MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X$ be a normal variety over an algebraically closed field $k$ of characteristic $p>0$ and $S$ be a prime Weil divisor on $X$ which is normal too. Now if $K_X+S$ is NOT $\mathbb{Q}$-Cartier, can I still say that $(K_X+S)|_S=K_S+\text{ Diff}(0)$ ?

I have looked at the following two books by Janos Kollar: (1) Flips and Abundance for Algebraic Threefolds and (2) Singularities of the Minimal Model Program.

On the 1st book, in Proposition $16.5$ he says the above relation will be true if $K_X+S$ is Cartier at all points $P\in S$ such that $\text{codim}_S\ P=1$.

In the 2nd book, Chapter $4$, Page $153$, in the definition of ''Different II'' he said that $K_X+S$ is $\mathbb{Q}$-Cartier at all codimension $1$ points of $S$ is enough for the adjunction formula. But then on a paragraph before the definition of ''Different II'' he says that the $\mathbb{Q}$-Cartierness of $K_X+S$ at codimension $1$ points of $S$ is automatically satisfied if $X$ and $S$ both are normal and I am unable to see this fact!

Since $X$ is normal, $\text{codim }X_{\text{sing}} \geq 2$, so if I choose a codim $1$ point $P\in S$ which is a codim $2$ point of X then I can not really say that $P\in X-X_{\text{sing}}$, so how can I conclude that $K_X+S$ is $\mathbb{Q}$-Cartier at $P$ ? I do realize that this $P$ lies the smooth locus of $S$ since $S$ is normal but that doesn't seem to be helping here!

share|cite|improve this question
What is a Devisor ? – Niels Apr 17 '13 at 11:46
That is just a typo! I mean Divisor, sorry about that! – Omprokash Apr 17 '13 at 14:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.