Dominant rational maps $X\dashrightarrow\mathbb P^1_k$ and $K(X)^\times$

Question

Let $X$ be an irreducible variety over an algebraically closed field $k$. I am trying to solve an exercise which states that the set of dominant rational maps $X\dashrightarrow \mathbb P_k^1$ is in one to one correspondence with the set $K(X)^\times$. Here $K(X)$ is the fraction field of $X$. But I am having serious doubt that the exercise is true, or put differently, I can't find a way to define such a bijection.

• According to lemma 9.23 in Görtz & Wedhorn there is a bijection between dominant rational maps $X\dashrightarrow \mathbb P_k^1$ and the sets $$Hom_{kAlg}(K(\mathbb P_k^1),K(X)) = Hom_{kAlg}(k(t),K(X))$$ where we may view $t$ as $t = x_0/x_1$. Since $K(X)$ is a field, we see that the dominant rational maps $X\dashrightarrow \mathbb P_k^1$ are in one to one correspondence with the elements of $K(X)$ which are transcental over $k$. But the transcendental elements are not the invertibles. E.g the function which is constant $1$ is invertible but not transcendental. Of course this does not prevent the sets to be in bijection (have the same cardinality), but it does make it seem as if the bijection, if it exists, must be fairly non-natural.

• Here is a second thing I tried. We may fix an affine open subspace $U$ of $X$. Let $\eta$ be the generic point of $X$. Then we have that $$K(X) =\mathcal O_{X,\eta} =\mathcal O_{U,\eta} = Frac(\mathcal O_X(U))$$ This means that I can write any $f\in K(X)$ as a quotient $f=g/s$ with $g,s\in \mathcal O_X(U)$ and $D(s)$ dense. If $f$ is invertible, then $D(g)$ is dense, and I can define a rational map $x\mapsto (g(x):s(x))$ into projective space. I also managed to show that the map does not depend on the choices made. But I see no reason for this map to be dominant. For example, if $f$ is constant $1$, then $(g(x):s(x))$ will be constant $(1:1)$.

• View $\mathbb P_k^1$ as the line $\mathbb A^1_k$ plus a point $\infty$ at infinity for a second. Then it seems to me that there is a bijection between rational maps $f:X\dashrightarrow \mathbb P_k^1$ for which $f^{-1}(\mathbb P^1_k \smallsetminus \{\infty, 0\})$ is non-empty, and invertible elements of $K(X)$.

Question: Is the exercise wrong, or am I stupid?

Answer 1

You are correct and the exercise is wrong. Dominant rational maps of $k$-varieties $X\to Y$ are in bijection with $k$-algebra homomorphisms $k(Y)\to k(X)$ as per your first bullet point, and since maps of fields are injective, we cannot ever have $t\in k(t)\cong k(\Bbb P^1)$ ever map to a constant. Therefore dominant rational maps from $X$ to $\Bbb P^1$ are the same as choices of nonconstant elements of $k(X)$.