Gate (math)

The goal - functor is a mathematical term from the branch of homological algebra . It is a bi-functor that occurs when examining the tensor product. Along with the Ext functor, it is one of the most important constructions of homological algebra.

Motivation using tensor products

We consider categories of modules over a ring . Is ${\displaystyle R}$

${\displaystyle 0\rightarrow X{\xrightarrow {\alpha }}Y{\xrightarrow {\beta }}Z\rightarrow 0}$

is a short exact sequence of left- modules and module morphisms and if it is a right- module, then tensing the above sequence from the left leads to an exact sequence ${\displaystyle R}$ ${\displaystyle A}$ ${\displaystyle R}$ ${\displaystyle A}$

${\displaystyle A\otimes _{R}X{\xrightarrow {\mathrm {id} _{A}\otimes \alpha }}A\otimes _{R}Y{\xrightarrow {\mathrm {id} _{A}\otimes \beta }}A\otimes _{R}Z\rightarrow 0}$

of Abelian groups , which in general cannot be continued with the zero object to the left to an exact sequence, i.e. it is generally not injective , or in short: the tensor functor is right- exact but generally not left-exact. ${\displaystyle \mathrm {id} _{A}\otimes \alpha }$

As an example, consider the short exact sequence

${\displaystyle 0\rightarrow \mathbb {Z} {\xrightarrow {\alpha }}\mathbb {Z} {\xrightarrow {\beta }}\mathbb {Z} _{2}\rightarrow 0}$

von modules, where and be the natural mapping from to the remainder class group . If you tensor this sequence , it is not injective, because it is ${\displaystyle \mathbb {Z} }$ ${\displaystyle \alpha (n):=2n}$ ${\displaystyle \beta }$ ${\displaystyle \mathbb {Z} }$ ${\displaystyle \mathbb {Z} _{2}=\{{\overline {0}},{\overline {1}}\}}$ ${\displaystyle \mathbb {Z} _{2}}$ ${\displaystyle \mathrm {id} _{\mathbb {Z} _{2}}\otimes \alpha }$

${\displaystyle (\mathrm {id} _{\mathbb {Z} _{2}}\otimes \alpha )({\overline {1}}\otimes 1)=\mathrm {id} _{\mathbb {Z} _{2}}({\overline {1}})\otimes \alpha (1)={\overline {1}}\otimes 2\cdot 1=2\cdot {\overline {1}}\otimes 1={\overline {0}}\otimes 1=0}$.

The factor 2 was shifted from the torsion-free group to the torsion group by means of a tensor operation and resulted in a 0 there. This is the typical reason why the injectivity of the morphism is lost in the transition to the tensored sequence. The lack of injectivity leads to the appearance of a nucleus and gives rise to the following definition. ${\displaystyle \mathbb {Z} }$ ${\displaystyle \mathbb {Z} _{2}}$ ${\displaystyle \alpha }$

definition

Let there be a right module and a left module. Be further ${\displaystyle A}$ ${\displaystyle R}$ ${\displaystyle B}$ ${\displaystyle R}$

${\displaystyle 0\rightarrow S{\xrightarrow {\mu }}P{\xrightarrow {\nu }}B\rightarrow 0}$

a short exact sequence with a projective module . Then one defines the Abelian group ${\displaystyle P}$

${\displaystyle \mathrm {Tor} (A,B):=\mathrm {ker} (A\otimes _{R}S{\xrightarrow {\mathrm {id} _{A}\otimes \mu }}A\otimes _{R}P)}$

and one can show that this definition does not depend on the chosen exact sequence with projective . This justifies the spelling without reference to this sequence. Sometimes you add the ring and write . ${\displaystyle 0\rightarrow S\rightarrow P\rightarrow B\rightarrow 0}$ ${\displaystyle P}$ ${\displaystyle \mathrm {Tor} (A,B)}$ ${\displaystyle R}$ ${\displaystyle \mathrm {Tor} ^{R}(A,B)}$

If there is a morphism, one can see from the commutative diagram ${\displaystyle \alpha \colon A\rightarrow A'}$

