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One part of your question asks for naturally occuring categories (which are triangulated, presumably, since triangulated categories are the categorical setting for earlier parts of the question) whichwhose $K_0$-groups are vector spaces over finite fields.

One source of such triangulated categories is stable module categories of quasi-Frobenius rings. Recall that the "stable module category" of a ring $R$ has, as objects, the $R$-modules, and as morphisms from $M$ to $N$, the set of equivalence classes of $R$-module morphisms $M\rightarrow N$, where two morphisms are said to be equivalent if their difference factors through a projective $R$-module. Recall also that a ring is "quasi-Frobenius" if its projective modules coincide with its injective modules. The stable module category of a quasi-Frobenius ring $R$ has a natural triangulation; I will write $stMod(R)$ for that triangulated category, restricted to its finitely generated modules, in order to avoid an Eilenberg swindle argument which would cause $K_0$ to vanish.

Consider the example $R = \mathbb{Z}/4\mathbb{Z}$. Every finitely-generated $R$-module is a direct sum of copies of $\mathbb{Z}/2\mathbb{Z}$ and $P = \mathbb{Z}/4\mathbb{Z}$, and $P$ is zero in the stable module category of $R$. Hence $K_0(stMod(R))$ must be a cyclic group generated by $[\mathbb{Z}/2\mathbb{Z}]$. However, we have the short exact sequence $$ 0 \rightarrow \mathbb{Z}/2\mathbb{Z}\rightarrow P \rightarrow \mathbb{Z}/2\mathbb{Z} \rightarrow 0$$ of $R$-modules, and consequently $[\mathbb{Z}/2\mathbb{Z}] + [\mathbb{Z}/2\mathbb{Z}] = [P] = 0$ in $K_0(stMod(R))$. So we have $K_0(stMod(R)) \cong \mathbb{Z}/2\mathbb{Z}$.

By a similar argument, for any prime $p$ and any positive integer $n$, we have $K_0(stMod(\mathbb{Z}/p^n\mathbb{Z})) \cong \mathbb{Z}/n\mathbb{Z}$. We also have $K_0(stMod(k[x]/x^n)) \cong \mathbb{Z}/n\mathbb{Z}$ for any field $k$. All that is really necessary for these kinds of $K_0$ calculations is that the ring $R$ be quasi-Frobenius, so that its stable module category is triangulated, and that $R$ is an Artinian principal ideal ring, so that its modules are all direct sums of cyclics.

Consequently, if $p$ is a prime and you want to recover $\mathbb{F}_p^n$ as $K_0$ of a triangulated category, then the triangulated category $stMod(R)$ will work, where $R$ is a Cartesian product of $n$ copies of $\mathbb{Z}/\ell^p\mathbb{Z}$ for any prime $\ell$ (or, alternatively, $n$ copies of $k[x]/x^p$ for any field $k$).


You also ask about constructing triangulated categories whose $K_0$ is a vector space over, for example, the rational numbers. This is much more difficult than the finite field case. It is a nontrivial task to construct a triangulated category whose $K_0$ is $\mathbb{Q}$, or even just $\mathbb{Z}[\frac{1}{2}]$; this is tied up in the general problem of figuring out how to modify a given triangulated category so that a prime becomes inverted in its $K_0$. The papers https://arxiv.org/abs/1610.07162 and https://arxiv.org/abs/1702.07466 are about this problem, and you can find positive results specifically about the $\mathbb{Q}$ case of your problem in the Barwick-Glasman-Hoyois-Nardin-Shah paper (the first arXiv link).

One part of your question asks for naturally occuring categories (which are triangulated, presumably, since triangulated categories are the categorical setting for earlier parts of the question) which are vector spaces over finite fields.

One source of such triangulated categories is stable module categories of quasi-Frobenius rings. Recall that the "stable module category" of a ring $R$ has, as objects, the $R$-modules, and as morphisms from $M$ to $N$, the set of equivalence classes of $R$-module morphisms $M\rightarrow N$, where two morphisms are said to be equivalent if their difference factors through a projective $R$-module. Recall also that a ring is "quasi-Frobenius" if its projective modules coincide with its injective modules. The stable module category of a quasi-Frobenius ring $R$ has a natural triangulation; I will write $stMod(R)$ for that triangulated category, restricted to its finitely generated modules, in order to avoid an Eilenberg swindle argument which would cause $K_0$ to vanish.

