2

In elementary treatments of algebraic varieties, a regular morphism between affine varieties is one whose components are all polynomials, and the coordinate ring as a ring of polynomial functions.

When defining the sheaf of regular functions on a variety $X$, then you have for any basic open set $D(f)$, the nonvanishing locus of a polynomial $f$, that $\mathcal{O}(D(f)) = \mathcal{O}(X)_f,$ which is the ring obtained by adjoining $1/f$ to the coordinate ring of polynomial functions. This is for example explained by user D_S here or the sources linked by Nils Matthes in this answer. We started by looking at polynomials, and then decided to include $1/f$. I understand including $1/f$ when we look for the rational functions, but the multiplicative inverse of a polynomial is definitely not polynomial. What is the justification for adding it, is there something stronger than "because it's nonzero on $D(f)$ therefore $1/f$ is defined. We add it because we can".

The proofs in those linked sources start out as defining a regular function as "locally a quotient of the form $g/h$ where $h$ doesn't vanish. So they appear to be just assuming the thing I am looking for justification of.

I'm not asking what inverting elements has to do with localization,which has some good answers on this site already. I liked Viktor Vaughn's answer on this thread and Jesko Hüttenhain's in this one.

KReiser
  • 74,746
ziggurism
  • 17,476
  • Is your question: how do we localize a ring? Or is it the link between localization and the sheaf $\mathcal{O}_f$? – julio_es_sui_glace Aug 29 '23 at 23:10
  • @julio_es_sui_glace yeah, what is my question exactly. how did we decide that the correct analogue of polynomial functions on $D(f)$ is actually polynomial functions over $f$, rather than, just polynomials? – ziggurism Aug 30 '23 at 00:23
  • the sidebar tagged this question as related: https://math.stackexchange.com/q/3420086/16490. i found some of the comments helpful. the right way to think of it isn't that regular means polynomial and $1/f$ counts as polynomial. instead, it's that regular means "locally rational", and it is a special result that locally rational means globally polynomial on an affine variety. or constant on a projective variety. polynomial is the exception, not the rule. and it's perhaps just for ease of pedagogy that some treatments will give the polynomial functions and morphisms a privileged position – ziggurism Aug 31 '23 at 15:19
  • Proposition 6.3.5 of Fulton's book Algebraic Curves, as pointed out by Nils Matthes in the question linked above, might be a good answer – ziggurism Sep 08 '23 at 04:45

2 Answers2

4

When you say

In elementary treatments of algebraic varieties, a regular morphism between affine varieties is one whose components are all polynomials, and the coordinate ring as a ring of polynomial functions.

you're assuming that a variety $X$ is given as embedded in some affine space $\Bbb A^n$, and when you say "polynomial", you mean the restriction of a polynomial function on $\Bbb A^n$ to $X$. But what if we took a different embedding? It turns out that all embeddings will give the same ring, so the ring of regular functions is independent of the embedding.

Let's now consider $1/f$: if $f=0$ on some but not all points of $X\subset\Bbb A^n$, then $D(f)\subset X \subset \Bbb A^n$ is not an affine variety, because it is not closed in $X$ (and hence not closed in $\Bbb A^n$). But there is a variety in $\Bbb A^{n+1}$ which looks very much like $D(f)$: if $I$ is the ideal of polynomials cutting out $X\subset\Bbb A^n$, then the zero locus of the ideal $(I,x_{n+1}f-1)$ in $\Bbb A^{n+1}$ is (1) a variety and (2) projects down to exactly $D(f)$ under the projection $\Bbb A^{n+1}\to\Bbb A^n$, $(x_1,\cdots,x_n,x_{n+1})\mapsto (x_1,\cdots,x_n)$. If we broadened our class of functions to "polynomial on some closed embedding", then we would still get everything we've developed so far and we could declare that these two objects are isomorphic, then apply our tools to a few more scenarios. In this sense, it's a natural extension of the theory you've already talked about.

(This is slightly out of order with how the history developed - there are already parametrizations using rational functions running around pretty early on, perhaps as early as Greek antiquity, depending on how generous you are. If you're not as generous, Riemann's work leads to some amount of rational geometry and considering "locally regular" functions in the 1800s when talking about Riemann surfaces - so certainly the idea of looking at functions which behave well locally is already floating around in the fields which kicked off much of the development of algebraic geometry. One can reasonably start from considering functions which are locally given by a quotient of polynomials and then prove that on an affine variety, the only functions which are regular everywhere are the polynomial functions, and I'd say this might be more reflective of how the field actually developed.)

KReiser
  • 74,746
  • $1/x$ is not polynomial on $\mathbb{A}^1$, but $V(xy-1)$ is isomorphic to $\mathbb{A}^1\setminus 0$ and it does have a polynomial function $y$ which is essentially $1/x$ or something? – ziggurism Aug 30 '23 at 15:50
  • 1
    the point is that the difference between a polynomial function and rational function (away from zeros of the denominator) are not intrinsic and depend on the embedding? – ziggurism Aug 30 '23 at 15:56
  • 1
    @ziggurism Yeah, that's pretty much it - if you have a rational function on a variety, you can find an open subset of the domain of that rational function with an embedding as an affine variety where the restriction of the rational function to that subset is given by a polynomial function. (Sometimes you'll have to be careful about gluing, which is why I had to add "open subset" instead of just "the domain".) – KReiser Aug 30 '23 at 16:53
  • Wait a second. Are $V(xy-1)$ and $D(1/x) = \mathbb{A}^1\setminus 0$ isomorphic? The map in one direction is $(x,y)\mapsto x$ which is polynomial. But in the other direction $x \mapsto (x,1/x)$ which is not polynomial. So they're not isomorphic in the sense of invertible polynomial morphisms? – ziggurism Sep 06 '23 at 15:44
  • 1
    @ziggurism From the post: "If we broadened our class of functions to 'polynomial on some closed embedding', then we would still get everything we've developed so far and we could declare that these two objects are isomorphic". One needs the full strength and idea of regular functions as locally the quotient of polynomials with nonvanishing denominator to say that these two are isomorphic. – KReiser Sep 06 '23 at 15:45
1

In addition to KReiser's nice answer, I want to add the following property we preserve by inverting $f$ on $D(f)$.

Let $X$ be a variety an affine variety and let $f \in \mathcal{O}(X) = A$ which is everwhere nonvanishing. Then, $1/f \in \mathcal{O}(X)$.

This is straightforward to prove: if $f$ is not invertible, then $f$ is in some maximal ideal $\mathfrak{m} \subset A$. It follows that $f$ vanishes at the point of $X$ corresponding to $\mathfrak{m}$.

The above property extends to any variety $X$ which is not necessarily affine: if $1/f$ is not invertible, choose some affine chart $U$ on which $(1/f)|_U$ is not invertible and apply the above argument.

Daniel
  • 6,379
  • sometimes the nonvanishing functions are just the nonzero constants, and you can't complain about including their inverses – ziggurism Aug 31 '23 at 00:31