Remark 23.3.6 (07GY)—The Stacks project

Remark 23.3.6. The forgetful functor $(A, I, \gamma ) \mapsto A$ does not commute with colimits. For example, let

\[ \xymatrix{ (B, J, \delta ) \ar[r] & (B'', J'', \delta '') \\ (A, I, \gamma ) \ar[r] \ar[u] & (B', J', \delta ') \ar[u] } \]

be a pushout in the category of divided power rings as discussed in Remark 23.3.5. Then in general the map $B \otimes _ A B' \to B''$ isn't an isomorphism. An explicit example is given by $(A, I, \gamma ) = (\mathbf{Z}, (0), \emptyset )$, $(B, J, \delta ) = (\mathbf{Z}/4\mathbf{Z}, 2\mathbf{Z}/4\mathbf{Z}, \delta )$, and $(B', J', \delta ') = (\mathbf{Z}/4\mathbf{Z}, 2\mathbf{Z}/4\mathbf{Z}, \delta ')$ where $\delta _2(2) = 2$ and $\delta '_2(2) = 0$. More precisely, using Lemma 23.5.3 we let $\delta $, resp. $\delta '$ be the unique divided power structure on $J$, resp. $J'$ such that $\delta _2 : J \to J$, resp. $\delta '_2 : J' \to J'$ is the map $0 \mapsto 0, 2 \mapsto 2$, resp. $0 \mapsto 0, 2 \mapsto 0$. Then $(B'', J'', \delta '') = (\mathbf{F}_2, (0), \emptyset )$ which doesn't agree with the tensor product.

Comments (3)

Comment #7050 by nkym on February 12, 2022 at 19:41

$\delta$ is not a divided power structure since $\delta_2(\delta_2(2)) = 2\neq 0 = \frac{4!}{(2!)^2 2!}\delta_4(2)$ .

Comment #7051 by Johan on February 12, 2022 at 21:25

Indeed, we should set $\delta_4(2) = 2$ and so on. A divided power structure $\delta$ does exist because we can apply Lemma 23.5.3 to the map $\delta_2 : I \to I$ mapping $2$ to $2$ and $0$ to $0$ . Thanks for pointing this out! I will fix this the next time I go through all the comments.

Comment #7243 by Johan on April 30, 2022 at 16:07

This has been fixed in this commit.

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