Bimonoid and Hopf annihilators

Suppose $H$ is a $K$ -bimonoid and $𝑉$ is a left $H$ -module. The bimonoid annihilator $H A n n \circ (𝑉)$ is the largest biïdeal contained in the annihilator $H A n n (𝑉)$ , and we call $𝑉$ bimonoid faithful iff $H A n n \circ (𝑉) = 0$ . When $H$ is a Hopf algebra we define Hopf annihilator and Hopf faithful similarly.¹

warning

In many typical cases, a biïdeal is automatically a Hopf ideal, and thus the distinction between these notion collapses.

Over a field

In case $H$ is a K-bimonoid, we have

H A n n \circ (𝑉) = ⋂ 𝑖 \in ℕ H A n n (𝑉 \otimes 𝑛) .

Proof

Let $𝐼$ denote the right hand side. To show that $H A n n \circ (𝑉) = 𝐼$ we must show that $𝐼$ is a biïdeal, and that any biïdeal contained in $H A n n (𝑉)$ is also contained in $𝐼$ .

First we show the former. As the intersection of ideals, $𝐼$ itself is an ideal. Since $𝐼 \subseteq H A n n (𝑉 \otimes 0) = k e r (𝜖)$ , we have $𝜖 (𝐼) = 0$ . To see that $Δ (𝐼) \subseteq 𝐼 \otimes H + H \otimes 𝐼$ , note that for all $𝑥 \in 𝐼$ and $𝑛 \in ℕ$ , we have $0 = 𝑥 ▹ 𝑉 \otimes 2 𝑛 = \sum (𝑥) (𝑥 (1) ▹ 𝑉 \otimes 𝑛) \otimes (𝑥 (2) ▹ 𝑉 \otimes 𝑛)$ and thus $𝑥 \in ℎ A n n (𝑉 \otimes 𝑛) \otimes H + H \otimes H A n n (𝑉 \otimes 𝑛)$ .² Therefore $𝐼$ is a coïdeal, and hence a biïdeal.

To show that any biïdeal $𝐽$ contained in $H A n n (𝑉)$ is also contained in $𝐼$ , we prove that any such $𝐽$ annihilates $𝑉 \otimes 𝑛$ for all $𝑛 \in ℕ$ . The $𝑛 = 1$ case is given, the $𝑛 = 0$ case follows from $𝜖 (𝐽) = 0$ . All other cases follow inductively by Annihilating coïdeal.

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Banica and Bichon call the Hopf case inner faithful. ↩
This is where we use the fact $𝕂$ is a field. ↩

Bimonoid and Hopf annihilators

Over a field

Footnotes