## 12.10 Serre subcategories

In [Chapter I, Section 1, Serre_homotopie_classes] a notion of a “class” of abelian groups is defined. This notion has been extended to abelian categories by many authors (in slightly different ways). We will use the following variant which is virtually identical to Serre's original definition.

Definition 12.10.1. Let $\mathcal{A}$ be an abelian category.

1. A Serre subcategory of $\mathcal{A}$ is a nonempty full subcategory $\mathcal{C}$ of $\mathcal{A}$ such that given an exact sequence1

$A \to B \to C$

with $A, C \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, then also $B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$.

2. A weak Serre subcategory of $\mathcal{A}$ is a nonempty full subcategory $\mathcal{C}$ of $\mathcal{A}$ such that given an exact sequence

$A_0 \to A_1 \to A_2 \to A_3 \to A_4$

with $A_0, A_1, A_3, A_4$ in $\mathcal{C}$, then also $A_2$ in $\mathcal{C}$.

In some references the second notion is called a “thick” subcategory and in other references the first notion is called a “thick” subcategory. However, it seems that the notion of a Serre subcategory is universally accepted to be the one defined above. Note that in both cases the category $\mathcal{C}$ is abelian and that the inclusion functor $\mathcal{C} \to \mathcal{A}$ is a fully faithful exact functor. Let's characterize these types of subcategories in more detail.

Lemma 12.10.2. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{C}$ be a subcategory of $\mathcal{A}$. Then $\mathcal{C}$ is a Serre subcategory if and only if the following conditions are satisfied:

1. $0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$,

2. $\mathcal{C}$ is a strictly full subcategory of $\mathcal{A}$,

3. any subobject or quotient of an object of $\mathcal{C}$ is an object of $\mathcal{C}$,

4. if $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ is an extension of objects of $\mathcal{C}$ then also $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$.

Moreover, a Serre subcategory is an abelian category and the inclusion functor is exact.

Proof. Omitted. $\square$

Lemma 12.10.3. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{C}$ be a subcategory of $\mathcal{A}$. Then $\mathcal{C}$ is a weak Serre subcategory if and only if the following conditions are satisfied:

1. $0 \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$,

2. $\mathcal{C}$ is a strictly full subcategory of $\mathcal{A}$,

3. kernels and cokernels in $\mathcal{A}$ of morphisms between objects of $\mathcal{C}$ are in $\mathcal{C}$,

4. if $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{A})$ is an extension of objects of $\mathcal{C}$ then also $A \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$.

Moreover, a weak Serre subcategory is an abelian category and the inclusion functor is exact.

Proof. Omitted. $\square$

Lemma 12.10.4. Let $\mathcal{A}$, $\mathcal{B}$ be abelian categories. Let $F : \mathcal{A} \to \mathcal{B}$ be an exact functor. Then the full subcategory of objects $C$ of $\mathcal{A}$ such that $F(C) = 0$ forms a Serre subcategory of $\mathcal{A}$.

Proof. Omitted. $\square$

Definition 12.10.5. Let $\mathcal{A}$, $\mathcal{B}$ be abelian categories. Let $F : \mathcal{A} \to \mathcal{B}$ be an exact functor. Then the full subcategory of objects $C$ of $\mathcal{A}$ such that $F(C) = 0$ is called the kernel of the functor $F$, and is sometimes denoted $\mathop{\mathrm{Ker}}(F)$.

Any Serre subcategory of an abelian category is the kernel of an exact functor. In Examples, Section 109.76 we discuss this for Serre's original example of torsion groups.

Lemma 12.10.6. Let $\mathcal{A}$ be an abelian category. Let $\mathcal{C} \subset \mathcal{A}$ be a Serre subcategory. There exists an abelian category $\mathcal{A}/\mathcal{C}$ and an exact functor

$F : \mathcal{A} \longrightarrow \mathcal{A}/\mathcal{C}$

which is essentially surjective and whose kernel is $\mathcal{C}$. The category $\mathcal{A}/\mathcal{C}$ and the functor $F$ are characterized by the following universal property: For any exact functor $G : \mathcal{A} \to \mathcal{B}$ such that $\mathcal{C} \subset \mathop{\mathrm{Ker}}(G)$ there exists a factorization $G = H \circ F$ for a unique exact functor $H : \mathcal{A}/\mathcal{C} \to \mathcal{B}$.

