# Function fields of irreducible varieties

The following might seem long, rambling and to contain more information than necessary. The problem is, I'm having a macro-understanding issue and feel like I need to tell you everything I think so that you can tell me why I'm wrong.

I'm learning Algebraic Geometry somewhat prematurely (without much knowledge of commutative algebra, category theory etc). The most general variety I'm dealing with are quasi-projective varieties, and I couldn't tell you the definition of a general variety (well, I could, but I wouldn't know what I was talking about).

As such I'm getting in a fine muddle over definitions and equivalences of definitions. In my notes, the function field $K(X)$ of an irreducible affine variety is defined to be the field of fractions of its coordinate ring $K[X]$. Okay, fine, I'm happy with that. I can see that under this definition, functions in the function field (which I think are called rational functions) look like quotients of polynomials, have (possibly) many different representations and are not (in general) defined everywhere. In fact, the only rational functions which are defined everywhere on $X$ (i.e. are regular on the whole of X) are those from the coordinate ring $K[X]$.

Now it's not too hard to see that, under this definition, $\displaystyle K(X) = \bigcup _{U \ \mathrm{Zariski \ open \ affine}} k[U]$.

My notes then tell me to "observe" that if $U \subseteq X$ is Zariski open and $X \backslash U$ is a hyperplane (so $U$ is also affine), then $k(U) = k(X)$. I cannot see why this is true. By the above equality, I can see that rational functions on $X$ are functions which are "locally polynomial", i.e. functions which are polynomial maps on some open affine neighbourhood in $X$. $X$ being irreducible implies any two such open neighbourhoods have non-empty intersection, which feels like a relevant observation but I don't know what to do with it. I'm then told that if $X$ is an arbitrary irreducible variety, we can define $K(X) = K(U)$ for $U$ any affine open neighbourhood in $X$

I'd much appreciate an enlightening statement that makes all this premature learning feel somewhat more worthwhile.

I've seen another definition of a function field for a projective variety $V$, namely that it's quotients of homogeneous polynomials (where numerator and denominator have the same degree), modulo the fairly obvious equivalence relation. Given the standard open affine cover of projective space, it's kind of easy to see that $K(V)$ is isomorphic to $K(U)$, where $U$ is one of the affine sets in the cover. By the relationship above, this means that $K(V)$ is isomorphic to $K(W)$ where $W$ is any open affine subset of $V$. So in this respect, a rational function here is something that is locally a polynomial map, which gives one direction of the supposed equivalence. However, it seems to me like these objects are more like examples of rational functions, rather than all rational functions; under the definition "$K(V)$ is space of functions which are locally polynomial maps", why must these things look like quotients of homogeneous polynomials?

Anything relevant would be great. Thanks for your time.

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I suppose the definitions in terms of quotients of polynomials just seem so much more useable, which is why I'm bothering to check that they're (sufficiently) correct –  Jonathan Apr 27 '12 at 19:06
I'll try to find time to write something actually helpful later, but I think what you want to say in the end is that $k[X] \subset k[U] \subset k(X)$. –  Dylan Moreland Apr 27 '12 at 19:13
@DylanMoreland Thanks. Can you embed $k[X]$ inside $k[U]$ just by restriction $f \mapsto f \big|_U$? I'm sort of uncomfortable with this, since I don't know what is formally happening here. I know elements in these rings are "considered" as functions $X \to k$ and $U \to k$, but they're really just cosets, right? Having established this sandwich, how do we get to the final result? Can you just "take" fields of fractions? –  Jonathan Apr 27 '12 at 19:21
Yes! That's usually how these arguments go. I just need to set aside time to translate everything over into these definitions of yours, which are good enough for many purposes (e.g. I'm sure that Silverman defines things this way in his book on Elliptic Curves). –  Dylan Moreland Apr 27 '12 at 19:34
About embedding: with this setup you might prefer to think of it as being the map on rings induced by the inclusion morphism $U \to X$. –  Dylan Moreland Apr 27 '12 at 20:12

If $X$ is a variety in $\mathbf A^n$, and $f \in k[x_1, \ldots, x_n]$, then the open subset $U = X - Z(f)$ is affine. It gets its structure from its correspondence with the variety in $\mathbf A^{n+1}$ defined by $I(X)$ and $f(x_1, \ldots, x_n)x_{n+1} - 1$. Thus, $k[U] = k[X]_{\bar f}$, where $\bar f$ is the image of $f$ in $k[X]$ and the subscript indicates localization. The inclusion morphism $U \to X$ corresponds to the standard homomorphism $k[X] \to k[X]_{\bar f}$. It would be good to work this out!
Sorry, but what does the subscript $f$ mean? –  Jonathan Apr 27 '12 at 22:07
@Jonathan It's the localization at the element $f$. –  Dylan Moreland Apr 27 '12 at 22:08
@Jonathan so I should ask: is this how you define $k[U]$? –  Dylan Moreland Apr 28 '12 at 13:31
I haven't come across localization before. But this looks like the same definition of $k[U]$; for an affine variety $V \subset \mathbb A^n$, define $k[V]$ to be $k[x_1, \ldots , x_n] / I(V)$. So in the case $V = X\backslash Z(f) \subset \mathbb A^n$, we have to embed $V$ inside $\mathbb A^{n+1}$ in order to make it affine (using the construction you've given) –  Jonathan Apr 28 '12 at 15:10