Let $R$ be a non-commutative ring. Then we can define the semigroup ring of $R$ over the free non-commutative semigroup $F^{\rightarrow}(A)$. Then this semigroup ring $RF^{\rightarrow}(A)$ consists of ordered words over the alphabet $A$ with coefficients in $R$. This mirrors the construction of the polynomial ring $R[x_1,x_2,...]$ in commutative algebra as a free commutative semigroup ring.
Elements of the free non-commutative semigroup ring $RF^{\rightarrow}(x,y,z)$ consist of ordered words with coefficients in $R$. For example, we might get a polynomial like $5xyx + 6xz + 7zx$. A term like $xyx$ is not the same as $x^2y$ and $xz$ and $zx$ are not the same as one another. This non-commutativity would naturally add a great deal of complexity to compuations over non-commutative rings.
The issue with this free semigroup ring, is that it is not truly the ring of polynomial functions over $R$. Consider $c_1,c_2,...$ to be constants and $x,y,z,...$ to be indeterminates. Then a coefficient may not commute with its indeterminate, and $c_1 x \not= xc_1$. So for polynomial functions over a non-commutative ring you ultimately get potentially nasty terms like $c_1 x c_2 c_3 y c_4 x c_5 z c_6$.
A term of a polynomial function in a non-commutative ring is an alternating sequence of coefficients and indeterminates, where the coefficients can be identities. So for example $c_1xyc_2yxc_3$ is a term except the coefficients between $x$ and $y$ and then between $y$ and $x$ are simply identities, but there is always the possibility of having coefficients between indeterminates.
The terms of the free semigroup ring are already complicated enough, but terms in the ring of polynomial functions with scattered coefficients are even more messy. This is a serious impediment to doing noncommutative algebraic geometry, so what is the big idea? It seems that noncommutative geometry has more to do with functional analysis and operator theory then polynomials.
Noncommutative spaces can instead be studied using concepts of functional analysis and operator theory like C*-algebras. In particular, the Gelfand representation related C* algebras to locally compact Hausdorff spaces. The study of noncommutative geometry based upon techniques from functional analysis is an interesting direction, albiet one that is very different from you might expect.
Showing posts with label ring theory. Show all posts
Showing posts with label ring theory. Show all posts
Friday, September 24, 2021
Wednesday, September 22, 2021
Rings of multivariable Laurent polynomials
The commutative group ring of the free commutative group $F^{\circ}(X)$ over a field $k$ is a ring extension of the commutative semigroup ring of the free commutative monoid $F(X)$ over $k$. Although, $F^{\circ}(X)$ is a natural semigroup extension of $F(X)$, rings of multivariable Laurent polynomials don't have the same level of importance as polynomial rings in commutative algebra.
A distinguishing property of ordinary polynomials is that they can be cast into functions, so that over affine space $\mathbb{A}^n$ they are functions $f: \mathbb{A}^n \to k$ and this works for any commutative ring. On the other hand, for Laurent polynomials to be cast into functions $f : \mathbb{A}^n \to k$ we need to have some notion of division to deal with negative degree exponents, so we need to work over a field.
In those cases when we take the commutative group ring of a commutative ring $R$ which is not a field, with respect to $F^{\circ}(X)$ then what we get is certainly still a commutative ring, it is just not a commutative ring of functions. Consider the commutative group ring $\mathbb{Z}(\mathbb{Z}^n,+)$ consisting of polynomials with integer exponents and integer coefficients. Then this is a valid commutative ring, but its elements are not functions because $\mathbb{Z}$ is not a field.
So in some sense we ought to have our free commutative group ring be over a field $k$. Then we get terms like $5\frac{x}{yz} + \frac{6y}{xz}$ consisting of fractional monomials, but then given any term like this we can add to get $\frac{5x^2 + 6y^2}{xyz}$ which is always a polynomial with a monomial in the denominator. So we see that multivariable Laurent polynomials are simply rational functions with monomials in the denominator, so we have an embedding. \[ kF^{\circ}(x,y,z) \subseteq k(x,y,z) \] The ring of Laurent polynomials is simply the localisation of the polynomial ring by the multiplicative set of monomials, and so they are merely a special case of rational functions. Now it is clear why we don't see Laurent polynomials so much in commutative alebra. They are simply part of the far more important and general concept of fields of rational functions $k(x,y,z)$.