Proof. Consider the set of arrows of $\mathcal{A}$ defined by the following formula

$S = \{ f \in \text{Arrows}(\mathcal{A}) \mid \mathop{\mathrm{Ker}}(f), \mathop{\mathrm{Coker}}(f) \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}) \} .$

We claim that $S$ is a multiplicative system. To prove this we have to check MS1, MS2, MS3, see Categories, Definition 4.27.1.

It is clear that identities are elements of $S$. Suppose that $f : A \to B$ and $g : B \to C$ are elements of $S$. There are exact sequences

$\begin{matrix} 0 \to \mathop{\mathrm{Ker}}(f) \to \mathop{\mathrm{Ker}}(gf) \to \mathop{\mathrm{Ker}}(g) \\ \mathop{\mathrm{Coker}}(f) \to \mathop{\mathrm{Coker}}(gf) \to \mathop{\mathrm{Coker}}(g) \to 0 \end{matrix}$

Hence it follows that $gf \in S$. This proves MS1. (In fact, a similar argument will show that $S$ is a saturated multiplicative system, see Categories, Definition 4.27.20.)

Consider a solid diagram

$\xymatrix{ A \ar[d]_ t \ar[r]_ g & B \ar@{..>}[d]^ s \\ C \ar@{..>}[r]^ f & C \amalg _ A B }$

with $t \in S$. Set $W = C \amalg _ A B = \mathop{\mathrm{Coker}}((t, -g) : A \to C \oplus B)$. Then $\mathop{\mathrm{Ker}}(t) \to \mathop{\mathrm{Ker}}(s)$ is surjective and $\mathop{\mathrm{Coker}}(t) \to \mathop{\mathrm{Coker}}(s)$ is an isomorphism. Hence $s$ is an element of $S$. This proves LMS2 and the proof of RMS2 is dual.

Finally, consider morphisms $f, g : B \to C$ and a morphism $s : A \to B$ in $S$ such that $f \circ s = g \circ s$. This means that $(f - g) \circ s = 0$. In turn this means that $I = \mathop{\mathrm{Im}}(f - g) \subset C$ is a quotient of $\mathop{\mathrm{Coker}}(s)$ hence an object of $\mathcal{C}$. Thus $t : C \to C' = C/I$ is an element of $S$ such that $t \circ (f - g) = 0$, i.e., such that $t \circ f = t \circ g$. This proves LMS3 and the proof of RMS3 is dual.

Having proved that $S$ is a multiplicative system we set $\mathcal{A}/\mathcal{C} = S^{-1}\mathcal{A}$, and we set $F$ equal to the localization functor $Q$. By Lemma 12.8.4 the category $\mathcal{A}/\mathcal{C}$ is abelian and $F$ is exact. If $X$ is in the kernel of $F = Q$, then by Lemma 12.8.3 we see that $0 : X \to Z$ is an element of $S$ and hence $X$ is an object of $\mathcal{C}$, i.e., the kernel of $F$ is $\mathcal{C}$. Finally, if $G$ is as in the statement of the lemma, then $G$ turns every element of $S$ into an isomorphism. Hence we obtain the functor $H : \mathcal{A}/\mathcal{C} \to \mathcal{B}$ from the universal property of localization, see Categories, Lemma 4.27.8. We still have to show the functor $H$ is exact. To do this it suffices to show that $H$ commutes with taking kernels and cokernels, see Lemma 12.7.2. Let $A \to B$ be a morphism in $\mathcal{A}/\mathcal{C}$. We may represent $A \to B$ as $fs^{-1}$ where $s : A' \to A$ is in $S$ and $f : A' \to B$ an arbitrary morphism of $\mathcal{A}$. Since $F = Q$ maps $s$ to an isomorphism in the quotient category $\mathcal{A}/\mathcal{C}$, it suffices to show that $H$ commutes with taking kernels and cokernels of morphisms $f : A \to B$ of $\mathcal{A}$. But here we have $H(f) = G(f)$ and the result follows from the fact that $G$ is exact. $\square$

Lemma 12.10.7. Let $\mathcal{A}$, $\mathcal{B}$ be abelian categories. Let $F : \mathcal{A} \to \mathcal{B}$ be an exact functor. Let $\mathcal{C} \subset \mathcal{A}$ be a Serre subcategory contained in the kernel of $F$. Then $\mathcal{C} = \mathop{\mathrm{Ker}}(F)$ if and only if the induced functor $\overline{F} : \mathcal{A}/\mathcal{C} \to \mathcal{B}$ (Lemma 12.10.6) is faithful.

