3
$\begingroup$

A full subcategory $\mathbb{B}$ of a category $\mathbb{A}$ is reflective if the inclusion functor has a left adjoint $R$. The left adjoin is called the reflector of $\mathbb{B}$, and the unit of the adjuction $\eta$ satisfies the next universal property. For all $A\in\mathbb{A}$, if there is a morphism $f\colon A\longrightarrow B$ with $B\in \mathbb{B}$ then there is an unique morphism $\alpha\colon T(A)\longrightarrow B $ such that $\alpha\circ\eta_A=f$.

As in the category of the fields all the morphisms are monomorphisms. The morphism can be though as field extension. In particular we have that $Gal(T(F)/F)=\{1\}$ for any field $F$. This condition seems (for me) very restrictive. That's why I think there is no reflective subcategories besides the total.

$\endgroup$
1
  • $\begingroup$ Your $TA$ is contained in every $B$ which contains $A$. In particular, it is contained in $A$. $\endgroup$ Commented Sep 17, 2016 at 20:17

1 Answer 1

6
$\begingroup$

The category of perfect fields $k$, i.e. those of characteristic $0$ or of characteristic $p>0$ such that $k=k^p$, is an example. The left adjoint to the inclusion is given by adjoining all $p^r$-th roots.

$\endgroup$
2
  • 2
    $\begingroup$ I would conjecture that this is basically the only example (more precisely, an arbitrary reflective subcategory is a separate choice of either all fields or perfect fields for each characteristic). $\endgroup$ Commented Sep 17, 2016 at 20:35
  • $\begingroup$ @Frank Murphy: Can this be closed? $\endgroup$
    – Hanno
    Commented Nov 27, 2017 at 23:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .