Birational affine schemes

Question

My question has to do mostly with the definition/terminology behind what a birational map is, in terms of affine schemes. For instance, if we have two affine schemes and a morphism between them say, $f: Spec(R) \rightarrow Spec(R^{\prime})$, then if there exists an open subset $U \subset Spec(R)$, such that the induced map $f|_U: U \rightarrow Spec(R^{\prime})$ is an isomorphism onto its image, is it correct to say that the two affine schemes are birational? If so, how should someone interpret this intuitively? If not could you please write/refer the definition/way to interpret the birationality in terms of (affine) schemes?

P.S. One comment to save some time of the potential user(s) who will put effort to write out their answer is, that in terms of affine varieties (meaning irreducible algebraic sets) I do know and understand what the above idea behind birational maps wants to construct.

Answer 1

No, a birational map is strictly weaker that a map that induces an isomorphism on a non-empty open set. As a slogan, a birational map of irreducible schemes is one that induces an isomorphism of generic fibers.

More precisely, if you assume that Spec $R$ and Spec $R'$ are both irreducible, then a map $f:$ Spec $R\to$ Spec $R'$ is a birational map if it induces an isomorphism on the local ring of the (unique) generic point. I.e. if $\eta,\eta'$ are the generic points of Spec $R$ and Spec $ R'$, the induced map $O_{Y,\eta'}\to O_{X,\eta}$ is an isomorphism.

Even if they are not irreducible, if you assume that they only have finitely many irreducible components with generic points $\eta_i$ and $\eta_i'$ respectively and the map $f$ maps these generic points bijectively and induces isomorphisms $O_{Y,\eta_i'}\to O_{X,\eta_i}$, then $f$ is a birational map.

If you assume that $Y$ is reduced (i.e. no non-zero nilpotents in the affine case) and the map $f$ is locally of finite type, then a birational map gives an isomorphism on a non-empty open (and hence dense) subscheme of $X$. I'm sure there's other (possibly weaker) conditions on $f$ that guarantee this but I can't think off the top of my head.