Okay, as long as my solution below stays incomplete, I'm giving the fix for even $n$: Read my reply until (4), and notice that, whenever $\left(t_1,t_2,...,t_n\right)$ is an $n$-tuple of points in counterclockwise position, then so is $\left(t_n,t_1,t_2,...,t_{n-1}\right)$, so (4) yields that the reals $\det\left(F^{\prime}\left(t_1\right),F^{\prime}\left(t_2\right),...,F^{\prime}\left(t_n\right)\right)$ and $\det\left(F^{\prime}\left(t_n\right),F^{\prime}\left(t_1\right),F^{\prime}\left(t_2\right),...,F^{\prime}\left(t_{n-1}\right)\right)$ have the same sign, which is absurd.

Okay, please wait a few more minutes... I think I've now solved the general case. BUT I need an additional condition: the function $F$ should be continuously differentiable (not just differentiable), and the derivatives at the endpoints should match as well.

I think I can solve the original problem for even $n$. Interested?

Goedel was not a mathematician?

I meant $w_i\not\in \left\{w_1,w_2,...,w_{i-1},w_{i+1},w_{i+2},...,w_n\right\}$ instead of $w_i\neq \left\{w_1,w_2,...,w_{i-1},w_{i+1},w_{i+2},...,w_n\right\}$.

That power series argument that Fulton-Harris do is just a fancy way to write out your Vandermonde argument. Both times, it's the same lemma: If $v_1$, $v_2$, ..., $v_n$ and $w_1$, $w_2$, ..., $w_n$ are elements of some field, and we have $\sum_{i=1}^n v_iw_i^k=0$ for every $k\in\mathbb N$, then $\sum_{i\in\left\{1,2,...,n\right\};\ w_i=x}v_i=0$ for every $x$ in the field, so that particularly, $v_i=0$ for every "unique" $i$ (that is, for every $i$ which satisfies $w_i\neq \left\{w_1,w_2,...,w_{i-1},w_{i+1},w_{i+2},...,w_n\right\}$).

Since I can't figure out a better place to ask this: Can anyone help me with two exercises from the above notes? In Problem 1.26 (c) I would like to know whether the field $k$ is assumed to be algebraically closed (no hints, please; just an answer to this question). In Problem 1.34 (d), am I right in assuming that the grading on $P_Q$ is given by $\mathrm{deg}p_i=0$ and $\mathrm{deg}a_h=1$ ? I really enjoy this text, by the way - it's concise and straight to the point (and not analysis-biased as Fulton-Harris).

I don't know what an affine closed set is when $k$ is not algebraically closed.

Ah, okay. Maybe this can be rescued by restating the condition that $k$ be algebraically closed in terms of schemes, but anyway this is just a rewording of my question.

Thanks. I'll look into these sources (EGA IV is a bit high for me, bur Bourbaki seems just right, and anyway I don't care for (3) as much as I care for (1)).

That's why I edited my above post to require $k$ to be algebraically closed (for (3)). Is it still wrong then?

I meant the assertion that the product of connected affine schemes is connected (I assume that you mean the fibred product, which is not a topological product - or am I mistaken here?). What you just showed is that the tensor product of rings corresponds to the fibred product of the corresponding affine schemes over the trivial scheme - but I know that.

Thanks, but how do you prove this true fact?