Primitive polynomial and some basic property of prime ideal in tensor category

This post continues the ideas from the previous blog:

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

Let $R$$R$ be a UFD, and consider the family of primitive polynomials in $R[X]$$R[X]$.

Definition.

A polynomial $f=a_n{X}^n+\cdots +a_0\in R[X]$$f = a_nX^n + \dots + a_0 \in R[X]$ is primitive if $gcd(a_0,\dots ,a_n)=1$$\gcd(a_0,\dots,a_n)=1$.

Equivalently, for every irreducible element $p\in R$$p\in R$, $f$$f$ is not in the kernel of the map

$${\pi }_p:R[X]⟶(R/(p))[X].$$

Hence the set of non‑primitive polynomials is

$$\bigcup _{p\text{ irreducible}}\ker ({\pi }_p).$$

Each $\ker ({\pi }_p)$$\ker(\pi_p)$ is a prime ideal.

Proposition. The product $fg$$fg$ is primitive if and only if both $f$$f$ and $g$$g$ are primitive.

Proof. Since $fg$$fg$ is non‑primitive precisely when $fg\in \ker ({\pi }_p)$$fg\in\ker(\pi_p)$ for some $p$$p$, and each $\ker ({\pi }_p)$$\ker(\pi_p)$ is prime, it follows that $f\in \ker ({\pi }_p)$$f\in\ker(\pi_p)$ or $g\in \ker ({\pi }_p)$$g\in\ker(\pi_p)$. Hence $fg$$fg$ is primitive if and only if neither factor lies in any $\ker ({\pi }_p)$$\ker(\pi_p)$. $◻$$\square$

Viewing the monoid $(R,\cdot )$$(R,\cdot)$ as a tensor category, the set of non‑primitive polynomials forms a prime ideal, and thus the primitive polynomials form a tensor‑closed subcategory. This completes the proof. $◻$$\square$

This example inspires a general result in tensor categories.

Proposition. Let $(T,\otimes ,I)$$(T,\otimes,I)$ be a tensor category, and let $\{J_n\}$$\{J_n\}$ be a family of ideals in $T$$T$. Then

$$\bigcup _n{J}_n$$

is also an ideal. Moreover, if each $J_n$$J_n$ is prime, then $\bigcup _n{J}_n$$\bigcup_n J_n$ is prime.

Proof. Trivial. $◻$$\square$

Recall that many pathological properties give rise to prime ideals. We say a property $p$$p$ of objects in $T$$T$ is bad if the class of objects satisfying $p$$p$ forms a prime ideal, denoted $(p)$$(p)$. The proposition above implies that if $p$$p$ and $q$$q$ are bad properties, then their join $p\vee q$$p\vee q$ is also bad. Thus, bad properties form a monoid under $\vee$$\vee$.