Ideals in Tensor Categories with applications to groups, rings and topology: the propagation of pathological properties

PrologueIdeals in Tensor CategorySolution of the Question via Ideals in Tensor CategoryPrime Ideal and Spec FunctorMaximal idealIdeals in ToposIdeals and Tensor-Closed SubcategoryProper Tensor Category and Tensor-Closed SubcategoryThe Duality between Tensor-Closed Subcategories and Prime IdealsGood and Bad Properties in Tensor category.Ideals and Tensor-Closed Subcategories in SetPrime Ideals and Tensor-Closed Subcategories in $\mathrm{Top}$Non-Compact Spaces as prime Ideal and Compact Spaces as a Tensor-Closed SubcategoryNon-Hausdorff Spaces as prime Ideal and Hausdorff Spaces as a Tensor-Closed SubcategoryNon-Connected/Path connected Spaces as prime Ideal and Connected/Path connected Spaces as a Tensor-Closed SubcategoryIdeals and Tensor-Closed Subcategories in $\mathrm{Grp}$Infinite Groups and Groups with Finite Subgroups as Prime Ideals Non commutative group as prime ideal and Abelian Group as Tensor-closed subcategory.Centre Non-Trivial group is a prime ideal and central trivial group as Tensor-closed subcategory.Nonsimple group is an idealNonsolvable group is an prime ideal.Ideals and Tensor-closed subcategory in PosACC and DCC as Tensor-closed subcategoryFinPos as Tensor-closed subcategoryIdeals and Tensor-Closed Subcategories in BoolThe category of atomless Boolean algebras forms a tensor-closed sub categoryThe category of infinite Boolean Algebras forms a prime ideals The category of reduced Boolean Algebras form an idealsThe category of Complete Boolean Algebra is a Tensor-closed subcategory and incomplete Boolean Algebra is a prime ideal.Ideals and Tensor-Closed Subcategories in CRing.Noetherian ring/Artinian ring are Tensor-Closed Subcategories.Reduced Ring are Tensor-Closed Subcategories.

 

Prologue

I encountered a question today asking whether an infinite group can have elements of finite order. This seems trivial since any group has an identity element. However, when considering the existence of nontrivial elements, I realized that this question is not about group structure but about the structure of the tensor category $(\mathrm{Grp},\times ,e)$$(\mathrm{Grp},\times,e)$.

Ideals in Tensor Category

Definition. Ideal in tensor category.(What is tensor cat? Click here)

Let $(T,\otimes ,I)$$(T,\otimes,I)$ be a tensor category and $T^{\prime }$$T'$ be a subcategory of $T$$T$ such that, if $t\in T$$t\in T$ is isomorphic to $t^{\prime }\in T^{\prime }$$t'\in T'$, then $t\in T^{\prime }$$t\in T'$. For any $t\in T,t^{\prime }\in T^{\prime }$$t\in T,t'\in T'$, we have $t\otimes t^{\prime }\in T^{\prime }$$t\otimes t'\in T'$. Then we call $T^{\prime }$$T'$ a left ideal of $T$$T$. Equivalently, this just says that for any $t\in T,t\otimes -$$t\in T, t\otimes-$ is a functor from $T^{\prime }$$T'$ to $T^{\prime }$$T'$. (Recall that scalar multiplication of an ideal is just an endomorphism of an abelian group.) Similarly, we could define a right ideal of $(T,\otimes ,I)$$(T,\otimes, I)$. If $T^{\prime }$$T'$ is a binary ideal, we simply say $T^{\prime }$$T'$ is an ideal. Every left/right ideal in a symmetric tensor category is an ideal.

The "intersection" of two ideals is still an ideal.

The intersection is non-empty since if $a\in I$$a\in I$ and $b\in J$$b\in J$, then $a\otimes b\in I\cap J$$a\otimes b\in I\cap J$, i.e., $I\otimes J\subseteq I\cap J$$I\otimes J\subseteq I\cap J$.

Remark. Tensor product of ideals.

In general, $I\otimes J:=\{a\otimes b\mid a\in I,b\in J\}$$I\otimes J:=\{a\otimes b \mid a\in I, b\in J\}$ is not an ideal, since it could contain some $c\cong i\otimes j$$c\cong i\otimes j$ but $c\notin I\otimes J$$c\notin I\otimes J$.

But if we define $I\otimes J:=\{c\in T\mid c\cong a\otimes b,a\in I,b\in J\}$$I\otimes J:=\{c\in T \mid c\cong a\otimes b, a\in I, b\in J\}$, then $I\otimes J$$I\otimes J$ is an ideal.