${\displaystyle {\begin{array}{cccccc}A\otimes _{R}S&{\xrightarrow {\mathrm {id} _{A}\otimes \mu }}&A\otimes _{R}P&{\xrightarrow {\mathrm {id} _{A}\otimes \nu }}&A\otimes _{R}B&\rightarrow 0\\\downarrow _{\alpha \otimes \mathrm {id} _{S}}&&\downarrow _{\alpha \otimes \mathrm {id} _{P}}&&\downarrow _{\alpha \otimes \mathrm {id} _{B}}&\\A'\otimes _{R}S&{\xrightarrow {\mathrm {id} _{A'}\otimes \mu }}&A'\otimes _{R}P&{\xrightarrow {\mathrm {id} _{A'}\otimes \nu }}&A'\otimes _{R}B&\rightarrow 0\end{array}}}$,

that the restriction of the core of to and so depicting a homomorphism defined. In this way we get a functor from the category of right- modules to the category of Abelian groups. ${\displaystyle \alpha \otimes \mathrm {id} _{S}}$ ${\displaystyle \mathrm {id} _{A}\otimes \mu }$ ${\displaystyle \mathrm {ker} (\mathrm {id} _{A'}\otimes \mu )}$ ${\displaystyle \alpha _{*}\colon \mathrm {Tor} (A,B)\rightarrow \mathrm {Tor} (A',B)}$ ${\displaystyle \mathrm {Tor} ^{R}(-,B)\colon {\mathfrak {rMod}}_{R}\rightarrow {\mathfrak {Ab}}}$ ${\displaystyle R}$

You can also swap the roles of and , that is, you start from the exact sequence of right- modules and show that you get a group that is naturally isomorphic to the above definition , which can therefore also be called or . Overall, you get a bi-functor like this ${\displaystyle A}$ ${\displaystyle B}$ ${\displaystyle 0\rightarrow S\rightarrow P\rightarrow A}$ ${\displaystyle R}$ ${\displaystyle \mathrm {ker} (S\otimes _{R}B\rightarrow P\otimes _{R}B)}$ ${\displaystyle \mathrm {Tor} (A,B)}$ ${\displaystyle \mathrm {Tor} ^{R}(A,B)}$

${\displaystyle \mathrm {Tor} ^{R}(-,-)\colon {\mathfrak {rMod}}_{R}\times {\mathfrak {lMod}}_{R}\rightarrow {\mathfrak {Ab}}}$

from the product of the category of the right modules over with the category of the left modules over to the category of the Abelian groups. ${\displaystyle R}$ ${\displaystyle R}$

The Tor functor is additive, that is, one has natural isomorphisms

${\displaystyle \mathrm {Tor} ^{R}(A\oplus A',B)\cong \mathrm {Tor} ^{R}(A,B)\oplus \mathrm {Tor} ^{R}(A',B)}$

${\displaystyle \mathrm {Tor} ^{R}(A,B\oplus B')\cong \mathrm {Tor} ^{R}(A,B)\oplus \mathrm {Tor} ^{R}(A,B')}$

for right modules and left modules . ${\displaystyle R}$ ${\displaystyle A,A^{'}}$ ${\displaystyle R}$ ${\displaystyle B,B^{'}}$

Abelian groups

If you choose as the base ring, you are in the category of the Abelian groups, because these are precisely the modules, and because of the commutativity of the base ring you do not have to differentiate between left and right modules. In this category there are certain simplifications and one finds a connection between the gate functor and the torsion of groups that gives it its name . ${\displaystyle \mathbb {Z} }$ ${\displaystyle \mathbb {Z} }$

Alternative description of gate (A, B)

In the case of Abelian groups and can be presented as follows by generators and relations . ${\displaystyle A}$ ${\displaystyle B}$ ${\displaystyle \mathrm {Tor} (A,B)}$

Let the set of producers be the set of all symbols with , and , with the module operation being written once on the left and once on the right for practical reasons, a distinction is not necessary, as mentioned above. Let the set of relations contain all expressions of the form ${\displaystyle {\mathcal {E}}}$ ${\displaystyle \langle a,m,b\rangle }$ ${\displaystyle a\in A,m\in \mathbb {Z} ,b\in B}$ ${\displaystyle am=0}$ ${\displaystyle mb=0}$ ${\displaystyle \mathbb {Z} }$ ${\displaystyle {\mathcal {R}}}$

${\displaystyle \langle a_{1}+a_{2},m,b\rangle =\langle a_{1},m,b\rangle +\langle a_{2},m,b\rangle ,\quad \langle a_{1},m,b\rangle ,\langle a_{2},m,b\rangle \in {\mathcal {E}}}$