Proof. We will use the results of Lemma 12.10.6 without further mention. The “only if” direction is true because the kernel of $\overline{F}$ is zero by construction. Namely, if $f : X \to Y$ is a morphism in $\mathcal{A}/\mathcal{C}$ such that $\overline{F}(f) = 0$, then $\overline{F}(\mathop{\mathrm{Im}}(f)) = \mathop{\mathrm{Im}}(\overline{F}(f)) = 0$, hence $\mathop{\mathrm{Im}}(f) = 0$ by the assumption on the kernel of $F$. Thus $f = 0$.

For the “if” direction, let $X$ be an object of $\mathcal{A}$ such that $F(X) = 0$. Then $\overline{F}(\text{id}_ X) = \text{id}_{\overline{F}(X)} = 0$, thus $\text{id}_ X = 0$ in $\mathcal{A}/\mathcal{C}$ by faithfulness of $\overline{F}$. Hence $X = 0$ in $\mathcal{A}/\mathcal{C}$, that is $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. $\square$

 By Definition 12.5.7 this means $\mathop{\mathrm{Im}}(A \to B) = \mathop{\mathrm{Ker}}(B \to C)$.

Comment #2613 by Xiaofa Chen on

Why is the functor G an exact functor? Is every exact sequence in the localization category isomorphisc to some standard exact sequence? I quite doubt that.

Comment #2615 by on

@Xiaofa: Do not understand your question. The only functor $G$ on the page is a functor which is assumed to be exact!

@Everybody: If you make a comment on a lemma, proposition, definition, etc, please, please go to the tags page of the thing you want to comment on and do not comment on the webpage of the section!

Comment #3824 by Xiaofa Chen on

Yes. It's my mistake. At that time, I just learned this theory and I took it for granted that it is universal among all functors which annihilate $\mathcal C$. Then I realized that I is universal among exact functors which annihilate $\mathcal C$. And I don't know how to delete a 'stupid comment' as that.

Comment #6301 by Mohammed on

In the definition, a serre subcategory of an abelian category must verifies $A, C \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ if and only if $B \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$

Comment #6413 by on

@#6301. I do not understand your comment.

Comment #6734 by Mohammed on

In the definition of a Serre subcategory, you said : .....with $A,C \in Ob(C)$, then also $B∈Ob(C)$ I think the right one is $A,C \in Ob(C)$, if and only if also $B∈Ob(C)$.

Comment #6737 by Bach on

@Mohammed: There are variants of this definition. His definition doesn't need to be wrong.

Comment #6741 by Mohammed on

In this case, any Quillen exact category (extension closed) is an abelian category? A counterexample is the subcategory of modules with finite projective dimension. It is a Serre subcategory in your sense (but I don't think it is a Serre subcategory), therefore it is abelian, which is not true in general. This is exactly Gabriel-Zisman localization. In their book the definition is compatible with yours. Thanks

Comment #6742 by Mohammed on

In this case, any Quillen exact category (extension closed) is an abelian category? A counterexample is the subcategory of modules with finite projective dimension. It is a Serre subcategory in your sense (but I don't think it is a Serre subcategory), therefore it is abelian, which is not true in general. This is exactly Gabriel-Zisman localization. In their book the definition is not compatible with yours. Thanks

Comment #6743 by on

@#6742: Look, the definition is almost literally the one from Serre's paper! See (I) on page 259 of the Serre paper cited in the introduction to this section.

Still, if you can find a mistake in a proof or you have a counter example to a lemma or proposition or theorem, then please let us know.

But beware, that if there is just some mismatch between what is being said here and in a different reference, then it may very well be the case that the definitions are different (and both references can be correct even though at first sight they appear incompatible). This happens a lot, so please check carefully.

Enjoy!