Since for any $t\in T$$t\in T$, $t\otimes c\cong t\otimes (a\otimes b)\cong (t\otimes a)\otimes b=a^{\prime }\otimes b$$t\otimes c\cong t\otimes (a\otimes b)\cong (t\otimes a)\otimes b=a'\otimes b$ since $t\otimes a\in I$$t\otimes a\in I$.

We say an ideal $T^{\prime }$$T'$ is proper if $T^{\prime }\ne T$$T'\ne T$. Obviously, $T^{\prime }$$T'$ is proper iff $I\notin \mathrm{Ob}(T^{\prime })$$I\notin\mathrm{Ob}(T')$.

Solution of the Question via Ideals in Tensor Category

Let us solve the question I mentioned. Consider the tensor category $(\mathrm{Grp},\times ,e)$$(\mathrm{Grp},\times ,e)$ and let $T^{\prime }$$T'$ be the category of groups that have nontrivial finite order elements (equivalently, groups that have nontrivial finite subgroups), and let $T^{″}$$T''$ be the category of infinite groups. It is easy to see that both $T^{\prime }$$T'$ and $T^{″}$$T''$ are ideals in $(\mathrm{Grp},\times ,e)$$(\mathrm{Grp},\times,e)$.

Then $T^{\prime }\cap T^{″}$$T'\cap T''$ is non-empty. That is, there exist some groups that are both infinite and have nontrivial finite subgroups.

You can construct it by considering $t^{\prime }\otimes t^{″}$$t'\otimes t''$. For example, $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}$$\mathbb Z/2\mathbb Z\times\mathbb Z$.

Examples of Ideals

$\mathrm{\infty }$$\infty$, the category of infinite sets, is an ideal of the tensor category $(\mathrm{Set},+,\mathrm{\varnothing })$$(\mathrm{Set,+,\empty})$. Similarly, this applies to the category of vector spaces or Banach spaces, where the category of infinite-dimensional vector spaces will form an ideal. You can also consider the category of groups and posets.

Let us consider the category of posets, and the category of non-totally ordered sets will be an ideal.

Let $(R,\cdot ,1)$$(R,\cdot,1)$ be a ring; then it is a tensor category. The usual ideals are ideals of the tensor category, but the converse is not true.

Prime Ideal and Spec Functor

Definition. Prime ideal.

Let $T^{\prime }$$T'$ be an ideal of $(T,\otimes ,I)$$(T,\otimes,I)$. We say that $T^{\prime }$$T'$ is a prime ideal if $a\otimes b\in T^{\prime }⟹a\in T^{\prime }$$a\otimes b\in T'\implies a\in T'$ or $b\in T^{\prime }$$b\in T'$.

Definition. The functor $\mathrm{Spec}(-)$$\mathrm{Spec(-)}$

Let $(T,\otimes ,I)$$(T,\otimes,I)$ be a tensor category; then define $\mathrm{Spec}(T)$$\mathrm{Spec}(T)$ to be the class of all the prime ideals of $T$$T$.

Proposition. Let $\mathcal{F}:(A,\otimes ,I)\to (B,\otimes ,I^{\prime })$$\mathcal F:(A,\otimes,I)\to(B,\otimes,I')$ be a functor such that $\mathcal{F}(a\otimes b)\cong \mathcal{F}(a)\otimes \mathcal{F}(b),\mathcal{F}(I)\cong I^{\prime }$$\mathcal F(a\otimes b)\cong\mathcal F(a)\otimes\mathcal F(b),\mathcal F(I)\cong I'$.

If $P$$P$ is a prime ideal in $B$$B$, then ${\mathcal{F}}^{-1}(P)$$\mathcal F^{-1}(P)$ is a prime ideal in $A$$A$.

Proof. Let us first prove that it is an ideal. In general, the pullback of any ideal(if it is non-empty) is still an ideal.

Let $J$$J$ be an ideal of $B$$B$. If $a\in {\mathcal{F}}^{-1}(B)$$a\in\mathcal F^{-1}(B)$ and $a\cong b$$a\cong b$, then $\mathcal{F}(a)\cong \mathcal{F}(b)$$\mathcal F(a)\cong\mathcal F(b)$ hence $\mathcal{F}(b)\in B$$\mathcal F(b)\in B$, hence $b\in {\mathcal{F}}^{-1}(B)$$b\in\mathcal F^{-1}(B)$.

For any $t\in A$$t\in A$ and $a\in {\mathcal{F}}^{-1}(B)$$a\in\mathcal F^{-1}(B)$, $\mathcal{F}(t\otimes a)\cong \mathcal{F}(t)\otimes \mathcal{F}(a)\in B$$\mathcal F(t\otimes a)\cong\mathcal F(t)\otimes\mathcal F(a)\in B$, hence $t\otimes a\in {\mathcal{F}}^{-1}(B)$$t\otimes a\in\mathcal F^{-1}(B)$.