Consider the example $R = \mathbb{Z}/4\mathbb{Z}$. Every finitely-generated $R$-module is a direct sum of copies of $\mathbb{Z}/2\mathbb{Z}$ and $P = \mathbb{Z}/4\mathbb{Z}$, and $P$ is zero in the stable module category of $R$. Hence $K_0(stMod(R))$ must be a cyclic group generated by $[\mathbb{Z}/2\mathbb{Z}]$. However, we have the short exact sequence $$ 0 \rightarrow \mathbb{Z}/2\mathbb{Z}\rightarrow P \rightarrow \mathbb{Z}/2\mathbb{Z} \rightarrow 0$$ of $R$-modules, and consequently $[\mathbb{Z}/2\mathbb{Z}] + [\mathbb{Z}/2\mathbb{Z}] = [P] = 0$ in $K_0(stMod(R))$. So we have $K_0(stMod(R)) \cong \mathbb{Z}/2\mathbb{Z}$.

By a similar argument, for any prime $p$ and any positive integer $n$, we have $K_0(stMod(\mathbb{Z}/p^n\mathbb{Z})) \cong \mathbb{Z}/n\mathbb{Z}$. We also have $K_0(stMod(k[x]/x^n)) \cong \mathbb{Z}/n\mathbb{Z}$ for any field $k$. All that is really necessary for these kinds of $K_0$ calculations is that the ring $R$ be quasi-Frobenius, so that its stable module category is triangulated, and that $R$ is an Artinian principal ideal ring, so that its modules are all direct sums of cyclics.

Consequently, if $p$ is a prime and you want to recover $\mathbb{F}_p^n$ as $K_0$ of a triangulated category, then the triangulated category $stMod(R)$ will work, where $R$ is a Cartesian product of $n$ copies of $\mathbb{Z}/\ell^p\mathbb{Z}$ for any prime $\ell$ (or, alternatively, $n$ copies of $k[x]/x^p$ for any field $k$).


You also ask about constructing triangulated categories whose $K_0$ is a vector space over, for example, the rational numbers. This is much more difficult than the finite field case. It is a nontrivial task to construct a triangulated category whose $K_0$ is $\mathbb{Q}$, or even just $\mathbb{Z}[\frac{1}{2}]$; this is tied up in the general problem of figuring out how to modify a given triangulated category so that a prime becomes inverted in its $K_0$. The papers https://arxiv.org/abs/1610.07162 and https://arxiv.org/abs/1702.07466 are about this problem, and you can find positive results specifically about the $\mathbb{Q}$ case of your problem in the Barwick-Glasman-Hoyois-Nardin-Shah paper (the first arXiv link).

One part of your question asks for naturally occuring categories (which are triangulated, presumably, since triangulated categories are the categorical setting for earlier parts of the question) whose $K_0$-groups are vector spaces over finite fields.

One source of such triangulated categories is stable module categories of quasi-Frobenius rings. Recall that the "stable module category" of a ring $R$ has, as objects, the $R$-modules, and as morphisms from $M$ to $N$, the set of equivalence classes of $R$-module morphisms $M\rightarrow N$, where two morphisms are said to be equivalent if their difference factors through a projective $R$-module. Recall also that a ring is "quasi-Frobenius" if its projective modules coincide with its injective modules. The stable module category of a quasi-Frobenius ring $R$ has a natural triangulation; I will write $stMod(R)$ for that triangulated category, restricted to its finitely generated modules, in order to avoid an Eilenberg swindle argument which would cause $K_0$ to vanish.

Consider the example $R = \mathbb{Z}/4\mathbb{Z}$. Every finitely-generated $R$-module is a direct sum of copies of $\mathbb{Z}/2\mathbb{Z}$ and $P = \mathbb{Z}/4\mathbb{Z}$, and $P$ is zero in the stable module category of $R$. Hence $K_0(stMod(R))$ must be a cyclic group generated by $[\mathbb{Z}/2\mathbb{Z}]$. However, we have the short exact sequence $$ 0 \rightarrow \mathbb{Z}/2\mathbb{Z}\rightarrow P \rightarrow \mathbb{Z}/2\mathbb{Z} \rightarrow 0$$ of $R$-modules, and consequently $[\mathbb{Z}/2\mathbb{Z}] + [\mathbb{Z}/2\mathbb{Z}] = [P] = 0$ in $K_0(stMod(R))$. So we have $K_0(stMod(R)) \cong \mathbb{Z}/2\mathbb{Z}$.

By a similar argument, for any prime $p$ and any positive integer $n$, we have $K_0(stMod(\mathbb{Z}/p^n\mathbb{Z})) \cong \mathbb{Z}/n\mathbb{Z}$. We also have $K_0(stMod(k[x]/x^n)) \cong \mathbb{Z}/n\mathbb{Z}$ for any field $k$. All that is really necessary for these kinds of $K_0$ calculations is that the ring $R$ be quasi-Frobenius, so that its stable module category is triangulated, and that $R$ is an Artinian principal ideal ring, so that its modules are all direct sums of cyclics.

