A subcategory $\mathbf A$ is reflective subcategory of $\mathbf B$ if for every $B\in\mathbf B$ there exists an $A_B\in\mathbf A$ and a $\mathbf B$-morphism $r_B \colon B \to A_B$ such that: for any $A\in\mathbf A$ and any $\mathbf B$-morphism $f \colon B \to A$ there exists unique $\mathbf A$-morphism $\overline f \colon A_B \to A$ such that $\overline f\circ r_B=f$.

Diagram - reflection

This is equivalent to saying that the inclusion functor $E \colon \mathbf A \to \mathbf B$ has a left adjoint.

Current revision of the Wikipedia article on reflective subcategories claims that

The category of fields is a reflective subcategory of the category of integral domains. The reflector is the functor which sends each integral domain to its field of fractions.

I don't think this is true - an easy counterexample is the homomorphism $f \colon \mathbb Z \to\mathbb Z_2$ given by $f(n)=n\mod 2$. This homorphism obviously cannot be extended to $\overline f \colon \mathbb Q \to \mathbb Z_2$.

The above claim would be true if we considered the categories of integral domains only with injective ring homomorhpisms.

Am I correct? Did I miss something there?

The same problem has already been mentioned on the talk page of this Wikipedia article, so this is more-or-less just a sanity check.

  • 3
    $\begingroup$ The article field of fractions states explicitly what the category of integral domains has to be here. $\endgroup$ Jun 6, 2012 at 10:16
  • $\begingroup$ Small typo in definition of $r_B$: domain should be $B$, not $A$. $\endgroup$ Jan 24, 2016 at 12:31
  • $\begingroup$ Thanks @JoshuaMeyers. Corrected now. $\endgroup$ Jan 24, 2016 at 14:50

1 Answer 1


Yes, you are correct... and the category of integral domains is defined to only have the injective homomorphisms as arrows. Unfortunately, this part of the definition sometimes seems to be left out when the category is introduced....


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