# fpqc

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 Jan22 accepted When do we have $\operatorname{depth}_{A} B = \operatorname{depth}_B B$? Jan22 comment When do we have $\operatorname{depth}_{A} B = \operatorname{depth}_B B$? Dear Youngsu, thanks for your answer. I do wonder now if there's a way of expressing depth in terms of local cohomology. Do you mind giving me a reference? Jan21 revised Hartshorne's Exercise II.5.1 - Projection formula added 340 characters in body Jan21 comment Hartshorne's Exercise II.5.1 - Projection formula I posted my solution above. Jan21 answered Hartshorne's Exercise II.5.1 - Projection formula Jan21 revised Hartshorne's Exercise II.5.1 - Projection formula added 20 characters in body Jan21 comment Hartshorne's Exercise II.5.1 - Projection formula There are some technicalities that are being glossed over in the grey box. In points 1,2 by Hom you mean global hom. In 3, you have sheaf hom. One is an abelian group the other a sheaf! Be careful! Jan21 comment When do we have $\operatorname{depth}_{A} B = \operatorname{depth}_B B$? @user115654 Sorry I forgot to add that, indeed that is my situation here. Jan21 revised When do we have $\operatorname{depth}_{A} B = \operatorname{depth}_B B$? added 45 characters in body Jan20 asked When do we have $\operatorname{depth}_{A} B = \operatorname{depth}_B B$? Jan19 revised Projections are finite morphisms deleted 40 characters in body Jan19 answered Projections are finite morphisms Jan19 comment Some questions on Hartshorne III Ex 6.8 Dear @MattE, thanks for writing your answer below. Just to be absolutely certain, am I right in saying (in my comment above) that the canonical rational section $1$ is sent to $g_i \in \mathcal{O}(U_i)$ on a trivialization $\mathcal{L}(D)|_{U_i} \cong \mathcal{O}|_{U_i}$? Thanks. Jan19 accepted Some questions on Hartshorne III Ex 6.8 Jan18 comment Prove $\ker {T^k} \cap {\mathop{\rm Im}\nolimits} {T^k} = \{ 0\}$ Hint: $\ker T^k = \ker T^{k+1} = \ldots$ when $k =n$ Jan18 revised Some questions on Hartshorne III Ex 6.8 edited title Jan18 revised Some questions on Hartshorne III Ex 6.8 added 113 characters in body Jan18 comment Some questions on Hartshorne III Ex 6.8 Now it is clear that ${g_i}_{P}$ is in the maximal ideal of $A_P$ where $P$ is a prime of codimension $1$ in $A$. But now this means that ${g_i}_Q$ is also in the maximal ideal of $A_Q$ for any prime $Q$ because $Q$ always contains a prime ideal of height $1$. I.e, we have shown that our canonical section $1$ has a zero at every point of $Z$. I would be grateful if you could point where my understanding is incorrect, or if I have made wrong deductions. Jan18 comment Some questions on Hartshorne III Ex 6.8 Dear @MattE, this is what I understand at the moment. Let $D = (f)_{\infty}$. Choose an affine cover $U_i$ of $X$ such that $D|_{U_i} = (g_i)$ where $g_i \in K(X)^\ast$. Here we use the locally factorial hypothesis. Then because $g_i$ has no poles on $U_i$, it is actually regular and the canonical section $1 \in H^0(X,\mathcal{L}(D))$ corresponds to $g_i \in \mathcal{O}(U_i)$. Now choose a point $z \in Z$, $z \in U_i$ for some $i$. As $Z \cap U_i \to U_i$ is a closed immersion into an affine, we can write $Z \cap U_i = \operatorname{Spec} A$ (continued) Jan17 revised Some questions on Hartshorne III Ex 6.8 added 11 characters in body