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I have been looking at Hartshorne III exercise 6.8 for nearly a week now and I don't seem to have a clue as to how to do it. In particular, I am stuck on part (a) which boils down to showing the following.

Let $X$ be integral, separated and locally factorial. Let $U$ be any open neighbourhood of a closed point $x$. Suppose that $Z := X- U$ is irreducible with generic point $\zeta$. Then there is a line bundle $\mathcal{L}$ and $s \in H^0(X,\mathcal{L})$ with $x \in X_s \subseteq U$.

Now I have convinced myself using separatedness of $X$ that we may choose a rational function $f$ with the property that $f \in \mathcal{O}_{X,x}$ and $f \notin \mathcal{O}_{X,\zeta}$. With this, we immediately deduce that for any $z \in Z$, $f \notin \mathcal{O}_{X,z}$ because we have an injection

$$ \mathcal{O}_{X,z} \hookrightarrow \mathcal{O}_{X,\zeta}.$$

Edit:

Ok here is an argument which I think works. It is not hard to see that $$H^0(X,\mathcal{L}(D)) = \{ f \in K(X)^\ast : div(f) + D \geq 0\} \cup \{0\},$$ i.e. all rational functions whose poles are no worse than $D$. Sitting inside of this is the canonical rational section $s$ corresponding to $1 \in K(X)^\ast$. Now I want to say that this $s$ has zeros wherever $f$ has a pole. When I look at a concrete example I can see this is true. If I look at $\Bbb{A}^1$ and $D= (f)_{\infty}$ for $f = 1/(x-1)(x-2)$, then $$\mathcal{L}(D) = \frac{1}{(x-1)(x-2)} \mathcal{O}_X$$

and the canonical section $1$ corresponds to $\frac{1}{(x-1)(x-2)} \times (x-1)(x-2).$ This "shows" wherever $f$ has a pole, $s$ has a zero. This is enough to conclude that $Z \subseteq V(s)$ because $f \notin \mathcal{O}_{X,z}$ for all $z \in Z$.

But now we are on an arbitrary scheme satisfying the conditions in the beginning, so:

My question is: How can we make rigorous the idea that wherever $f$ has a pole $s$ has a zero? Also, where do we use the locally factorial hypothesis? Is it just to make that $\mathcal{L}(D)$ be invertible?

<p><strong>Added:</strong> Can someone elaborate on what exactly this "canonical section" $1 \in H^0(X,\mathcal{L}(D))$ is?</p>
  • Dear Benjamin, I think there are some typos in your question. E.g. in the first piece of blocked text, you probably mean $X_s \subset U$ (not $X_s \subset \mathcal L$). In the second, you probably mean $x \in X_s$ (not $s \in X_s$)? Cheers, – Matt E Jan 17 '14 at 03:59
  • Dear @MattE, I have made the necessary corrections to my question. I have also edited it after thinking about the problem a little more, and I have reformulated my original question. I would be very grateful if you could answer my post. Regards, –  Jan 17 '14 at 06:03
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    Dear Benjamin, Part of what you want seems to be precise statements (which is good!), but unfortunately I haven't thought enough to give them yet. The basic idea is that since $Z$ is irred. and $x \not\in Z$, we can find a "divisor" containing $Z$ (just choose it so that it contains $\xi$) that doesn't contain $x$. A priori "divisor" just means something vague like codimension 1 closed subscheme (the point being that $Z$ might have higher codimension, but we can always put it inside something codim'n one), but we want it to actually mean Cartier divisor. Giving an effective Cartier ... – Matt E Jan 17 '14 at 06:52
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    ... divisor is the same thing as giving a line bundle with a regular section $s$ e.g. see here. The Cartier divisor doesn't contain $x$ precisely when $x \in X_s$. We will use the "locally factorial'' hypothesis to promote our initial vague notion of divisor to the precise notion of Cartier divisor.

    We will use "integral" to know that any non-zero section is regular.

    But of course you have to sort all this out into an actual argument, and I haven't quite seen how to write it down yet. Cheers,

