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Following question: Why and how to see that a connected, smooth, projective curve $C$ (so a so a $1$-dimensional, proper $k$-scheme) is rational if it is unirational.

Remark: unirational means that there exist a dense morphism $phi: mathbb{P}^n to C$

Especially that would mean that every unirational normal curve have a rational point.

Remark #2: This question arises from the answer given https://mathoverflow.net/questions/319483/extend-group-action-of-mathbba1-g-to-projective-line

but it seems to me that my question above about the step in the given answer would be too low for a further MO-thread.

This result is called Lüroth's theorem; see, e.g., this Math.SE question. For an elementary proof, you can look at Theorem 1.3 in this book by Ojanguren.

For more canonical references, you can look at Jacobson's Basic Algebra II, p. 522, or Hartshorne's Algebraic Geometry, Ex. IV.2.5.5.

Below is a version of Ojanguren's argument that I wrote down for a course taught by Mircea Mustaţă. See Lecture 1, Theorem 2.1 in his notes, or Theorem 1.6 in my LiveTeXed notes from the course.

Proof.
We show that if $k subseteq K subseteq k(t)$ is a sequence of field
extensions such that $operatorname{trdeg}_k K = 1$ and $t$ is transcendental over $k$, then $K simeq k(X)$ for a transcendental element $X$ over $k$. Since
$operatorname{trdeg}_k K = 1$, it is enough to find some $a in K$ such that
$K = k(a)$.

First, the second extension $K subseteq k(t)$ must be algebraic, so
$t$ is algebraic over $K$. Let
$$
f(X) = X^n + a_1X^{n-1} + cdots + a_n in K[X]
$$
be the minimal polynomial of $t$ over $K$. Since $t$ is
transcendental over $k$, we cannot have all $a_i in k$, and so there is some
$i$ such that $a_i in K smallsetminus k$. We will show that in this case, $K
= k(a_i)$. We know that the $a_i in K subseteq k(t)$,
and so we may write
$$
a_i = frac{u(t)}{v(t)},
$$
where $u,v in k[t]$ are relatively prime, and where at least one of them of
positive degree by the assumption that $a_i notin k$. Now consider the following polynomial:
$$
F(X) = u(X) - a_iv(X) in k(a_i)[X] subseteq K[X].
$$
Since $F(t) = 0$, we have that $f(X) mid F(X)$ in $K[X]$ by minimality of
$f(X)$, and so
begin{equation}
u(X) - a_iv(X) = f(X)g(X).tag{1}label{eq:0105star}
end{equation}
where $g in K[X]$. We then claim the following:

Claim. $g in K$.

Showing the Claim would conclude the proof, for then we have a sequence of
extensions
$$
k hookrightarrow k(a_i) hookrightarrow K hookrightarrow k(t)
$$
where $t$ is the root of a degree $n$ polynomial in $k(a_i)[X]$, hence
$[k(t) : K] ge n = [k(t) : k(a_i)]$, which implies $K = k(a_i)$.

To prove the Claim, the first step is to get rid of all denominators: by
multiplying eqref{eq:0105star} by $v(t)$ and the product of all denominators in the coefficients of $f$ and $g$, we
get a relation
$$
c(t)bigl(u(X)v(t) - v(X)u(t)bigr) = f_1(t,X)g_1(t,X),
$$
where $c(t) in k[t]$ is nonzero and $f_1(t,X),g_1(t,X) in k[t,X]$ are obtained from $f$ resp. $g$ by
multiplication by an element in $k[t]$. Note that $k[t,X]$ is a UFD, hence every prime factor of $c(t)$ divides either $f_1(t,X)$ or $g_1(t,X)$.
Thus, we can get rid of $c(t)$ by successively dividing by prime factors of $c(t)$
to get a relation
begin{equation}
u(X)v(t) - v(X)u(t) = f_2(t,X)g_2(t,X),tag{2}label{eq:0105rel3}
end{equation}
where now $f_2(t,X),g_2(t,X) in k[t,X]$ are obtained from $f,g$ by
multiplication by a nonzero element in $k(t)$. The trick is now to look at the
degrees in $t$ on both sides. First,
$$
deg_tbigl(u(X)v(t) - v(X)u(t)bigr) le maxbigl{deg u(t),deg
v(t)bigr}.
$$
Letting $f_2(t,X) = gamma_0(t)X^n + cdots + gamma_n(t)$, we see that
$$
frac{gamma_i(t)}{gamma_0(t)} = a_i(t) = frac{u(t)}{v(t)},
$$
where we recall that $u(t),v(t)$ were relatively prime. This implies that in fact,
$$
deg_tbigl(f_2(t,X)bigr) ge maxbigl{deg u(t),deg v(t)bigr}.
$$
By looking at degrees in $t$ on both sides of the relation
eqref{eq:0105rel3}, we have that $deg_t(g_2(t,X)) = 0$, hence $g_2 in k[X]$.

Now we show that $g_2 in k$. For sake of contradiction, suppose that
$g_2 notin k$. Then, there is a root $gamma in overline{k}$ such that
$g_2(gamma) = 0$, which implies that $u(gamma)v(t) = v(gamma)u(t)$ by eqref{eq:0105rel3}. But
since $u(t)$ and $v(t)$ are relatively prime, and are not both constants, we
must have $u(gamma) = 0 = v(gamma)$. This contradicts that $u,v$ are
relatively prime, and so $g_2 in k$.
Finally, since $g in K[X]$ and $g = g_2 cdot (text{element of
$k(t)$})$, we have that $g in K$. $blacksquare$