Comment #6744 by Mohammed on

Thank you for the explanations. But tell how can you prove that if $C$ is Serre subcategory, then any subobject or quotient of an object of $C$ is an object of $C$? (Lemma 02MP (3)) For the example, just take the full subcategory of finitely generated modules with finite projective dimension. Thank you very much.

Comment #6745 by on

Dear Mohammed, are you asking me to prove Lemma 12.10.2? A counter question: What did you try yourself in order to prove Lemma 12.10.2? For example, can you prove that a Serre subcategory (as defined in this section) of an abelian category contains $0$? If yes, how did you do it?

Comment #6747 by on

OK, Mohammed, just to make sure we are on the same page. Your example isn't a Serre subcategory in general (in the sense defined in this section). For example if $A$ is an Artinian local ring but not a field, then any $A$-module of finite projective dimension is free (this is a great exercise in commutative algebra -- I highly recommend it). But the subcategory $\mathcal{C}$ of free modules isn't a Serre subcategory of the category of all modules for almost any ring $A$. For example, if $a \in A$ is a nonzero nonunit, then $A \to A/(a) \to 0$ is exact and $A$ and $0$ are in $\mathcal{C}$ but $A/(a)$ is not.

This is also a hint to the proof of Lemma 12.10.2. Also, I think you never looked at the comments on the page of Definition 12.10.1 as I suggested which would have clarified everything for you a while ago.

Comment #6748 by Mohammed on

Dear Johan, Since $C$ is a full subcategory of an abelian category $A$, $Hom_{C}(X,Y) = Hom_{A}(X,Y)$ for all $X, Y \in C$. Now, by the definition of the zero morphism, given $X,Y \in C$ we can find a sequence $X \rightarrow 0 \rightarrow Y$ in $A$, but since this latter is exact in the middle and by the defintion of Serre subcategory, $X Y \in C$, $0$ also must be in $C$. Now, let's go back to a concrete coutreexample : Let $**mod-R**$ be the category of finitely generated modules over a artinian ring $**R**$ and let $**proj-R**$ be the full subcategory of $**mod-R**$ of projective modules. Clearly this is a Serre subcategory of $**mod-R**$. Indeed, if we have an exact sequence $0 \rightarrow X \rightarrow Z \rightarrow Y \rightarrow 0$ such that $X, Y in **proj-R**$. Sine $Y$ is projective, the latter exact sequence split, therefore $Z$ is a projective module. Now, in Lemma 0754 (3) fails. Indeed, a submodule of a projective module is not always projective, unless the ring $**R**$ is hereditary. As a conclusion, the category $**proj-R**$ is not always abelian, but it is a Serre subcategory in your sense. Thank you!

Comment #6749 by on

Dear Mohammed, In your argument that proj-R is a Serre subcategory you are only considering exact sequences $X \to Y \to Z$ which extend to short exact sequences $0 \to X \to Y \to Z \to 0$ (note the added zeros!!!). But in Definition 12.10.1 we have to check the condition for all exact sequences $X \to Y \to Z$ (not necessarily short exact).

Hence proj-R is not a counter example because it isn't a Serre subcategory in the sense of Definition 12.10.1 as I said in my comment #6747. As far as I can tell from your comments we agree on what a Serre subcategory is. It seems that you think Definition 12.10.1 is formulated wrong. Can you please read the comments on Definition 12.10.1 and/or read the proof of Lemma 12.10.2 in Serre's paper. He has exactly the same definition and exactly the same lemma.

Hope you agree now!

Clearly, I have to change the wording of the definition as it causes too many people to get confused even though it is almost word for word the same as what Serre said!

Comment #6750 by Mohammed on

Dear Johan, Thank you for the example and yes, I saw the definition that you sent me at the beginning. Maybe I see the difference now. I was working with exact sequences of this form $0\rightarrow X \rightarrow Z \rightarrow Y \rightarrow 0$, while in your case, you are working with exact sequence of this form $X \rightarrow Z \rightarrow Y$, maybe this is why there is a misunderstanding. I will check Serre paper. Thank you for your time.

Comment #6751 by Mohammed on

Dear Johan, I just saw your last message. Yes I see the difference now. Thank you very much. Mohammed

Comment #6752 by on

OK, great! Happy we agree now. I will clarify the definition the next time I go through all the comments.

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