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Let $A$ be a ring. Assume it has some nice properties if necessary, e.g., $A$ is a Dedekind domain.

Let $\mathbf{P}^1_A$ be the projective line over $A$. I want to show that $-2 [\infty]$ is a canonical divisor on $\mathbf{P}^1_A$.

How do I do this?

I know how to do this when $A=k$ is a field. In this case, the identity morphism $\mathbf{P}^1_k\to \mathbf{P}^1_k$ is a rational function $f$ and div $df$ is a canonical divisor. One easily computes that div $df = -2 [\infty]$.

I might actually be wrong...In this case, how do I compute a canonical divisor for $\mathbf{P}^1_A$. What if $A$ is a Dedekind domain?

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Over an arbitrary basis, the canonical divisor on $X\to S$ is defined by the sheaf of differential $\Omega_{X/S}^1$. If you write $X$ (the projective line) as union of $U=\mathrm{Spec}A[T]$ and of $V=\mathrm{Spec}A[T^{-1}]$, then $$\Omega^1_{U/S}=A[T]dT, \quad \Omega^1_{V/S}=A[1/T]d(1/T).$$ On the intersection, we have $T^{-2}dT= -d(1/T)$, so $T^{-2}dT$ is a global section of $O_X(2)\otimes\Omega^1_{X/S}$ and we check that this global section induces an isomorphism $O_X\simeq O_X(2)\otimes \Omega^1_{X/S}$. So $$\Omega^1_{X/S}\simeq O_X(-2).$$

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Just to elaborate: with these charts we see that $-2 [\infty]$ is indeed a canonical divisor on $X$. –  Harry Nov 24 '12 at 10:30
    
@Harry, Yes thanks ! –  user18119 Nov 24 '12 at 18:04

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