The Grothendieck group and the definition of integers
event_availablePosted·updateUpdated·
In the previous post, we examined the definition of a monoid; as a prime example, the natural numbers defined in set theory form a commutative monoid under addition. In this post, we introduce a method for obtaining an abelian group from a commutative semigroup.
First, considering the category \(\Ab\) of abelian groups, any abelian group can be viewed as a commutative monoid by forgetting the information about its inverses, so there exists a forgetful functor \(U: \Ab \rightarrow \cMon\). This functor is known to have a left adjoint \(K:\cMon \rightarrow \Ab\), and writing out this adjunction yields the formula
\[\Hom_\Ab(K(M), G)\cong\Hom_\cMon(M, U(G))\]
. That is, given a commutative monoid \(M\) and a monoid homomorphism \(M\rightarrow U(G)\), we must be able to obtain a unique group homomorphism \(K(M)\rightarrow G\).
Universal mapping problem
Now we must show the existence of the left adjoint \(K\) described above. More generally, following our main reference [Bou], we examine the process of obtaining an abelian group from a commutative semigroup. Meanwhile, using the unit of the adjunction, we can write out the property that \(K\) must satisfy.
An abelian group \(K(S)\) and a semigroup homomorphism \(\eta_S:S\rightarrow K(S)\) are a pair satisfying the following property.
(Universal mapping problem) Whenever an arbitrary abelian group \(G\) and an arbitrary semigroup homomorphism \(f:S\rightarrow G\) are given, there exists a unique group homomorphism \(\bar{f}:K(S)\rightarrow G\) such that \(f=\bar{f}\circ\eta_S\) holds.
Interpreting this intuitively, \(K(S)\) can be thought of as the smallest abelian group obtained from \((S,+)\).
\(K(S)\) satisfying the above property is unique up to isomorphism.
Proposition 1 If an abelian group \(H\) and a semigroup homomorphism \(\eta_S'\) satisfy the above universal mapping problem, then \(K(S)\cong H\) holds.
Proof
First, consider the following diagram.
Then by the universal property, there exists \(\bar{\eta}_S': K(S)\rightarrow H\) such that \(\eta_S'= \bar{\eta}_S'\circ\eta_S\). Meanwhile, from the following diagram
using the universal property for \(H\), there exists \(\bar{\eta}_S:H\rightarrow K(S)\) such that \(\eta_S=\bar{\eta}_S\circ\eta_S'\). Then
and again by the universal property, since \(f\) satisfying \(f\circ \eta_S'=\eta_S'\) is unique, \(f=\id_H=\bar{\eta}_S'\circ \bar{\eta}_S\) holds. Or, in the language of diagrams, since the map \(H\rightarrow H\) making the following diagram commute is unique, we must have \(\id_H=\bar{\eta}_S'\circ \bar{\eta}_S\).
Similarly, one can show that \(\id_{K(S)}=\bar{\eta}_S\circ \bar{\eta}_S'\) holds, and therefore \(K(S)\cong H\) holds.
On the other hand, if \(S\) was already an abelian group, \(K(S)\) should turn out to be \(S\) itself without the need to add other elements.
Proposition 2 If \(S\) is an abelian group, the abelian group \(K(S)\) satisfying the above universal mapping problem satisfies \(K(S)\cong S\).
Proof
Since \(S\) and \(\id_S\) trivially satisfy the universal property, by the preceding Proposition 1, any abelian group satisfying the universal property must be isomorphic to \(S\).
The two propositions above show that \(K(S)\) satisfying the universal mapping problem is the abelian group we are seeking, but they do not show that \(K(S)\) actually exists.
Definition of \(K(S)\)
The reason why \(S\) cannot be an abelian group is that an identity element and inverses for arbitrary elements might not exist. Intuitively, this can be resolved by adding \(0\) and negative numbers.
For a given commutative semigroup \((S,+)\), consider the product semigroup \(S\times S\). (§Algebraic Structures, ⁋Example 5) If we think of the second component of \(S\times S\) as negative numbers, the equation
\[(a_1, b_1)+(a_2, b_2)=(a_1+a_2, b_1+b_2)\]
can be thought of as if representing
\[(a_1+a_2)-(b_1+b_2)=(a_1-b_1)+(a_2-b_2)\]
.
Of course, even if we choose \(a\) and \(b\) differently, the difference \(a-b\) can be the same value, so we define an equivalence relation \(R\) on \(S\times S\) as follows:
\[(a_1, b_1)\equiv (a_2, b_2)\pmod{R}\iff a_1+b_2+c=a_2+b_1+c\text{ for some $c\in S$}\]
First, we must show that this relation is an equivalence relation.
Lemma 3 The relation \(R\) defined above is an equivalence relation compatible with the operation on the product semigroup \(S\times S\).
Proof
First, we show that \(R\) is an equivalence relation. For any \((a,b)\in S\times S\),
\[a+b+c=a+b+c\]
holds for any \(c\in S\), so \((a,b)\equiv(a,b)\). Suppose \((a_1,b_1)\equiv (a_2,b_2)\). That is, for some \(c\in S\),
\[a_1+b_2+c=a_2+b_1+c\]
holds. But this is precisely the condition for \((a_2,b_2)\equiv (a_1,b_1)\), so \(R\) is symmetric. Finally, suppose \((a_1,b_1)\equiv(a_2,b_2)\) and \((a_2,b_2)\equiv (a_3,b_3)\). Then for some \(c\), \(c'\),
so \((a_1,b_1)\equiv(a_3,b_3)\) holds. That is, \(R\) is an equivalence relation.
