Morphisms of Function Fields do NOT induce Rational Maps

We have seen that dominant rational maps between integral schemes induce morphisms of function fields.  The converse however, is not true.  Morphisms of function fields of two integral schemes do not always induce rational maps.

Example. Let $X = \text{Spec} k[x]$ and $Y = \text{Spec} k(x)$. Then their function fields are canonically isomorphic (both being $k(x)$).

Claim. There is no rational map $X \dashrightarrow Y$.

Proof. Suppose we have a rational map $f: U \to Y$ for some open dense subset $U$ of $X$. Shrinking $U$ if necessary (as all nonempty subsets of $X$ are dense), assume that $U = D(f)$. Thus $f$ induces a morphism $\varphi: k(x) \to k[x, 1/f]$.   Note that on stalks, this morphism should be identity $\varphi_{\eta}: k(x) \to k(x)$, so $\varphi$ must be identity on $k[x]$. This is impossible.
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For integral finite type $k$-schemes (in particular, irreducible varieties), morphisms of function fields do indeed determine rational maps between the schemes.

Proposition. Let $X$ be an integral $k$-scheme and $Y$ be an integral finite type $k$-scheme. Let $\varphi: K(Y) \hookrightarrow K(X)$ be a $k$-homomorphism. Then there exists a dominant $k$-rational map $\phi: X \dashrightarrow Y$ inducing $\varphi$.

Proof. The map $\phi$ is equivalent to a map from a non-empty open subset of $X$ to a subset $Y$ sending the generic point $\eta_X$ to $\eta_Y$. Moreover, as function fields of nonempty open subsets of $X$ (resp. $Y$) are canonically isomorphic to $K(X)$ (resp. $K(Y)$), we can assume $Y = \text{Spec} B$ and $X = \text{Spec} A$ are affine. We have a morphism $\varphi: B \hookrightarrow K(Y) \hookrightarrow K(X)$.

Warning: Clearly $\varphi$ induces a morphism $\text{Spec} \varphi(B) \to \text{Spec} B$. However we do NOT have a way to take $U = \text{Spec}\varphi(B)$ as an open subset of $X$.

So instead, we want to find some $A_g \subset K(X)$ such that $\varphi: B \hookrightarrow A_g$. This would induce a map a map $D(g) \to Y$.

Why can we find such an $A_g$? This is because $Y$ is a $k$-scheme of finite type. Thus $B = k[b_1, \ldots, b_n]$ for some $b_i$'s. Let $\varphi(b_i) = a_i/f_i$. Let $g = f_1 \cdots f_n$. Then as $\varphi$ is a $k$-homomorphism, the image of $B$ indeed lies in $A_g$.
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In particular if both $X$ and $Y$ are integral finite type $k$-schemes then a rational map $X \dashrightarrow Y$ that induces isomorphism $K(Y) \cong K(X)$ must be birational.