The Stacks project

In this tag and also 42.57.1, is it $\prod_{p\ge0}A^p(X)\otimes \mathbb Q$ instead of $A^*(X)\otimes \mathbb Q$ or somehow we can prove that the Chern classes are finite polynomials?

It would be helpful to cite equation (0AG2) when saying "the induced map $M_{0} \to \Gamma(X,\mathcal{F})$ is identity" which is confusing as written. Instead, replace it with "the induced map $M_{0} \to \Gamma(X, \mathcal{M}) \to \Gamma(X, \mathcal{F})$ (see tag(0AG2)) is identity"

And similarly this one can follow from other compatibility properties of first Chern classes.

I think this proposition can follow from other compatibility properties of Gysin homomorphisms and 42.35.1.

I think we can have a better proof for this one if we use double induction on rank of $\mathcal E_1$ and $\mathcal E_2$, we can use splitting principle to reduce the invertible sheaf case, which is treated before.

Good point: we do not need to shrink and in fact we do not want to shrink as we want our finite etale covering to be over $X$. Thanks!

I think we have to add a bar over $U$ in the second line of the proof:

you proved that $\eta$ is in $U$ for each $U$ open affine in $X$, then $X=\bar{\{\eta\}}=\bar{U}$, then each open affine set is dense in $X$.

Since open affine are basis for the topology, each open non empty set is dense in $X$, then each open non empty set in $U$ is dense in $U$ and we conclude that $U$ is irreducible for each $U$ open (why do we require affine?) set of $X$.

You write "After possibly shrinking further we may assume we may assume $\mu$ defines a a commutative $\mathcal{O}_X$-algebra structure on $\mathcal{F}$ compatible with the given algebra structures on $\mathcal{F}_n$. "

Is it clear wy we might need to shrink?

Indeed. The rest of the sentence does make sense.

In the proof we only do this reduction when we already have $W$ as a subset and $T$ is affine. By Definition 34.9.1 the covering can be refined by a standard one, say $\{V_j \to T\}$ and $V_j$ maps to $T_{i(j)}$ over $T$. Then the pullback of $W$ to $V_j$ is also the pullback of $W_{i(j)}$ and hence open. Maybe this should be added?

In the first paragraph it reads "This is not a contradiction with Theorem 97.16.1 as the 1-morphism $(Sch/U)_{fppf} \rightarrow [U/R]$ is not representable by algebraic spaces in general, ..."

Maybe I misunderstood something, but I thought $[U/R]$ having diagonal representable by algebraic spaces (as claimed earlier in the same paragraph) implies that for any algebraic space $V$, any morphism $V \rightarrow [U/R]$ is representable by algebraic spaces? (Lemma 94.10.11(7)).

Could you please tell me how to reduce it to the case that $\{f_i:T_i\to T\}$ is a standard fpqc covering?

Because $\sum \sigma(a_{ij})y_j$ is in $M$ by the first conclusion applied to $x_i \otimes 1$. But I think we need to rewrite the proof as it is too succint.

Some typos. Part (2) of statement: "then so does $\lim_{\mathcal{I}}M$" should be "then so does $\lim_{\mathcal{I^{opp}}}M$" 4th paragraph of proof: 2th/3rd line should read "... we may construct a cocone in $\mathcal{I} \times \omega$ for the diagram $\mathcal{F} \rightarrow \mathcal{I} \times \omega$." 4th line: the order of the maps is reversed, it should be "$f_i = f_{i'} \circ f$".

It seems to me AB5* should state 'cofiltered limits' are exact (since by Lemma 4.12.5 limits of directed inverse systems coincide with cofiltered limits).

Could you please tell me how to deduce that $n$ is in the image of $A\otimes_R M\to N$? I cannot see it from the fact that $\sum\sigma(a_i)x_i\in M$ and $\sum _ i a_ i (\sum _ j \sigma (a_{ij}) y_ j) = \sum a_ i x_ i = n$.

I think you reversed the unit and counit at the beginning.

In the last formula $c(\mathcal F)$ should be replaced by $c(\mathcal F\otimes \mathcal G)$

In Definition 01B2, require $U$ to contain $x$.

Just a very tiny suggestion: in the proof of 7.10.10 (2), it will be more understandable if you use $s_i|_{U_i\times_U U_{i'}}=s_{i'}|_{U_i\times_U U_{i'}}$ instead of $s_i|_{U_i\times_U U_{j}}=s_{j}|_{U_i\times_U U_{j}}$, because you used label $j$ again later, and your $j$ here actually corresponds to later's $i'$. But again, the proof is perfect!