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.


2 Answers 2


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.

  • 1
    $\begingroup$ Very nice! I knew that these sort of problems could be solved by taking derivatives, or by using algebraic geometry, and that these were somehow the same method, but I never saw someone explicitly relate the two. $\endgroup$ Commented Sep 23, 2010 at 19:34
  • $\begingroup$ Thanks very much! I'm now hungry to learn algebraic geometry :) $\endgroup$
    – shirker
    Commented Sep 24, 2010 at 2:51
  • 1
    $\begingroup$ Why would $-x^{-2}\phi(1/x)$ be a polynomial? $\endgroup$
    – B.A
    Commented Apr 7, 2017 at 22:12
  • $\begingroup$ @B.A, I had the same question, so decided to add some more details to Robin's answer, thanks to Wu. Let $x = 1/t$, and then we see $$\phi(1/t) = \frac{g'(1/t)}{3f(1/t)^2-1} = \frac{f'(1/t)}{2g(1/t)}.$$ Now let $g_1(t) = g(1/t), f_1(t) = f(1/t)$. We have $$\phi_1(t) := -t^{-2}\phi(1/t) = \frac{g_1'(t)}{3f_1(t)^2-1} = \frac{f_1'(t)}{2g_1(t)}.$$ Therefore, $g_1(x)$ and $f_1(x)$ is another pair of functions satisfying the equation in the definition of $\phi$, but we have proved previously that then the quotient $\phi_1$ should be a polynomial, which is impossible. $\endgroup$
    – froyooo
    Commented Apr 17, 2020 at 4:59

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.


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