${\displaystyle \langle a,m,b_{1}+b_{2}\rangle =\langle a,m,b_{1}\rangle +\langle a,m,b_{2}\rangle ,\quad \langle a,m,b_{1}\rangle ,\langle a,m,b_{2}\rangle \in {\mathcal {E}}}$

${\displaystyle \langle a,mn,b\rangle =\langle am,n,b\rangle ,\quad \langle am,n,b\rangle \in {\mathcal {E}}}$

${\displaystyle \langle a,mn,b\rangle =\langle a,m,nb\rangle ,\quad \langle a,m,nb\rangle \in {\mathcal {E}}}$

Then one can show that the group presented by is too isomorphic. To construct an image, let us use a short exact sequence with a projective module and a generator. Choose with . Then and because of the accuracy there is exactly one with . You can show that it doesn't depend on the choice . There ${\displaystyle \langle {\mathcal {E}}|{\mathcal {R}}\rangle }$ ${\displaystyle \mathrm {Tor} (A,B)}$ ${\displaystyle \langle {\mathcal {E}}|{\mathcal {R}}\rangle \rightarrow \mathrm {Tor} (A,B)}$ ${\displaystyle 0\rightarrow S{\xrightarrow {\mu }}P{\xrightarrow {\nu }}B\rightarrow 0}$ ${\displaystyle \mathbb {Z} }$ ${\displaystyle P}$ ${\displaystyle \langle a,m,b\rangle }$ ${\displaystyle p\in P}$ ${\displaystyle \nu (p)=b}$ ${\displaystyle \nu (mp)=mb=0}$ ${\displaystyle s\in S}$ ${\displaystyle \mu (s)=mp}$ ${\displaystyle a\otimes s}$ ${\displaystyle p}$

${\displaystyle (\mathrm {id} _{A}\otimes \mu )(a\otimes s)=a\otimes \mu (s)=a\otimes mp=am\otimes p=0\otimes p=0}$,

lies at the core of and thus by definition in . The construction presented therefore defines a map that can be shown to be a group isomorphism. ${\displaystyle a\otimes s}$ ${\displaystyle \mathrm {id} _{A}\otimes \mu }$ ${\displaystyle \mathrm {Tor} (A,B)}$ ${\displaystyle \langle {\mathcal {E}}|{\mathcal {R}}\rangle \rightarrow \mathrm {Tor} (A,B)}$

Characterization of torsion-free groups

For an Abelian group , the following statements are equivalent: ${\displaystyle A}$

• ${\displaystyle A}$is torsion-free, i.e. contains no elements of finite order apart from 0 .
• ${\displaystyle \mathrm {Tor} (A,B)=0}$for all Abelian groups . ${\displaystyle B}$
• For all injective group homomorphisms is also injective. ${\displaystyle \beta \colon B\rightarrow C}$ ${\displaystyle \mathrm {id} _{A}\otimes \beta \colon A\otimes _{\mathbb {Z} }B\rightarrow A\otimes _{\mathbb {Z} }C}$
• Every exact sequence of Abelian groups is converted back into an exact sequence by tensing with . ${\displaystyle A}$

In particular , if one of the groups is the same or is. ${\displaystyle \mathrm {Tor} (A,B)=0}$ ${\displaystyle \mathbb {Z} }$ ${\displaystyle \mathbb {Q} }$

Finally generated abelian groups

${\displaystyle \mathrm {Tor} (A,B)}$can be completely calculated for finitely generated Abelian groups. According to the main theorem about finitely generated Abelian groups , such groups are direct sums of cyclic groups , so that, due to the additivity of the Tor functor, can only be determined for cyclic groups. If one of the groups is the same , then only the case of finite cyclic groups is and remains. Be the cyclic group of order . Then follows ${\displaystyle \mathrm {Tor} (A,B)}$ ${\displaystyle \mathbb {Z} }$ ${\displaystyle \mathrm {Tor} (A,B)=0}$ ${\displaystyle \mathbb {Z} _{n}}$ ${\displaystyle n}$

${\displaystyle \mathrm {Tor} (\mathbb {Z} _{n},B)\cong \{b\in B;nb=0\}}$

and from this, if one denotes the greatest common factor of and with : ${\displaystyle m}$ ${\displaystyle n}$ ${\displaystyle (m,n)}$