Now let us prove that for a prime ideal $P$$P$, ${\mathcal{F}}^{-1}(P)$$\mathcal F^{-1}(P)$ is prime as well.

$$\begin{array}{}\text{(1)}& \mathcal{a}\otimes b\in {\mathcal{F}}^{-1}(P)⟹\mathcal{F}(a)\otimes \mathcal{F}(b)\in P⟹\mathcal{F}(a)\in P\vee \mathcal{F}(b)\in P⟹a\in {\mathcal{F}}^{-1}(P)\vee b\in {\mathcal{F}}^{-1}(P)\end{array}$$

Hence $\mathrm{Spec}(-):{\mathrm{Tensor}}^{\mathrm{op}}\to \mathrm{Class}$$\mathrm{Spec}(-):\mathrm{Tensor^{op}}\to\mathrm{Class}$ is a functor.

Example.

Let $T$$T$ be the category of finite sets, and consider the objects of $T^{\prime }$$T'$ to be the sets such that $|S|\in p\mathbb{Z}$$|S|\in p\mathbb Z$.

Then $a\times b\in T^{\prime }⟺|a\times b|\in p\mathbb{Z}⟺|a|\times |b|\in p\mathbb{Z}⟺a\in T^{\prime }\vee b\in T^{\prime }$$a\times b\in T'\iff |a\times b|\in p\mathbb Z\iff |a|\times|b|\in p\mathbb Z\iff a\in T'\vee b\in T'$.

Let $FG$$FG$ be the category of finite groups, and let $P$$P$ be the subcategory of $FG$$FG$ containing all finite groups such that $|G|\in p\mathbb{Z}$$|G|\in p\mathbb Z$. This is also a prime ideal.

Similarly, you can consider the category of finite posets where $|\mathcal{P}|\in p\mathbb{Z}$$|\mathcal P|\in p\mathbb Z$...

Since the forgetful functor $F:\mathcal{C}\to T$$F:\mathcal C\to T$​​ preserves product.

Maximal ideal

Let $(T,\otimes ,I)$$(T,\otimes,I)$ be a non-trivial tensor category, then $\{a\in T,a\ncong I\}$$\{a\in T,a\ncong I\}$​ is a prime ideal, we call it maximal ideal since it contains all the ideals (it looks like local ring huh)

Hence $\mathrm{Spec}(T)=\mathrm{\varnothing }⟺T$$\mathrm{Spec}(T)=\empty\iff T$​ is trivial.

Ideals in Topos

Definition. Ideals in topos.

Let $T$$T$ be a topos; then $(T,\times ,\ast )$$(T,\times,*)$ forms a tensor category, where $\ast$$*$ is the final object. We say $T^{\prime }$$T'$ is an ideal of the topos $T$$T$ if $T^{\prime }$$T'$ is an ideal of $(T,\times ,\ast )$$(T,\times,*)$ as a tensor category, and $T^{\prime }$$T'$ is closed under the coproduct $+$$+$.

I think we should use $(T,+,\times ,\ast )$$(T,+,\times,*)$ when we are talking about ideals in a topos (by the way, it looks like a semiring).

In a topos, we have the product-exponential adjunction (tensor-hom adjoint). Hence for any $t\in T$$t\in T$, $t\times (a+b)\cong t\times a+t\times b$$t\times(a+b)\cong t\times a+t\times b$.

You can imagine $T^{\prime }$$T'$ as a sub-$T$$T$-module of $T$$T$. (Recall that an ideal of a ring $R$$R$ is just a sub-$R$$R$-module of $R$$R$.)

For an $R$$R$-module $M$$M$, we have

$$\begin{array}{}\text{(2)}& r\in R⟼r\cdot -\in {\mathrm{End}}_{\mathrm{Ab}}(M)\end{array}$$

Similarly,

$$\begin{array}{}\text{(3)}& t\in T⟼t\times -\in {\mathrm{End}}_{\mathrm{CAT}}(T^{\prime })\end{array}$$

Example.

Consider $T=\mathrm{Set}$$T=\mathrm{Set}$, which is a topos, and let $T^{\prime }=\mathrm{\infty }$$T'=\infty$, the category of infinite sets.

It is easy to see that $(\mathrm{\infty },+,\times )$$(\infty,+,\times)$ is an ideal of $(\mathrm{Set},+,\times ,\ast )$$(\mathrm{Set},+,\times,*)$.

