I have heard that the canonical divisor can be defined on a normal variety X since the smooth locus has codimension 2. Then, I have heard as well that for ANY algebraic variety such that the canonical bundle is defined:


$$\mathcal{K}=\mathcal{O}\_{X,-\sum D\_i}$$


where the $D\_i$ are representatives of all divisors in the Class Group.


I want to prove that formula or I want to find a reference for that formula, or I want someone to rephrase it in a similar way if they heard about it.


Why do I want to prove it? Well, I use the definition that something is Calabi Yau if its canonical bundle is 0. In the case of toric varieties, $\sum D\_i$~0 if all the primitive generators for the divisors lie on a hyperplane. Then the sum is 0 and therefore the toric variety is Calabi-Yau.


Can someone confirm or fix the above formula? I do not ask for a debate on when something is Calabi-Yau, I handle that OK, I just ask whether the above formula is correct. A reference would be enough. I have little access to references at the moment.



The canonical line bundle of a normal variety

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> **Edit** (11/12/12): I added an explanation of the phrase "this is essentially equivalent to $X$ being $S\_2$" at the end to answer aglearner's question in the comments.
>  [See also [here](https://mathoverflow.net/questions/45347/why-does-the-s2-property-of-a-ring-correspond-to-the-hartogs-phenomenon/45354#45354) and [here](https://mathoverflow.net/questions/45347/why-does-the-s2-property-of-a-ring-correspond-to-the-hartogs-phenomenon/45616#45616)]
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Dear Jesus,


I think there are several problems with your question/desire to define a canonical divisor on *any* algebraic variety. 


First of all, what is *any* algebraic variety? Perhaps you mean a quasi-projective variety (=reduced and of finite type) defined over some (algebraically closed) field.


OK, let's assume that $X$ is such a variety. Then what is a *divisor* on $X$? Of course, you could just say it is a formal linear combination of *prime divisors*, where a prime divisor is just a codimension 1 irreducible subvariety. 


OK, but what if $X$ is not equidimensional? Well, let's assume it is, or even that it is irreducible. 


Still, if you want to talk about divisors, you would surely want to say when two divisors are *linearly equivalent*. OK, we know what that is, $D\_1$ and $D\_2$ are linearly equivalent iff $D\_1-D\_2$ is a *principal divisor*.


But, what is a principal divisor? Here it starts to become clear why one usually assumes that $X$ is normal even to just talk about divisors, let alone defining the canonical divisor. In order to define principal divisors, one would need to define something like the *order of vanishing* of a regular function along a prime divisor. It's not obvious how to define this unless the local ring of the general point of any prime divisor is a DVR. Well, then this leads to one to want to assume that $X$ is $R\_1$, that is, regular in codimension $1$ which is equivalent to those local rings being DVRs.


OK, now once we have this we might also want another property: If $f$ is a regular function, we would expect, that the zero set of $f$ should be 1-codimensional in $X$. In other words, we would expect that if $Z\subset X$ is a closed subset of codimension at least $2$, then if $f$ is nowhere zero on $X\setminus Z$, then it is nowhere zero on $X$. In (yet) other words, if $1/f$ is a regular function on $X\setminus Z$, then we expect that it is a regular function on $X$. This in the language of sheaves means that we expect that the push-forward of $\mathscr O\_{X\setminus Z}$ to $X$ is isomorphic to $\mathscr O\_X$. Now this is essentially equivalent to $X$ being $S\_2$.


So we get that in order to define *divisors* as we are used to them, we would need that $X$ be $R\_1$ and $S\_2$, that is, *normal*. 


Now, actually, one can work with objects that behave very much like divisors even on non-normal varieties/schemes, but one has to be very careful what properties work for them.


As far as I can tell, the best way is to work with *Weil divisorial sheaves* which are really reflexive sheaves of rank $1$. On a normal variety, the sheaf associated to a Weil divisor $D$, usually denoted by $\mathcal O\_X(D)$, is indeed a reflexive sheaf of rank $1$, and conversely every reflexive sheaf of rank $1$ on a normal variety is the sheaf associated to a Weil divisor (in particular a reflexive sheaf of rank $1$ on a regular variety is an invertible sheaf) so this is indeed a direct generalization. One word of caution here: $\mathcal O\_X(D)$ may be defined for Weil divisors that are not Cartier, but then this is (obviously) not an invertible sheaf.


Finally, to answer your original question about canonical divisors. Indeed it is possible to define a canonical divisor (=Weil divisorial sheaf) for all quasi-projective varieties. If $X\subseteq \mathbb P^N$ and $\overline X$ denotes the closure of $X$ in $\mathbb P^N$, then the dualizing complex of $\overline X$ is 
$$
\omega\_{\overline X}^\bullet=R{\mathscr H}om\_{\mathbb P^N}(\mathscr O\_{\overline X}, \omega\_{\mathbb P^N}[N])
$$
and the canonical *sheaf* of $X$ is
$$
\omega\_X=h^{-n}(\omega\_{\overline X}^\bullet)|\_X=\mathscr Ext^{N-n}\_{\mathbb P^N}(\mathscr O\_{\overline X},\omega\_{\mathbb P^N})|\_X
$$ 
where $n=\dim X$. (Notice that you may disregard the derived category stuff and the dualizing complex, and just make the definition using $\mathscr Ext$.) Notice further, that 
if $X$ is normal, this is the same as the one you are used to and otherwise it is a reflexive sheaf of rank $1$.


As for your formula, I am not entirely sure what you mean by "where the $D\_i$ are representatives of all divisors in the Class Group". For toric varieties this can be made sense as in Josh's answer, but otherwise I am not sure what you had in mind.



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> (Added on 11/12/12):
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**Lemma** A scheme $X$ is $S\_2$ if and only if for any $\iota:Z\to X$ closed subset of codimension at least $2$, the natural 
map $\mathscr O\_X\to \iota\_\*\mathscr O\_{X\setminus Z}$ is an isomorphism.


**Proof**
Since both statements are local we may assume that $X$ is affine.
Let $x\in X$ be a point and $Z\subseteq X$ its closure in $X$. If $x$ is a codimension at most $1$ point, there is nothing to prove, so we may assume that $Z$ is of codimension at least $2$.


Considering the exact sequence (recall that $X$ is affine):
$$
0\to H^0\_Z(X,\mathscr O\_X) \to H^0(X,\mathscr O\_X) \to H^0(X\setminus Z,\mathscr O\_X) \to H^1\_Z(X,\mathscr O\_X) \to 0
$$
shows that $\mathscr O\_X\to \iota\_\*\mathscr O\_{X\setminus Z}$ is an isomorphism
if and only if 
$H^0\_Z(X,\mathscr O\_X)=H^1\_Z(X,\mathscr O\_X)=0$ the latter condition is equivalent to 
$$
\mathrm{depth}\mathscr O\_{X,x}\geq 2,
$$
which given the assumption on the codimension is exactly the condition that $X$ is $S\_2$ at $x\in X$. $\qquad\square$


