A new user who doesn't have enough reputation to comment (Max Briest) wants to know whether there is a typo in the third short exact sequence, where the kernel ought to be $\mathcal{S}^\vee \otimes \mathcal{S}^\perp/\mathcal{S}$.

@Ben123 Hmm, "no good deed goes unpunished". Don't ask me about the technical details of the nLab software, but tikzcd became integrated with it long after I wrote that diagram up. Meanwhile, due to further technical reasons which I won't explain, I am currently unable to edit my articles there. Finally, I use notation that I am accustomed to and that I find satisfying. In particular, I'm not going to bend over backwards to try to suit everyone's taste. Perhaps your time is better spent grasping the essence of the shorter, condensed version that I thoughtfully supplied?

@KevinCarlson Oh, this is a nice observation -- I'll remember this. Thanks!

I have trouble following the reasoning pretty much by the first display line. However, the answer to the question "Does the following statement necessarily hold? 'Since $m^2 - p^k$ is not a square, then it is between two (consecutive) squares.' " is yes, and the proof is very easy. Since the natural numbers are well-ordered, there is a least square $M^2$ greater than this $N := m^2 - p^k$. Then $(M-1)^2 \leq N$; otherwise, $M^2$ wouldn't have been the least square. $(M-1)^2 < N$ since $N$ is assumed not to be a square. So $(M-1)^2 < N < M^2$.

A flag has been raised with respect to the large number of edits. The trouble is that each edit bumps the post to the front page, at the expense of other posts also vying for attention, and this is annoying to the community. Please try to condense as many edits as you can into a single revision.

@SaúlRM Yes indeed, it's generally better to give an answer than just to answer in the comments, if the question belongs on the site. (I'd say this question is borderline, in the sense that it would have been widely considered acceptable for MO years ago, but not as much anymore.) Despite the incongruity, I wouldn't say you did anything wrong. What I would consider wrong is to both vote to close and to answer, but you didn't do that. Hope this all makes sense!

It's slightly incongruous to say that a question belongs somewhere else, and then to answer it as though it belongs here.

Renan, I'm only just seeing this question. If you're looking for some sort of synthetic description of morphisms, like "morphisms in the category of rings are defined to be functions between their underlying sets that preserve the ring operations", then there is no such description. The objects and morphisms of Lawvere's category are constructed formally by recursively applied rules, just like types and terms and formulas for a first-order theory are built up recursively.

@DenisT Yes indeed, $\hat{\mathbb{Z}}$ can also be thought of as the algebraic double dual: $$\hat{\mathbb{Z}} \cong \hom(\hom(\mathbb{Z}, \mathbb{Q}/\mathbb{Z}), \mathbb{Q}/\mathbb{Z}).$$ One pleasant thing about $\hat{\mathbb{Z}} \cong \hom(\mathbb{Q}/\mathbb{Z}, \mathbb{Q}/\mathbb{Z})$ is that the ring multiplication on $\hat{\mathbb{Z}}$ comes from the composition of endomorphisms $\mathbb{Q}/\mathbb{Z} \to \mathbb{Q}/\mathbb{Z}$.

You can write the moderators at [email protected], as mentioned here. The only other way I can think of is to open a chat room. (Last time I tried to create a private chat room, it didn't work.)

Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on MathOverflow Meta, or in MathOverflow Chat. Comments continuing discussion may be removed.

I don't know, but I'll give you an upvote anyway.