Ideals and Tensor-Closed Subcategory

In some tensor categories, prime ideals correspond to "bad" properties, while tensor-closed subcategories correspond to "good" properties. ($a\otimes b$$a\otimes b$ is good iff $a$$a$ is good and $b$$b$ is good.)

For example, let $(\mathrm{Ring},\times ,0)$$(\mathrm{Ring,\times,0})$ be the category of rings; then all non-commutative rings form an prime ideal, but $(\mathrm{CRing},\times ,0)$$(\mathrm{CRing},\times,0)$, the category of commutative rings, is a tensor-closed subcategory.

In $(R-\mathrm{Mod},\times ,0)$$(R-\mathrm{Mod},\times,0)$, all modules such that $\mathrm{tor}(M):=\{m\in M\mid \mathrm{\exists }r\in R\setminus \{0\},rm=0\}\ne 0$$\mathrm{tor}(M):=\{m\in M \mid \exists r\in R\setminus\{0\}, rm=0\} \neq 0$ form an prime ideal, but all torsion-free modules form a tensor-closed subcategory.

These "bad" properties, combined with the absorption property of ideals, tend to propagate through tensor products.

We will define what is good and what is bad for a tensor category.

We will see this pattern recur in $(\mathrm{Top},\times ,\ast )$$(\mathrm{Top},\times,*)$​ and other categories.

Proper Tensor Category and Tensor-Closed Subcategory

Definition. Let $(T,\otimes ,I)$$(T,\otimes,I)$ be a tensor category. A sub-tensor category $S$$S$ is proper if

$$\begin{array}{}\text{(4)}& a\in S,a\cong b⟹b\in S.\end{array}$$

Definition. Let $(T,\otimes ,I)$$(T,\otimes,I)$ be a tensor category. We say a sub-tensor category $S$$S$ of $T$$T$ is tensor-closed if

$$\begin{array}{}\text{(5)}& a\otimes b\in S⟺a\in S\text{ and }b\in S,a\in S,a\cong b⟹b\in S.\end{array}$$

The Duality between Tensor-Closed Subcategories and Prime Ideals

Let $(T,\otimes ,I)$$(T,\otimes,I)$ be a tensor category.

If all objects satisfying property $P$$P$ form a proper sub-tensor category and all objects satisfying property $\mathrm{\neg }P$$\neg P$ form an ideal, then we have $a\otimes b$$a \otimes b$ in the sub-tensor category if and only if $a$$a$ and $b$$b$ are both in it. If $a$$a$ or $b$$b$ is not in the sub-tensor category, it belongs to the ideal, which has the absorption property. Additionally, $a\otimes b$$a \otimes b$ belongs to the ideal if and only if $a$$a$ or $b$$b$ is in it. Thus, the ideal is a prime ideal.

Proposition. If $S$$S$ is a proper tensor-closed subcategory and $S^c:=\{a\in T\mid a\notin S\}$$S^c:=\{a\in T \mid a\notin S\}$ is an ideal, then $S$$S$ is tensor-closed and $S^c$$S^c$ is prime.

Proof.

Let $a\otimes b\in S$$a\otimes b\in S$. If $a\notin S$$a\notin S$, i.e., $a\in S^c$$a\in S^c$, then $a\otimes b\in S^c$$a\otimes b\in S^c$, so $S$$S$ is tensor-closed.

If $a\otimes b\in S^c$$a\otimes b\in S^c$, then $a\in S^c$$a\in S^c$ or $b\in S^c$$b\in S^c$. Otherwise, $a\in S$$a\in S$ and $b\in S$$b\in S$ implies $a\otimes b\in S$$a\otimes b\in S$. $◻$$\square$

Corollary. In $(\mathrm{Set},\times ,\ast )$$(\mathrm{Set},\times,*)$, $(\mathrm{\infty },\times )$$(\infty,\times)$ is a prime ideal, and the category of finite sets is a tensor-closed subcategory.

Proposition. $S$$S$ is a proper tensor-closed subcategory $⟺$$\iff$ $S^c:=\{a\in T\mid a\notin S\}$$S^c:=\{a\in T \mid a\notin S\}$ is a prime ideal.

Proof.

Let $S$$S$ be a proper tensor-closed subcategory, and let $a\in S^c$$a\in S^c$. If $a\cong b$$a\cong b$, then $b\in S^c$$b\in S^c$, since if $b\in S$$b\in S$, then $a\in S$$a\in S$.

For any $t\in T$$t\in T$, $t\otimes a\in S^c$$t\otimes a\in S^c$, since if $t\otimes a\in S$$t\otimes a\in S$, then $a\in S$$a\in S$. Also, $a\otimes b\in S^c⟹a\in S^c$$a\otimes b\in S^c\implies a\in S^c$ or $b\in S^c$$b\in S^c$. Hence, $S^c$$S^c$ is a prime ideal. Conversely, let $P$$P$ be a prime ideal, then $P^c$$P^c$ is a proper tensor-closed subcategory.

