(x)$ by $f(1/x)$ and $g(1/x)$. Then $\phi(x)$
is replaced by $-x^{-2}\phi(1/x)$ so that is also a polynomial, which
it isn't. So we get the required contradiction.
Added. The above was composed in a rush, and I cut a few corners,
so some extra details. We have $f'(x)=\textrm{polynomial}/w(x)^3$ and so
$$\phi(x)=\frac{f'(x)}{g(x)}=\frac{\textrm{polynomial}}{v(x)}.$$
Similarly
$$\phi(x)=\frac{\textrm{polynomial}}{3u(x)^2-w(x)^4}.$$
If $x-a$ divides the denominator of $\phi(x)$ then $v(a)=0=u(a)^2-w(a)^4$.
We can't have $w(a)=0$ as then $x-a$ would divide $u(x)$ and $w(x)$
contrary to $u(x)/v(x)^2$ being in lowest terms. Hence $f(a)$ and $g(a)$
make sense and we get $g(a)=3f(a)^2-1=0$.
This is really a geometric argument. On the elliptic curve
$$E:\quad z^2=t^3-t$$
the "invariant differential"
$$\omega=\frac{dt}{2z}=\frac{dz}{3t^2-1}$$
would have no poles on the (projective curve) $E$. But were $E$ rational
then $\omega=\phi(x)dx$ has no poles on the projective line; but every
nonzero differential on the projective line has a pole. The above is just
a naive version of this argument, which works for all elliptic curves.
A: By definition a purely transcendental extension of $\mathbf{Q}$ is the function field of a rational variety over $\mathbf{Q}$, namely the fraction field of a polynomial ring $\mathbf{Q}[x_1,\ldots, x_n]$ for some $n$. Call your field $K$. Obviously the transcendence degree of $K$ over $\mathbf{Q}$ is 1, so it's enough to show that $K$ is not isomorphic to a rational function field in 1 variable, i.e. to $\mathbf{Q}(t)$. Now by inspection $K$ is the function field of the elliptic curve $E:y^2 = x^3-x$ over $\mathbf{Q}$. It's a standard fact, shown in various books concerned with algebraic curves, that $K\cong \mathbf{Q}(t)$ if and only if $E$ is birational to the projective line $\mathbf{P}^1_{\mathbf{Q}}$. This is not the case, because (for example) the genus of $E$ is 1 (say by the degree-genus formula for plane curves), the genus of $\mathbf{P}^1$ is zero, and the genus is a birational invariant of a curve. 
I don't know how to solve this problem with "elementary" (non-geometric) methods off the top of my head; hopefully someone else can explain how to do this.
