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I was recently reminded about this old question on math.stackexchange.

Let $\operatorname{Mod}R$ be the category of (right) modules for a ring $R$. The questioner mistakenly thought that the Freyd-Mitchell embedding theorem implied that for every ring $R$ there was another ring $S$ and a full embedding $(\operatorname{Mod}R)^{\text{op}}\to\operatorname{Mod}S$, and asked for an explicit description of such an $S$.

I realized (see this answer) that general facts about locally presentable categories imply that, if Vopěnka's principle is true, then there is no such embedding (for nonzero $R$).

Is there a proof that doesn't assume Vopěnka's principle? Or could it be that this fact genuinely depends on very strong large cardinal assumptions?

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  • $\begingroup$ I'm accepting Jiří Rosický's excellent answer, as although it doesn't definitively answer the specific questions I asked, it greatly clarified for me the state of the art. I'm still interested in any other contributions, though. $\endgroup$ Commented Jan 19, 2021 at 16:09
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    $\begingroup$ Assuming (M), every concrete category is boundable (see A. Pultr and V. Trnková, Combinatorial, algebraic and topological representations of groups, semigroups and categories, North-Holland 1980). Hence, every $(\bf{Mod}$ $R$)$^{\text{op}}$ is boundable. $\endgroup$ Commented Jan 25, 2021 at 10:26

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Assuming (M) (= there is only a set of measurable cardinals), the category $\bf{Vec}$ of vector spaces (over every field) has a small dense subcategory. This is an old result of Isbell (see also https://arxiv.org/pdf/1812.10649.pdf). Hence $\bf{Vec}$$^{\text{op}}$ is boundable, i.e., it can be fully embedded to a category of algebras (I do not know whether to a category of modules). Hence the set-theoretical strength of "no opposite category of modules is boundable" lies between $\neg$(M) and VP.

Moreover, the existence of a full embedding of $\bf{Ab}^{op}$ to $\bf{Mod}$ $R$ implies WVP (weak Vopěnka's principle) is false. Indeed, the accessible category of graphs with monomorphisms can be fully embedded to the category $\bf{Gra}$ of graphs. A. J. Przezdziecki https://arxiv.org/pdf/1104.5689.pdf constructed an embedding $G:\bf{Gra}\to\bf{Ab}$ such that $\bf{Ab}$$(GX,GY)$ is the free abelian group on $\bf{Gra}$$(X,Y)$. This yields $R$-modules $A_i$ indexed by ordinals such that the only morphisms $A_i\to A_j$, $i<j$ are $0$ and there are non-zero morphisms $A_j\to A_i$ for every $i<j$. Let $X_i\subseteq A_i$ consist of elements $x_i\neq 0$ such that for every $i<j$ there is $y_j\in A_j$ and $f:A_j\to A_I$ with $x_i=f(y_j)$. We have $X_i\neq\emptyset$ for every $i$. Otherwise, every element $0\neq z\in A_i$ would have a height $k$, i.e., the smallest ordinal such that $z$ is not in the image of any $f:A_k\to A_i$. Since these heights are arbitrarily large and $A_i$ is a set, we get a contradiction. Add to $\bf{Ab}$ a unary relation and interpret it on $A_i$ as $X_i$. Homomorphisms $A_j\to A_i$, $i<j$ preserve these relations but $0:A_i\to A_j$, $i<j$ do not. Hence sWVP (semiweak Vopěnka's principle) does not hold. Following the recent result of T. M. Wilson https://arxiv.org/pdf/1909.09333.pdf, WVP does not hold.

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    $\begingroup$ Very interesting, thank you! Does this rely on very special properties of vector spaces, or would you expect a similar statement to be true for arbitrary module categories? $\endgroup$ Commented Jan 17, 2021 at 9:27
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    $\begingroup$ It seems that it depends on the double dualization on vector spaces. It would be interesting to know what happens for abelian groups. I could show that WVP (weak VP) suffices for $\bf{Ab}$ not being boundable but I do not see that it really depends on set theory. $\endgroup$ Commented Jan 17, 2021 at 10:02
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    $\begingroup$ Thanks again. Is it also true that WVP is sufficient to prove that $(\operatorname{Mod}R)^\text{op}$ is not boundable for any ring $R$? $\endgroup$ Commented Jan 17, 2021 at 12:38
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    $\begingroup$ No, it works for $\bf{Ab}$ only and uses the result of A. J. Przezdziecki. I will add the argument to my answer. $\endgroup$ Commented Jan 18, 2021 at 9:00
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    $\begingroup$ Thanks very much for the details. I think the first sentence of the second paragraph should say "implies WVP is false" rather than "implies WVP"? $\endgroup$ Commented Jan 18, 2021 at 9:39

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