I cannot follow the end of the proof of proposition III.9.7. The situation is as follows. We have a map $f:X\to Y$ of schemes with $Y$ integral, regular of dimension 1. The proposition says that $f$ is flat if and only if every associated point of $X$ maps to the generic point of $Y$. It's the if-part I cannot follow. Hartshorne takes $x\in X$ and assumes $y:=f(x)$ is a closed point. Then $\mathcal{O}_{Y,y}$ is a DVR so we can choose a uniformizing parameter $t$. We want to show $\mathcal{O}_{X,x}$ is torsion free so to get a contradiction we assume $t$ is a zero divisor on $\mathcal{O}_{X,x}$. This means $f^\#(t)$ lands in some associated prime $P\subset \mathcal{O}_{X,x}$ of $(0)$ and $P$ corresponds to an associated point $x'\in X$. Hartshorne claims that $f(x')=y$ which would be a contradiction but I cannot figure out why $x'$ could not map to the generic point of $Y$. Any help is appreciated!
Hartshorne proposition III.9.7
algebraic-geometrycommutative-algebrasheaf-theory
Related Solutions
Coherent has a finiteness assumption which can be fail very easily. Consider the natural projection $\pi:\Bbb A^1_k\to \operatorname{Spec} k$. Then $\pi_*(\mathcal{O}_{\Bbb A^1_k})$ is an infinite-dimensional vector space which is not coherent.
The rank of $\mathcal{F}$ at a point $x\in X$ is $\dim_{k(x)} \mathcal{F}_x\otimes k(x)$, and when we talk about rank of a sheaf, we usually mean at the generic point. Since there is a neighborhood of the generic point where normalization is an isomorphism, we see that the rank of $f_*\mathcal{O}_X$ is 1.
A normal variety is singular in codimension two, which means that a normal curve is smooth. The same method you used to show that the (proper, birational) blowdown map from a smooth curve to your node has two points in the fiber over the node should work here.
Consider the indicator functions of the distinct points in the fiber (the function which is $1$ on that point and $0$ on the other point of the fiber). There's no $\mathcal{O}_Y$ relation between them (no function on $Y$ can tell the two points apart since they both map to the same point of $Y$), so they have to be independent.
In a line bundle, any stalk of a nonvanishing section should be a generator of the stalk, so there should be an $\mathcal{O}_Y$ relation between the two sections. This contradicts 4.
Writing $D=\sum n_iD_i$ for $D_i$ prime, we have $\pi^*D=\sum n_i\pi^*D_i$. But $\pi^*D_i=\pi^{-1}(D_i)$, and $\pi(\pi^{-1}(D_i))=D_i$, hence the image of $D$ does not contain the generic point of $X$, and all the $\pi^{-1}D_i$ are divisors of type I.
If $f=g/h$ in lowest terms for $g,h\in K[t]$ so that at least one is in $K[t]\setminus K$, then the valuation of $f$ at a maximal ideal containing a zero of whichever $g,h$ is in $K[t]\setminus K$ is not zero. Such a maximal ideal is a point in the fiber of $X\times\Bbb A^1$ over the generic point of $X$, so the divisor of $f$ contains a type II divisor.
Yes, we already know we're surjective on type I divisors: if $D\subset X\times\Bbb A^1$ is a prime divisor of type 1, then $\pi(D)\subset X$ is a prime divisor which maps to $\pi^{-1}(\pi(D))=D$.
Your statement is not super clear, but the point is that if $D$ is a prime divisor of type II, then it must intersect the fiber of $X\times\Bbb A^1\to X$ over the generic point of $X$ by definition of a type II divisor. As this fiber is exactly $\operatorname{Spec} K[t]$ and the divisor cannot contain the generic point of $\operatorname{Spec} K[t]$ (this is the generic point of $X\times\Bbb A^1$), we get a collection of closed points in $\operatorname{Spec} K[t]$. If we had multiple points in this, we'd have multiple irreducible components of our divisor, so there's only one point, and thus we get a prime divisor on $\operatorname{Spec} K[t]$. (Alternatively, any prime divisor on a scheme $X$ has a unique point with $\dim \mathcal{O}_{X,p}=1$ and no points with $\dim = 0$.)
Take $X=\operatorname{Spec} k[x]$ and $f=xt$. Then the divisor associated to $f$ on $\Bbb A^1\times\Bbb A^1=\Bbb A^2$ is the union of the axes, which is a type one divisor plus a type two divisor. There are worse examples, but this is one very accessible one; conversely, you might be able to fix this most of the time. Digging in to the stuff required to make it just a type II divisor is more work than you need to do here because we already know we're surjective on type I divisors, and so we don't have to figure it out to finish the proof.
Part 5 could make an interesting problem to think about further (when can one guarantee the divisor associated to $f$ is just a type II divisor?), but it's not necessary to finish this proof.
Best Answer
Are you asking why $f(x')=y$? Suppose $f:U=\operatorname{Spec} B \to V=\operatorname{Spec}A $ where $U\subset X, V\subset Y,A$ is an integral domain and $y\in V$. Since $y$ is a closed point, it corresponds to a maximal ideal $m$ of $A$. Since $dimA=1$ and $A$ is an integral domain, $(0)\subsetneq m$. We have the following commutative diagram: $$\require{AMScd} \begin{CD} A @>{\phi} >> B;\\ @VVV @VVV \\ A_{m} @>{f^{\#}}>> B_{n}; \end{CD}$$
Suppose $nB_{n}$ is not torsion free, then by definition $f^{\#}t$ is a zero divisor where $mA_{m}=(t)$ . By prop 1.11 and the remark below prop 4.7 of Atiyah and MacDonald's Commutative Algebra, $f^{\#}t$ is contained in some associated prime ideal $\tilde{\mathfrak{p}}$ of $B_{n}$. Let $\mathfrak{p}=\tilde{\mathfrak{p}}\bigcap B$, then $\phi^{-1}(\mathfrak{p})=m$ since $f^{\# -1}(\tilde{\mathfrak{p}})=(t)$. Hence $f(x')=y$, where $x'$ is the point corresponding to $\mathfrak{p}$. Contradiction.