Consequently, if $p$ is a prime and you want to recover $\mathbb{F}_p^n$ as $K_0$ of a triangulated category, then the triangulated category $stMod(R)$ will work, where $R$ is a Cartesian product of $n$ copies of $\mathbb{Z}/\ell^p\mathbb{Z}$ for any prime $\ell$ (or, alternatively, $n$ copies of $k[x]/x^p$ for any field $k$).


You also ask about constructing triangulated categories whose $K_0$ is a vector space over, for example, the rational numbers. This is much more difficult than the finite field case. It is a nontrivial task to construct a triangulated category whose $K_0$ is $\mathbb{Q}$, or even just $\mathbb{Z}[\frac{1}{2}]$; this is tied up in the general problem of figuring out how to modify a given triangulated category so that a prime becomes inverted in its $K_0$. The papers https://arxiv.org/abs/1610.07162 and https://arxiv.org/abs/1702.07466 are about this problem, and you can find positive results specifically about the $\mathbb{Q}$ case of your problem in the Barwick-Glasman-Hoyois-Nardin-Shah paper (the first arXiv link).

Source Link
user164898
user164898

One part of your question asks for naturally occuring categories (which are triangulated, presumably, since triangulated categories are the categorical setting for earlier parts of the question) which are vector spaces over finite fields.

One source of such triangulated categories is stable module categories of quasi-Frobenius rings. Recall that the "stable module category" of a ring $R$ has, as objects, the $R$-modules, and as morphisms from $M$ to $N$, the set of equivalence classes of $R$-module morphisms $M\rightarrow N$, where two morphisms are said to be equivalent if their difference factors through a projective $R$-module. Recall also that a ring is "quasi-Frobenius" if its projective modules coincide with its injective modules. The stable module category of a quasi-Frobenius ring $R$ has a natural triangulation; I will write $stMod(R)$ for that triangulated category, restricted to its finitely generated modules, in order to avoid an Eilenberg swindle argument which would cause $K_0$ to vanish.

Consider the example $R = \mathbb{Z}/4\mathbb{Z}$. Every finitely-generated $R$-module is a direct sum of copies of $\mathbb{Z}/2\mathbb{Z}$ and $P = \mathbb{Z}/4\mathbb{Z}$, and $P$ is zero in the stable module category of $R$. Hence $K_0(stMod(R))$ must be a cyclic group generated by $[\mathbb{Z}/2\mathbb{Z}]$. However, we have the short exact sequence $$ 0 \rightarrow \mathbb{Z}/2\mathbb{Z}\rightarrow P \rightarrow \mathbb{Z}/2\mathbb{Z} \rightarrow 0$$ of $R$-modules, and consequently $[\mathbb{Z}/2\mathbb{Z}] + [\mathbb{Z}/2\mathbb{Z}] = [P] = 0$ in $K_0(stMod(R))$. So we have $K_0(stMod(R)) \cong \mathbb{Z}/2\mathbb{Z}$.

By a similar argument, for any prime $p$ and any positive integer $n$, we have $K_0(stMod(\mathbb{Z}/p^n\mathbb{Z})) \cong \mathbb{Z}/n\mathbb{Z}$. We also have $K_0(stMod(k[x]/x^n)) \cong \mathbb{Z}/n\mathbb{Z}$ for any field $k$. All that is really necessary for these kinds of $K_0$ calculations is that the ring $R$ be quasi-Frobenius, so that its stable module category is triangulated, and that $R$ is an Artinian principal ideal ring, so that its modules are all direct sums of cyclics.

Consequently, if $p$ is a prime and you want to recover $\mathbb{F}_p^n$ as $K_0$ of a triangulated category, then the triangulated category $stMod(R)$ will work, where $R$ is a Cartesian product of $n$ copies of $\mathbb{Z}/\ell^p\mathbb{Z}$ for any prime $\ell$ (or, alternatively, $n$ copies of $k[x]/x^p$ for any field $k$).


You also ask about constructing triangulated categories whose $K_0$ is a vector space over, for example, the rational numbers. This is much more difficult than the finite field case. It is a nontrivial task to construct a triangulated category whose $K_0$ is $\mathbb{Q}$, or even just $\mathbb{Z}[\frac{1}{2}]$; this is tied up in the general problem of figuring out how to modify a given triangulated category so that a prime becomes inverted in its $K_0$. The papers https://arxiv.org/abs/1610.07162 and https://arxiv.org/abs/1702.07466 are about this problem, and you can find positive results specifically about the $\mathbb{Q}$ case of your problem in the Barwick-Glasman-Hoyois-Nardin-Shah paper (the first arXiv link).