Since $a\otimes b\in P^c⟺a\otimes b\notin P⟺a\notin P$$a\otimes b\in P^c \iff a\otimes b\notin P \iff a\notin P$ and $b\notin P⟺a\in P^c$$b\notin P \iff a\in P^c$ and $b\in P^c$$b\in P^c$. Also, $a\in P^c$$a\in P^c$, $a\cong b⟹b\in P^c$$a\cong b \implies b\in P^c$. $◻$$\square$

You can use the first proposition to prove the second as well.

Corollary. Let $F$$F$ be a functor between two tensor categories and preserve tensor product and $TC$$TC$ be a tensor-closed subcategories. Then $F^{-1}(TC)$$F^{-1}(TC)$ is a tensor-closed subcategories.

Proposition. The intersection of two tensor-closed subcategories is still a tensor-closed subcategory; hence the union of two prime ideals is still a prime ideal.

Proof. This is straightforward. This proposition allows you to combine desirable properties, such as compactness and Hausdorffness, in topological spaces.

By the duality of prime ideals and tensor-closed subcategories, we can also define $\mathrm{Spec}(-)$$\mathrm{Spec}(-)$​​ via tensor-closed subcategories.

Easy to see that union of two prime ideals are prime and intercetion of two Tensor-Closed Subcategories are Tensor-Closed Subcategories.

Good and Bad Properties in Tensor category.

Definition. We say a property for a tensor category is a good property if all the object satisfies this property form a Tensor-closed subcategory. We say a property for a tensor category is a bad property if all the object satisfies this property form a prime ideal.

A good property is really pure, since if $a\otimes b$$a\otimes b$ is good then both $a$$a$ and $b$$b$​ are good. A bad property can actually be "redeemed": if a bad object can be decomposed into tensor products, then removing the bad factors will result in a good object. This reveals a profound connection between good/bad properties and structural decomposition theorems.

Ideals and Tensor-Closed Subcategories in Set

Let $\mathfrak{N}$$\mathfrak N$ be the category of set such that $\mathrm{Card}(S)\ge n$$\mathrm{Card}(S)\ge n$, then $\mathfrak{N}$$\mathfrak N$ is an ideal but it is not a prime ideal.

Let ${\mathrm{\aleph }}_K$$\aleph_K$ be the category of set such that $\mathrm{Card}(S)\ge {\mathrm{\aleph }}_K$$\mathrm{Card} (S)\ge \aleph_K$​, then it is a prime ideal. Since

$$\begin{array}{}\text{(6)}& |X\times Y|\ge {\mathrm{\aleph }}_k⟺|X|·|Y|\ge {\mathrm{\aleph }}_k⟺|X|\ge {\mathrm{\aleph }}_k\vee |Y|\ge {\mathrm{\aleph }}_k\end{array}$$

Theorem: For a cardinal $k$$k$, the following are equivalent:

1. $K$$K$ is an infinite cardinal

2. The class of all sets with cardinality $\ge k$$\ge k$ forms a prime ideal

Notice that for any concrete categiry $\mathcal{C}$$\mathcal C$ over $\mathrm{Set}$$\mathrm{Set}$, the forgetful functor preserve product, hence

${\mathcal{F}}^{-1}({\mathrm{\aleph }}_K)$$\mathcal F^{-1}(\aleph_K)$ will be always prime ideal. Hence everytime you have a bad property respect to an ideal $J$$J$, $J\cap {\mathcal{F}}^{-1}({\mathrm{\aleph }}_K)\ne \mathrm{\varnothing }$$J\cap\mathcal F^{-1}(\aleph_K)\ne\empty$​.

Also, $\mathrm{\varnothing }$$\empty$ forms a prime ideals, it looks like $(0)$$(0)$ ideal is prime, i.e. integral domain.

Prime Ideals and Tensor-Closed Subcategories in $\mathrm{Top}$$\mathrm{Top}$

Now let us think about $(\mathrm{Top},\times ,\ast )$$(\mathrm{Top},\times,*)$. Topological spaces can have various properties, such as compactness, connectedness, path connectedness, Hausdorffness, separability, and irreducibility.

To construct an ideal, we need to make sure that the identity element is not in the ideal. In $\mathrm{Top}$$\mathrm{Top}$, the identity object is $\ast$$*$, the one-point space.

The space $\ast$$*$ is connected, path-connected, irreducible, compact, Hausdorff, separable, and second-countable.

Thus, we can consider the non-compact spaces.