${\displaystyle \mathrm {Tor} (\mathbb {Z} _{m},\mathbb {Z} _{n})\cong \mathbb {Z} _{(m,n)}}$,

but which can also be derived directly from the definition with the resolution . Therewith is determined for finitely generated abelian groups. ${\displaystyle 0\rightarrow \mathbb {Z} {\xrightarrow {a\mapsto ma}}\mathbb {Z} \rightarrow \mathbb {Z} _{m}}$ ${\displaystyle \mathrm {Tor} (A,B)}$

Gate as a derivative of the tensor functor

A more general definition can be obtained through

${\displaystyle \mathrm {Tor} _{n}^{R}(A,B):=L_{n}(-\otimes _{R}B)(A)\cong L_{n}(A\otimes _{R}-)(B)}$

as the -th left derivative of the tensor functor. If the basic ring is given by the context, it is left out in the name and simply writes . So you get a sequence of bi-functors ${\displaystyle n}$ ${\displaystyle R}$ ${\displaystyle \mathrm {Tor} _{n}(A,B)}$

${\displaystyle \mathrm {Tor} _{n}^{R}(-,-)\colon {\mathfrak {rMod}}_{R}\times {\mathfrak {lMod}}_{R}\rightarrow {\mathfrak {Ab}}}$.

If one uses projective resolutions for the computation of , one sees that coincides with the above defined -function. ${\displaystyle \mathrm {Tor} _{n}^{R}(A,B)}$ ${\displaystyle \mathrm {Tor} _{1}^{R}(A,B)}$ ${\displaystyle \mathrm {Tor} }$

The following long exact sequences are obtained from the general theory, which show how the gate functor compensates for the lack of left-hand accuracy of the tensor functor.

If there is a short exact sequence of right- modules and a left- module, you have a long exact sequence ${\displaystyle 0\rightarrow A\rightarrow A^{'}\rightarrow A^{''}\rightarrow 0}$ ${\displaystyle R}$ ${\displaystyle B}$ ${\displaystyle R}$

${\displaystyle \ldots \rightarrow \mathrm {Tor} _{2}(A^{''},B)\rightarrow \mathrm {Tor} _{1}(A,B)\rightarrow \mathrm {Tor} _{1}(A^{'},B)\rightarrow \mathrm {Tor} _{1}(A^{''},B)}$

${\displaystyle \rightarrow A\otimes _{R}B\rightarrow A^{'}\otimes _{R}B\rightarrow A^{''}\otimes _{R}B\rightarrow 0}$.

If there is a short exact sequence of left- modules and a right- module, you have a long exact sequence ${\displaystyle 0\rightarrow B\rightarrow B^{'}\rightarrow B^{''}\rightarrow 0}$ ${\displaystyle R}$ ${\displaystyle A}$ ${\displaystyle R}$

${\displaystyle \ldots \rightarrow \mathrm {Tor} _{2}(A,B^{''})\rightarrow \mathrm {Tor} _{1}(A,B)\rightarrow \mathrm {Tor} _{1}(A,B^{'})\rightarrow \mathrm {Tor} _{1}(A,B^{''})}$

${\displaystyle \rightarrow A\otimes _{R}B\rightarrow A\otimes _{R}B^{'}\rightarrow A\otimes _{R}B^{''}\rightarrow 0}$.

Individual evidence

1. ^ PJ Hilton and U. Stammbach: A course in homological algebra. 2nd edition, Springer-Verlag, Graduate Texts in Mathematics, 1997, ISBN 0-387-94823-6 , Chapter III.8: The Functor Tor
2. ^ Saunders Mac Lane : Homology , Springer Grundlehren der Mathematischen Wissenschaften, Volume 114 (1967), Chap. V, § 6, "Torsion Products of Groups"
3. ^ Saunders Mac Lane: Homology , Springer Grundlehren der Mathematischen Wissenschaften, Volume 114 (1967), Chap. V, theorem 6.2
4. ^ Saunders Mac Lane: Homology , Springer Grundlehren der Mathematischen Wissenschaften, Volume 114 (1967), Chap. V, § 6, "Torsion Products of Groups"
5. ^ PJ Hilton and U. Stammbach: A course in homological algebra. 2nd edition, Springer-Verlag, Graduate Texts in Mathematics, 1997, ISBN 0-387-94823-6 , Chapter IV.11: The Functor ${\displaystyle \mathrm {Tor} _{n}^{\Lambda }}$