I have a question about a point in the proof of Theorem 1.52 from Janos Kolloar's Lecture on Resolution of Singularities (page 33):
THEOREM 1.52 (Weak embedded resolution, I). Let $S_0$ be a smooth projective surface over a perfect field $k$ and $C_0 \subset S_0$ a reduced projective curve. After finitely many steps, the weak embedded resolution algorithm (1.42) stops with a smooth curve $C_m \subset S_m$. (we consider 1.42 as black box)
Proof. We look at the intersection number $C \cdot (C +K_S)$ and prove two properties:
$$(1.52.1) \ \ \ \ \ \ C_{i+1} \cdot (C_{i+1} +K_{S_{i+1}}) < C_i \cdot (C_i +K_{S_i}) $$
$$(1.52.2) \ \ \ \ \ \ C_i \cdot (C_i +K_{S_i}) \text{ is bounded from below } $$
The proof of (1.52.1) is a straightforward local computation using (1.53). (black box)
Now the proof of (1.52.2) is interesting. The author uses two methods and I not understand an argument of Method 1:
Method 1 using duality theory. If you know enough duality theory (say, as in [Har77, Sec.III.8]) then you know that $\mathcal{S}_S(C + K_S) \vert _C $ is isomorphic to the dualizing sheaf $\omega_c$. Then we claim that
$$deg \ \omega_C \ge (-2) \cdot \# \{ \text{ irreducible geometric components of } C \}$$
If $C$ is irreducible over an algebraically closed field $k$, pick two smooth points $p, q \in C$. From
$$ 0 \to \mathcal{O}_C(-p-q) \to \mathcal{O}_C \to k_p + k_q \to 0 $$
we conclude that $h^1(C,\mathcal{O}_C(-p-q)) > 0$. (that because $h^0(C, \mathcal{O}_C)=1$ ($C$ irreducible & $k$ alg closed) & $\dim_k(k_p + k_q)=2$ ).
Since $H^1(C,\mathcal{O}_C(-p-q))$ is dual to $H^0(C, \mathcal{O}_C(p+q)) \otimes \omega_C) = Hom(\mathcal{O}_C(-p-q), \omega_C)$ by Serre-duality & definition of $\omega_C$, we conclude that $ deg \ \omega_C \ge deg \ \mathcal{O}_C(-p-q))=-2$ (???)
Now the author says what I not understand:
(It is here that we use that $C$ is reduced, and hence $\omega_C$ has no nilpotents. Otherwise, the homomorphism $\mathcal{O}_C(-p-q) \to \omega_C$ could not be used to bound the degree.)
Question: This last argument that $C$ reduced allows to bound to degree I not understand.
What does the author mean by this argument? And which role plays that $C$ is reduced? My first guess was that he intend to use the aditive property of degree: Let
$$ 0 \to \mathcal{F}_1 \to \mathcal{F}_2 \to \mathcal{F}_3 \to 0 $$
exact sequence of coherent sheaves on $C$, then $deg \ \mathcal{F}_2 = deg \ \mathcal{F}_1 + deg \ \mathcal{F}_3 $
To apply this we need to assure that $\mathcal{O}_C(-p-q) \to \omega_C$ is injective. Why the fact that $\omega_C$ has no nilpotents imply it? It seems that the author possibly confuses here "nilpotents" with "torsion elements".
The question is which problems can occure making the morphism $\mathcal{O}_C(-p-q) \to \omega_C$ not injective? It's a local question and $\mathcal{O}_C(-p-q) $ is locally invertible, $\omega_C$ is coherent.
Do we know more about the dualizing $\omega_C$ ? If the curve $C$ smooth then it's invertible. We not assumed smoothness. Is $\omega_C$ nevertheless at least locally free?
That is we asking when a non zero $R$-module map $\varphi: R \to M$ with $M$ coherent can be not injective? It is determíned by $1_R \to m \in M$. Thus the injectiveness of $\varphi$ holds if $m$ is not a torsion element.
Why the author refers to assumtion on reducedness of $C$?
Can anybody explain the argument behind?