Non-Compact Spaces as prime Ideal and Compact Spaces as a Tensor-Closed Subcategory

The Tychonoff theorem states that if $X$$X$ and $Y$$Y$ are both compact spaces, then $X\times Y$$X\times Y$ is also compact.

Also compact is a topological invarance. Thus, $(\mathrm{ComTop},\times ,\ast )$$(\mathrm{ComTop},\times,*)$ is a proper subcategory of $\mathrm{Top}$$\mathrm{Top}$.

Now consider all the non-compact spaces. We claim that they form an ideal.

Let $Y$$Y$ be a noncompact space and consider the epimorphism ${\pi }_Y:X\times Y\to Y$$\pi_Y:X\times Y\to Y$, then $Y$$Y$ is noncompact, but we know that for a continuous function $f:X_1\to X_2$$f:X_1\to X_2$, $X_1$$X_1$ is compact implies $f(X_1)$$f(X_1)$ is compact. Hence it is a contradiction.

We will use this trick again and again.

Therefore, all non-compact spaces form a prime ideal, and $(\mathrm{ComTop},\times ,\ast )$$(\mathrm{ComTop},\times,*)$ is a tensor-closed subcategory.

Non-Hausdorff Spaces as prime Ideal and Hausdorff Spaces as a Tensor-Closed Subcategory

Hausdorff spaces are closed under the product. Let $(x,y)\ne (x^{\prime },y^{\prime })\in X\times Y$$(x,y)\ne (x',y')\in X\times Y$.

Without loss of generality, let us assume that $x\ne x^{\prime }$$x\ne x'$. Then there exist disjoint open neighborhoods $U$$U$ and $U^{\prime }$$U'$ of $x$$x$ and $x^{\prime }$$x'$, respectively.

Then $(x,y)\in U\times Y$$(x,y)\in U\times Y$, $(x^{\prime },y^{\prime })\in U^{\prime }\times Y$$(x',y')\in U'\times Y$, and $(U\times Y)\cap (U^{\prime }\times Y)=\mathrm{\varnothing }$$(U\times Y)\cap(U'\times Y)=\emptyset$. Hence the Hausdorff spaces form a tensor-closed subcategory.

Let $E$$E$ be a non-Hausdorff space. Then for any topological space $F$$F$, $E\times F$$E\times F$ is not Hausdorff.

Since every subspace of a Hausdorff space is Hausdorff, and $E\times \ast \cong E$$E\times *\cong E$ since $\ast$$*$ is the identity object, we have $E\cong E\times \ast \subseteq E\times F$$E\cong E\times *\subseteq E\times F$, which is not Hausdorff. This implies that $E\times F$$E\times F$ is not Hausdorff.

Thus, all non-Hausdorff spaces form a prime ideal, and $(\mathrm{Haus},\times ,\ast )$$(\mathrm{Haus},\times,*)$ is a tensor-closed subcategory.

Non-Connected/Path connected Spaces as prime Ideal and Connected/Path connected Spaces as a Tensor-Closed Subcategory

Let $X$$X$ and $Y$$Y$ be two connected spaces. Then $X\times Y$$X\times Y$ is connected.

We only need to prove that $X\times Y$$X\times Y$ has one connected component.

Let $x_0\in X$$x_0\in X$ and consider the fiber ${\pi }_X^{-1}(x_0)$$\pi_X^{-1}(x_0)$. We have ${\pi }_X^{-1}(x_0)\cap {\pi }_Y^{-1}(y)=(x_0,y)\ne \mathrm{\varnothing }$$\pi_X^{-1}(x_0)\cap\pi_Y^{-1}(y)=(x_0,y)\ne\emptyset$.

Also, ${\pi }_X^{-1}(x)=\{x\}\times Y\cong Y$$\pi_X^{-1}(x)=\{x\}\times Y\cong Y$ and ${\pi }_Y^{-1}(y)=X\times \{y\}\cong X$$\pi_Y^{-1}(y)=X\times\{y\}\cong X$, which are connected.

Thus, ${\pi }_X^{-1}(x)$$\pi_X^{-1}(x)$ and ${\pi }_Y^{-1}(y)$$\pi^{-1}_Y(y)$ are in the same connected component since they intersect non-trivially.

Hence, all the fibers lie in the same connected component, and their union should lie in the connected component as well, but the union is $X\times Y$$X\times Y$ itself. Therefore, $X\times Y$$X\times Y$​ is connected.

Path connected is pretty easy.

Now let us prove that all non-connected/path connected spaces form an ideal.

Let $Y$$Y$ be a non-connected/path connected spaces, then ${\pi }_Y:X\times Y\to Y$$\pi_Y:X\times Y\to Y$ is a continuous map, but $Y$$Y$ is not non-connected/path connected, hence $X\times Y$$X\times Y$ is not.