Now we must show that \(R\) is compatible with the operation on \(S\times S\). To this end, suppose \((a_1, b_1)\equiv(a_1',b_1')\) and \((a_2, b_2)\equiv (a_2',b_2')\). We must show that \((a_1+a_2, b_1+b_2)\equiv(a_1'+a_2', b_1'+b_2')\). From the given conditions, there exist suitable \(c_1\), \(c_2\) such that
hold. Now, adding the two equations,
\((a_1+a_2)+(b_1'+b_2')+(c_1+c_2)=(a_1'+a_2')+(b_1+b_2)+(c_1+c_2)\)
holds, so by definition \((a_1+a_2, b_1+b_2)\equiv(a_1'+a_2', b_1'+b_2')\pmod{R}\) holds, and therefore \(R\) is compatible with the operation on \(S\times S\).
Therefore, \((S\times S)/R\) becomes a commutative semigroup. Let this be \(K(S)\).
Lemma 4 \(K(S)\) is an abelian group.
Proof
It suffices to show that \(K(S)\) has an identity element and inverses. Since we think of \((a,b)\) as \(a-b\), the identity element will be \((a,a)\), and the inverse of \((a,b)\) will be \(-(a-b)=b-a\), that is, \((b,a)\). Let us prove this.
First, for any \(c\in S\), we show that \([(c,c)]\) is an identity element. For any \([(a,b)]\in K(S)\),
\[[(a,b)]+[(c,c)]=[(a+c, b+c)]\]
holds. But since
\[(a+c)+b+d=(b+c)+a+d\]
holds for any \(d\in S\), we have \((a+c, b+c)\equiv (a,b)\), and therefore \([(a+c, b+c)]=[(a,b)]\) holds. By commutativity, \([(c,c)]+[(a,b)]=[(a,b)]\) also naturally holds, so \([(c,c)]\) is an identity element of \(K(S)\).
Meanwhile, for any \([(a,b)]\in K(S)\),
\[[(a,b)]+[(b,a)]=[(a+b,a+b)]\]
so by the preceding argument, \([(a,b)]+[(b,a)]\) is an identity element of \(K(S)\), and the same is true for \([(b,a)]+[(a,b)]\). Therefore, the inverse of any element of \(K(S)\) exists, so \(K(S)\) has the structure of a group.
Then \(K(S)\) is the abelian group we were looking for. That is, \(K(S)\) satisfies the above universal mapping problem.
Proposition 5 For a commutative semigroup \((S, +)\), the abelian group \(K(S)\) constructed as above and the natural semigroup homomorphism \(\eta_S:S\rightarrow K(S)\) satisfy the universal property.
Proof
First, let us think about what the natural semigroup homomorphism from \(S\) to \(K(S)\) should be. Since we treat \((a,b)\) as \(a-b\) in \(K(S)\), we can see that \(a\) corresponds in \(K(S)\) to \((a+b)-b\), that is, \([(a+b, b)]\). Therefore, let us define \(\eta_S\) by \(a\mapsto[(a+a, a)]\). Of course, choosing any \(b\) and defining it by \(a\mapsto[(a+b,b)]\) yields the same value.
To prove the universal property, let an arbitrary abelian group \(G\) and a semigroup homomorphism \(f:S\rightarrow G\) be given.
First, if \(\bar{f}:K(S)\rightarrow G\) satisfying the given property exists, \(\bar{f}\) must be unique. This is because for any \([(a,b)]\),
so the function values at each element are uniquely determined.
Now, taking a hint from the uniqueness proof, let us define \(\bar{f}([(a,b)])\) to be \(f(a)-f(b)\). First, this is well-defined. That is, if \((a_1,b_1)\equiv(a_2,b_2)\), then \(f(a_2)-f(b_2)=f(a_1)-f(b_1)\) holds. Since \((a_1,b_1)\equiv(a_2,b_2)\), there exists some \(c\in S\) such that \(a_1+b_2+c=a_2+b_1+c\), and therefore
Finally, that \(\bar{f}\) satisfies the given condition \(f=\bar{f}\circ\eta_S\) is clear upon calculation.
In this way, we have obtained the abelian group \(K(S)\) we wanted. In particular, we can define the integers in a rigorous manner.
Definition 6 For the monoid \((\mathbb{N},+)\), the abelian group obtained through the above process is written as \((\mathbb{Z},+)\).
Monoid of fractions
In the discussion above, we obtained \(K(S)\) by adding inverses of all elements of \(S\). Meanwhile, looking at Definition 6, what we actually do is add inverses only for the elements of the subset \(\mathbb{N}\setminus\{0\}\) of \(\mathbb{N}\). This can also be obtained by slightly modifying the discussions above; we omit the proofs and examine only the process.
Consider a commutative monoid \(E\), a subset \(S\) of \(E\), and the submonoid \(S'\) of \(E\) generated by \(S\). We also consider the operation of \(E\) to be written multiplicatively. If we define the following relation on \(E\times S'\):
\[(a,p)\equiv (b,q)\pmod{R}\iff aqs=bps\text{ for some $s\in S'$}\]
this relation is an equivalence relation compatible with the operation on \(E\times S'\), and therefore \((E\times S')/R\) becomes a monoid.
Definition 7 The monoid \((E\times S')/R\) obtained as above is called the monoid of fractions of \(E\) with denominator \(S\) and is denoted by \(E_S\). The element of this monoid having \((a,p)\) as a representative is denoted by \(a/p\).
In this case, since \(E\) is a monoid, unlike the discussion above, it has an identity element \(1\). Then the canonical morphism corresponding to the homomorphism \(\eta_S\) in Proposition 5 can be explicitly thought of as
\[\epsilon:E\rightarrow E_S;\quad a\mapsto a/1\]
.
References
[Bou] Bourbaki, N. Algebra I. Elements of Mathematics. Springer. 1998.
No comments yet.