1
$\begingroup$

Currently I am reading the book "Hopf Algebras. An Introduction" by S. Dascalescu, C. Nastasescu, S. Raianu. There is a Definition 1.5.20 on page 44 (boldface is mine):

An algebra $A$ is called proper (or residually finite dimensional) if $i_A:A \to A^{\circ*}$ is injective, it is called weakly reflexive if $i_A$ is injective, and it is called reflexive if $i_A$ is bijective.

This definition is stated as if being proper and weakly reflexive are two different things, but according to their definition these two concepts are equivalent. I'm wondering if there is a typo in the text. Or the two concepts are actually identical?

Here $A$ is assumed to be a unital, associative algebra over some field $k$. $$A^\circ = \Big\{ f \in A^* : \exists I : \text{Ideal of} \; A \; \text{such that } \; I \subset \ker f \; \text{and}\; \dim_k \frac{A}{I} < \infty \Big\}$$ is a finite-dual coalgebra and $i_A$ is a canonical map defined by $i_A(a)(f) = f(a)$. The ideal $I$ in the above definition is assumed to be two-sided.

I don't want to learn wrong definitions, so I will be grateful if somebody more experienced than me can tell me if the definition is correct.

$\endgroup$
5
  • $\begingroup$ Do you assume $A$ commutative? could you mention the book's author? $\endgroup$
    – YCor
    Commented Oct 15, 2019 at 5:27
  • $\begingroup$ No I don't assume $A$ to be commutative. However, I assume $A$ to be unital and associative algebra over some field $k$. $\endgroup$
    – Nik Bren
    Commented Oct 15, 2019 at 5:29
  • 5
    $\begingroup$ Guessing it’s a typo and that the second “injective” was intended to read “surjective.” $\endgroup$ Commented Oct 15, 2019 at 5:47
  • $\begingroup$ @QiaochuYuan Thank you. I think I can use your guess as a working hypothesis. $\endgroup$
    – Nik Bren
    Commented Oct 15, 2019 at 6:02
  • $\begingroup$ The relevant page (which I have also edited into the post): books.google.com/books?id=pBJ6sbPHA0IC&pg=PA44 . $\endgroup$
    – LSpice
    Commented Oct 15, 2019 at 7:33

0

You must log in to answer this question.