Ideals and Tensor-Closed Subcategories in $\mathrm{Grp}$$\mathrm{Grp}$

Infinite Groups and Groups with Finite Subgroups as Prime Ideals

We already see that the category of infinite groups and the category of groups with finite subgroups form ideals in $(\mathrm{Grp},\times ,e)$$(\mathrm{Grp},\times,e)$​. Indeed, both of them are prime ideals.

Non commutative group as prime ideal and Abelian Group as Tensor-closed subcategory.

Easy to see that $(\mathrm{Ab},\times ,e)$$(\mathrm{Ab},\times,e)$​​ is a tensor-closed subcategory.

Centre Non-Trivial group is a prime ideal and central trivial group as Tensor-closed subcategory.

Lemma. Let $Z(-):\mathrm{Grp}:\to \mathrm{Ab}$$Z(-):\mathrm{Grp}:\to\mathrm{Ab}$, then $Z(-)$$Z(-)$ preserve product.

Proof. $(h,k)\in Z(H\times K)⟹\mathrm{\forall }{h}^{\prime }\in H,(h,k)(h^{\prime },e)=(h^{\prime },e)(h,k)$$(h,k)\in Z(H\times K)\implies \forall h'\in H,(h,k)(h',e)=(h',e)(h,k)$ hence $h\in Z(H)$$h\in Z(H)$. Similarly $k\in Z(K)$$k\in Z(K)$.

Hence $Z(H\times K)\subseteq Z(H)\times Z(K)$$Z(H\times K)\subseteq Z(H)\times Z(K)$. The converse is obviously. $◻$$\square$

Let us consider the category of centre non-trivial group, whcih is a prime ideal. Let $\mathcal{A}$$\mathcal A$ be the category of all abelian group wihout $e$$e$, then it is a prime ideal since all the trivial group is a Tensor-closed subcategory. Then the category of centre non-trivial group is just the $Z^{-1}(\mathcal{A})$$Z^{-1}(\mathcal A)$​​. Hence a prime ideal.

Nonsimple group is an ideal

Consider all the nonsimple groups, obviously it is an ideal but not prime ideal.

Nonsolvable group is an prime ideal.

Let $G$$G$ be a solvable group, then the subgroup of $G$$G$ and quotient group of $G$$G$ are solvable as well.

Hence nonsolvable group form an ideal since $G$$G$ is a quotient group of $G\times H$$G\times H$.

Also, solvable group form an sub tensor category.

Ideals and Tensor-closed subcategory in Pos

Let $(\mathrm{Pos},\times ,I)$$(\mathrm{Pos},\times,I)$ be the category of Partial order set, here $I=\{\ast \}$$I=\{*\}$.

For the definition of Accending Chain Condition and Decending Chain Condition, click here.

ACC and DCC as Tensor-closed subcategory

Let $P,Q$$P,Q$ be two poset satisfies ACC, then $P\times Q$$P\times Q$ satisfies ACC as well, since for

$$\begin{array}{}\text{(7)}& (a_1,b_1)\le ....\le (a_n,b_n)...\end{array}$$

You can get $a_1\le ...\le a_i\le ...$$a_1\le...\le a_i\le...$ and $b_1\le ...\le b_j\le ...$$b_1\le...\le b_j\le...$ the first sequnece will satble after $a_n$$a_n$ and the second sequence will stale under $m$$m$. Then pick $N=max\{n,m\}$$N=\max\{n,m\}$...

If $P\times Q$$P\times Q$ satisfies ACC, then $(a_1,b)\le (a_2,b)\le ...$$(a_1,b)\le(a_2,b)\le...$ will be stable, hence $P$$P$ satisfies ACC, similarly for $Q$$Q$.

By duality, we get the result for DCC. Hence non ACC/DCC Poset will be a prime ideal.

FinPos as Tensor-closed subcategory

Obviously.

Ideals and Tensor-Closed Subcategories in Bool

Let us consider $(\mathrm{Bool},\times ,0)$$(\mathrm{Bool},\times,0)$​ to be the tensor cat.

Definition. Let $B$$B$ be a Boolean algebra, we call $a\ne 0$$a\ne 0$ an atom of $B$$B$ if $0\le a^{\prime }\le a⟹a^{\prime }=a$$0\le a'\le a\implies a'=a$.