     – Matt E Jan 17 '14 at 06:56
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    P.S. Assembling the pieces should be pretty straightforward, though; hopefully when I'm not as tired I'll write a proper answer, if you haven't sorted it out yourself by then. – Matt E Jan 17 '14 at 06:57
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    P.P.S. Your ideas and the examples your thinking about are good. If you look at the answer I linked, you will see it is very similar to your definition of $\mathcal L(D)$ and taking $1$ to be the section vanishing along $D$ (indeed, it is the same mathematics just phrased slightly differently). What you are missing is seeing that $D$ is being cut out (locally) by an equation $g = 0$, so that (locally) $\mathcal L(D) = \frac{1}{g} \mathcal O$, and hence is invertible. This is where locally factorial will come in. – Matt E Jan 17 '14 at 07:00
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    P.P.P.S. Coming at it from your point of view, via rational functions, try using locally factorial to make the denominator of $f$ (i.e. $f_{\infty}$) well-defined (locally); at the moment it isn't well defined (since in a non-UFD, "the" denominator of a rational function isn't well-defined). Once you do this, you'll see that $D$ really is a Cartier divisor, and everything should click into place. – Matt E Jan 17 '14 at 07:02
  • Dear @MattE, thanks so much for your feedback! I am a little tired now (due to some sleepless nights thinking about this!) but I will think about it in the morning (right now it's at night in Canberra). I have to say though that the concepts of divisors/ line bundles are some of the hardest to get your head around! –  Jan 17 '14 at 09:38
  • @MattE Quick question: I am a little confused by this canonical section $1 \in H^0(X,\mathcal{L}(D))$. As a function, it doesn't have zeros anywhere but viewed as a section something else happens. Can you clarify on this? Thanks, –  Jan 17 '14 at 14:02
  • Dear @MattE, this is what I understand at the moment. Let $D = (f){\infty}$. Choose an affine cover $U_i$ of $X$ such that $D|{U_i} = (g_i)$ where $g_i \in K(X)^\ast$. Here we use the locally factorial hypothesis. Then because $g_i$ has no poles on $U_i$, it is actually regular and the canonical section $1 \in H^0(X,\mathcal{L}(D))$ corresponds to $g_i \in \mathcal{O}(U_i)$. Now choose a point $z \in Z$, $z \in U_i$ for some $i$. As $Z \cap U_i \to U_i$ is a closed immersion into an affine, we can write $Z \cap U_i = \operatorname{Spec} A$ (continued) –  Jan 18 '14 at 03:46
  • Now it is clear that ${g_i}_{P}$ is in the maximal ideal of $A_P$ where $P$ is a prime of codimension $1$ in $A$. But now this means that ${g_i}_Q$ is also in the maximal ideal of $A_Q$ for any prime $Q$ because $Q$ always contains a prime ideal of height $1$. I.e, we have shown that our canonical section $1$ has a zero at every point of $Z$. I would be grateful if you could point where my understanding is incorrect, or if I have made wrong deductions. –  Jan 18 '14 at 03:48
  • Dear Benjamin, Regarding ``the canonical section $1$'', I have added an answer about this, which I hope will help. Otherwise, you are right that the section $1$ vanishes along $Z$. Indeed, that's the whole point: in more geometric, less function-theoretic, terms, you have chosen an effective Cartier divisor that contains $Z$ and doesn't contain $x$. (This disivor is actually the locus of poles of $f$, i.e. $(f)_{\infty}$.) Regards, – Matt E Jan 19 '14 at 00:07
  • Dear @MattE, thanks for writing your answer below. Just to be absolutely certain, am I right in saying (in my comment above) that the canonical rational section $1$ is sent to $g_i \in \mathcal{O}(U_i)$ on a trivialization $\mathcal{L}(D)|{U_i} \cong \mathcal{O}|{U_i}$? Thanks. –  Jan 19 '14 at 11:44
  • Dear Benjamin, Yes (up to an invertible element, which can be taken to be $1$ if you choose $g_i$ appropriately; note that writing $D_{| U_i} = (g_i)$ only determines $g_i$ up to a unit). Cheers, – Matt E Jan 19 '14 at 14:29

1 Answers1

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Regarding the section $1$ of of $\mathcal L(D)$, I have discussed this here, but it won't hurt to recall some of the ideas again.

To understand this, I think it helps to first consider how to convert an arbitrary line bundle (with a section) into one of the form $\mathcal L(D)$. So, start with a line bundle $\mathcal L$ and a (regular --- but this is equivalent to non-zero when the base is integral) section $s$, cutting out an effective Cartier divisor $D$ as its zero locus.

Now suppose that $s'$ is any other section of $\mathcal L$. The ratio $s'/s$ is then a rational function on the base $X$, and furthermore it has poles contained in $D$ (because away from $D$ the section $s$ doesn't vanish, and so the ratio $s'/s$ is actually a regular function, not just a rational one).

Conversely, if $f$ is a rational function whose poles are contained in $D$, then $f s$ is a well-defined section of $\mathcal L$ (the possible poles of $f$ are cancelled out by the fact that $s$ vanishes along $D$).

Thus we get an isomorphism between the space of sections of $\mathcal L$, and the space of rational functions whose poles are contained in $D$. Furthermore, the original section $s$ maps to the rational function $1$.

Conversely, if we start with an effective Cartier divisor $D$, we can make it the zero section of a line bundle by reversing the above construction: we are going to declare that the sections of the line bundle are the rational functions that have poles contained in $D$ (this actually makes sense, because we can first of all restrict our attention to some open subset $U \subset X$ just by intersecting $D$ with $U$ and talking about rational functions on $U$ with poles contained in $U \cap D$, so we really are describing a sheaf).

That we get an invertible sheaf follows from the fact that $D$ is a Cartier divisor. And now a little thought shows that the rational function $1$, when we think of it as a section of the bundle $\mathcal L(D)$ we have just constructed, does vanish precisely along $D$.

Incidentally, I think this is one of the really amazing constructions in algebraic geometry: that beginning just with a divisor, which is a very specific, localized object, you can produce the much more amorphous line bundle $\mathcal L(D)$ out of it, which together with the associated section $s$ gives back $D$, and whose other sections describe all the ways of deforming $D$ (in its linear equivalence class). And conversely, given something amorphous like a line bundle, by choosing a section you can convert it into something much more specific and hands-on: a particular divisor $D$.

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