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EXAMPLES OF SOME PROPERTIES OF

PETER J. HAINE

Exercise ([2, Exer. 1.5.xi]). Consider the functors Ab ↪ Grp (inclusion), → Ab (forgetting the multiplication), (−)× ∶ Ring → Grp (taking the of units), Ring → (dropping the multiplicative ), Fld ↪ Ring (inclusion), Mod푅 → Ab (forgetful). Determine which functors are full, which are faithful, and which are essentially surjective. Do any define an equivalence of categories? In order to not state this many times, we remark that by definition it follows that an inclusion 퐶′ ↪ 퐶 of a 퐶′ into a 퐶 is faithful and injective on objects, and full just in case 퐶′ is a full subcategory. We use the characterization that functors which provide equivalences of categories are fully faithful and essentially surjective on objects to classify the functors that are equivalences, and say nothing more. Below is an enumerated list describing the functors in question. (a) The inclusion functor Ab ↪ Grp is fully faithful buy not essentially surjective on objects, hence not an equivalence of categories. The inclusion is fully faithful since Ab is obviously a full subcategory of Grp, however it is not essentially surjective on objects since not every group is abelian. (b) The 푈∶ Ring → Ab is faithful, but neither full nor essentially surjective on objects, hence not an equivalence of categories. To see that 푈 is faithful, notice that for any ring 휙∶ 푅 → 푅′, is a homomorphism of the underlying group, but respects additional structure. Thus, in particular, for ring to be distinct, they must already be distinct as homomorphisms of underlying groups, which additionally respect the multiplicative structure. Hence the map of hom-sets is injective. A basic fact from algebra is that there exists no ring homomorphisms 퐙/푛 → 퐙. We know that 푈(퐙/푛) is the cyclic group 퐙/푛 and 푈(퐙) is the infinite cyclic group 퐙, and there does exist a 퐙/푛 → 퐙, the trivial homomorphism, hence 푈 is not full. To see that 푈 is not essentially surjective on objects, we show that not every can have a ring structure on put on it. Fix a prime 푝 and consider the Prüfer 푝-group 푛 퐙(푝∞) ≔ 퐙[1/푝]/퐙 ≅ { 푒2휋푖푚/푝 | 푚, 푛 ∈ 퐍 } , (so that 퐙(푝∞) can be regarded as the group of 푝th -power roots of unity in the unit circle). Every element of 퐙(푝∞) has finite order, though there is no upper bound of the order of elements. Suppose, for the sake of contradiction, that there exists a unital ring 푅 with 푈(푅) ≅ 퐙(푝∞). Then the multiplicative identity 1 ∈ 푅 has finite additive order 푛, from which it follows that for all 푟 ∈ 푅, 푛푟 = 푛(1 ⋅ 푟) = (푛 ⋅ 1)푟 = 0 ⋅ 푔 = 0 ,

Date: January 3, 2018. 1 2 PETER J. HAINE

which shows that all elements of 푈(푅) have order at most 푛, contradicting the fact that 퐙(푝∞) has elements of arbitrarily large order. (c) The functor (−)× ∶ Ring → Grp is neither full, faithful, nor essentially surjective on ob- jects, hence not an equivalence of categories To see that (−)× is not full, notice that since 퐙 is the initial ring, for any ring 푅 there is a unique 퐙 → 푅. However 퐙× ≅ 퐙/2, and the group of units of 퐅푝, for an odd prime 푝 is 퐙/(푝 − 1). However, there are two group homomorphisms 퐙/2 → 퐙/(푝 − 1), the trivial homomorphism and the homomorphism that sends the nontrivial element of 퐙/2 to the unique element of 퐙/(푝 − 1) of order 2, hence the map of hom-sets is not surjective. To see that (−)× is not faithful, suppose 푘 a and consider the 푘[푡]. It is a well-known fact that the automorphisms of 푘[푡] that fix 푘 are the affine maps 푡 ↦ 푎푡 + 푏, where 푎 ∈ 푘×. Since the group of units of 푘[푡] is simply 푘×, any two auto- fixing 푘 induce the identity homomorphism on 푘×, so the map Ring(푘[푡], 푘[푡]) → Grp(푘×, 푘×) is not injective in general as there are usually many affine transformations of 푘[푡] (even for 퐅2). A paper of Pearson and Schneider from 1970 shows that not every cyclic group is isomorphic to the group of units of some ring [1, Cor. 3], though this is nontrivial. In particular, the cyclic group 퐙/5 is not isomorphic to 푅× for any ring 푅. This proves that (−)× is not essentially surjective on objects (d) The inclusion functor Ring ↪ Rng obtained by dropping the multiplicative unit is faithful but neither full nor essentially surjective on objects, hence not an equivalence of categories. The inclusion is not full since the zero homomorphism is not a homomorphism in Ring but is a homomorphism in Rng: since 퐙 is the initial ring, there is one homomor- phism 퐙 → 푅 in Ring, but there are two homomorphisms 퐙 → 푅 in Rng. It is not essentially surjective on objects since a ring with no multiplicative unit clearly is not isomorphic to a ring with a multiplicative unit. (e) The inclusion functor Fld ↪ Ring is fully faithful, but not essentially surjective on objects, hence not an equivalence of categories. Since a field homomorphism is simply a ring homomorphism between fields, the category Fld forms a full subcategory of Ring. Since not every ring is isomorphic to a field, the inclusion is not essentially surjective on objects. (f) In the forgetful functor 푈∶ Mod푅 → Ab is faithful and essentially surjective on objects, but need not be full—moreover, the fullness of 푈 depends on the underlying ring 푅. Hence, in general, the forgetful functor is not an equivalence of categories. To see that 푈 is faithful, notice that for any 푅- homomorphism 푀 → 푀′, is a homomorphism of the underlying group, but respects additional structure. Thus, in particular, for 푅-module homomorphisms to be distinct, they must already be distinct as homomorphisms of underlying groups and in addition respect the module structure. Hence the map of Hom-sets is injective. To see that 푈 is essentially surjective on objects, it suffices to notice that given any ring 푅 and abelian group 퐴, there is a trivial 푅-module 퐴푅 given by setting 푟푎 = 푎 for 푟 ∈ 푅 and 푎 ∈ 퐴. Then 푈(퐴푅) = 퐴. Now notice that 푈 is not necessarily full. Treating a ring 푅 as an 푅-module, since an endomorphism 휙∶ 푅 → 푅 is uniquely determined by 휙(1) we have that End푅(푅) ≅ 푅. In general, however, group homomorphisms are only determined by the of EXAMPLES OF SOME PROPERTIES OF FUNCTORS 3

one group element when the group is free, so in general the endomorphisms ring of the underlying group is not isomorphic to 푅, though obviously this depends on the underlying ring 푅. When 푅 = 퐙, this is true, and, moreover, since abelian groups are precisely 퐙-modules, it is obvious that the “forgetful” functor 푈∶ Mod퐙 → Ab is, in particular, an equivalence of categories.

References 1. K. R. Pearson and J. E. Schneider, Rings with a cyclic group of units, Journal of Algebra 16 (1970), 243–251. 2. Emily Riehl, in context, Aurora: Dover Modern Math Originals, Dover Publications, 2016.