The category of atomless Boolean algebras forms a tensor-closed sub category

Let $A,B$$A,B$ be two atomless Boolean algebras, then $A\times B$$A\times B$ has non atom since $(0,0)\le (a,0)\le (a,b)$$(0,0)\le (a,0)\le (a,b)$. p\

If $A$$A$ has atom $a$$a$, then $(0,0)\le (a^{\prime },0)\le (a,0)⟹(a^{\prime },0)=(a,0)$$(0,0)\le(a',0)\le(a,0)\implies (a',0)=(a,0)$. Hence $A\times B$$A\times B$ is atomless iff both $A,B$$A,B$ are atomless.

Hence Boolean algebra with atom is a prime ideal.

The category of infinite Boolean Algebras forms a prime ideals

Since it is pull back the $\mathrm{\infty }$$\infty$ in $\mathrm{Set}$$\mathrm{Set}$ respect to the forgetful functor $F:\mathrm{Bool}\to \mathrm{Set}$$F:\mathrm{Bool}\to\mathrm{Set}$.

The category of reduced Boolean Algebras form an ideals

Definition. Let $B$$B$ be a Boolean Algebra, we say $B$$B$ is reduced if $B$$B$ can be write as union of two proper sub Boolean Algebra. That is, $B=A\cup A^{\prime },A\subset B,A^{\prime }\subset B$$B=A\cup A',A\sub B,A'\sub B$.

Then it is an ideals, since reduce is an invariance under isomorphism, and for any reduced Boolean algebra $B$$B$ and a Boolean algebra $C$$C$, $C\times B=C\times A\cup C\times A^{\prime }$$C\times B=C\times A\cup C\times A'$. Hence it is reduce as well.

The category of Complete Boolean Algebra is a Tensor-closed subcategory and incomplete Boolean Algebra is a prime ideal.

Let us prove that category of Complete Boolean Algebra is a Tensor- closed subcategory. If $A\times B$$A\times B$ is complete then $A$$A$ is complete since $\underset{i\in I}{\bigwedge }(a_i,0)=(\underset{i\in I}{\bigwedge }{a}_i,0)\in A\times B,\underset{i\in I}{\bigvee }(a_i,0)=(\underset{i\in I}{\bigvee }{a}_i,0)\in A\times B$$\bigwedge_{i\in I}(a_i,0)=(\bigwedge_{i\in I} a_i,0)\in A\times B,\bigvee_{i\in I}(a_i,0)=(\bigvee_{i\in I} a_i,0)\in A\times B$. Similarly $B$$B$​ is complete. Hence complete Boolean A;gebra is a Tensor- closed subcategory, hence incomplete Boolean Algebra is a prime ideal.

Hence there exists some incomplete, reduce, atom existed, non ACC non DCC Boolean algebra.

Ideals and Tensor-Closed Subcategories in CRing.

Let $(\mathrm{CRing},\times ,e)$$(\mathrm{CRing},\times,e)$​ be the tensor category.

Noetherian ring/Artinian ring are Tensor-Closed Subcategories.

Then Noetherian ring/Artinian ring are Tensor-Closed Subcategories.

Let $\mathcal{I}:\mathrm{CRing}\to \mathrm{Pos}$$\mathcal I:\mathrm{CRing}\to\mathrm{Pos}$ to be the sub ideals functor. That is,

$$\begin{array}{}\text{(8)}& \mathcal{I}(R)=\{I\subseteq R:I\text{ is an ideal of }R\}\end{array}$$

For a ring homomorphism $f:R\to S$$f:R\to S$ and $I\in \mathcal{I}(R),(f(I))\in \mathcal{I}(S)$$I\in\mathcal I(R),(f(I))\in\mathcal I(S)$. Where $(f(I))$$(f(I))$ is the ideal generated by $f(I)$$f(I)$.

The order is the inclusion, then a ring homomorphism will preserve the order.

Lemma. $\mathcal{I}(R\times S)\cong \mathcal{I}(R)\times \mathcal{I}(S)$$\mathcal I(R\times S)\cong\mathcal I(R)\times\mathcal I(S)$.

Proof. Let $J\subseteq R\times S$$J\subseteq R\times S$ be an ideal then $J=({\pi }_R(J),{\pi }_S(J))$$J=(\pi_R(J),\pi_S(J))$, conversely, let $I\subseteq R,I^{\prime }\subseteq S$$I\sube R,I'\subseteq S$, then $I\times I^{\prime }$$I\times I'$ is an ideal of $R\times S$$R\times S$. $◻$$\square$

Hence Noetherian ring/Artinian ring are ${\mathcal{I}}^{-1}(\mathrm{ACC})/{\mathcal{I}}^{-1}(\mathrm{DCC})$$\mathcal I^{-1}(\mathrm{ACC})/\mathcal I^{-1}(\mathrm{DCC})$​, hence they are Tensor-Closed Subcategories.

Reduced Ring are Tensor-Closed Subcategories.

Obviously.