Just as we don't tend to restricted localisations of the integers like the dyadic rationals that much, we won't see rings of multivariable Laurent polynomials showing up as much. In general, we always want to deal with the largest localisation of a domain, which is its field of fractions.
So although the group completion of the free commutative semigroup $F(X)$ is an important and natural concept in commutative semigroup theory, it doesn't have the same role and level of importance in commutative algebra, as determined by commutative semigroup rings. This demonstrates that not everything in commutative algebra is a consequence of commutative semigroup rings, but the use of semigroup rings is still an infinitely powerful technique in the construction of commutative rings.
A distinguishing property of ordinary polynomials is that they can be cast into functions, so that over affine space $\mathbb{A}^n$ they are functions $f: \mathbb{A}^n \to k$ and this works for any commutative ring. On the other hand, for Laurent polynomials to be cast into functions $f : \mathbb{A}^n \to k$ we need to have some notion of division to deal with negative degree exponents, so we need to work over a field.
In those cases when we take the commutative group ring of a commutative ring $R$ which is not a field, with respect to $F^{\circ}(X)$ then what we get is certainly still a commutative ring, it is just not a commutative ring of functions. Consider the commutative group ring $\mathbb{Z}(\mathbb{Z}^n,+)$ consisting of polynomials with integer exponents and integer coefficients. Then this is a valid commutative ring, but its elements are not functions because $\mathbb{Z}$ is not a field.
So in some sense we ought to have our free commutative group ring be over a field $k$. Then we get terms like $5\frac{x}{yz} + \frac{6y}{xz}$ consisting of fractional monomials, but then given any term like this we can add to get $\frac{5x^2 + 6y^2}{xyz}$ which is always a polynomial with a monomial in the denominator. So we see that multivariable Laurent polynomials are simply rational functions with monomials in the denominator, so we have an embedding. \[ kF^{\circ}(x,y,z) \subseteq k(x,y,z) \] The ring of Laurent polynomials is simply the localisation of the polynomial ring by the multiplicative set of monomials, and so they are merely a special case of rational functions. Now it is clear why we don't see Laurent polynomials so much in commutative alebra. They are simply part of the far more important and general concept of fields of rational functions $k(x,y,z)$.
Just as we don't tend to restricted localisations of the integers like the dyadic rationals that much, we won't see rings of multivariable Laurent polynomials showing up as much. In general, we always want to deal with the largest localisation of a domain, which is its field of fractions.
So although the group completion of the free commutative semigroup $F(X)$ is an important and natural concept in commutative semigroup theory, it doesn't have the same role and level of importance in commutative algebra, as determined by commutative semigroup rings. This demonstrates that not everything in commutative algebra is a consequence of commutative semigroup rings, but the use of semigroup rings is still an infinitely powerful technique in the construction of commutative rings.
Monday, September 20, 2021
Centralizers of division rings
We have that centralizers of commuting graphs determine subsemigroups and subrings, and corresponding to this there is the fact that centralizers of groups are subgroups and centralizers of division algebras are not simply subrings but division subalgebras. This can be used for example to determine the maximal subfields of a division algebra.
Subalgebras of division rings:
Let $R$ be a division ring, then $R$ as a ring is associated to a lattice of subrings. A meet subsemilattice of this lattice of subrings is the lattice of division subrings $Sub(R)$ of $R$, which consists of subrings that are also inverse closed with respect to non-zero elements.
The special case of centralizers:
By the theory of commutativity necessary subrings, we know that the centralizers of $R$ are subrings. The centralizers of $R$ together form a lattice: the centralizer lattice of the commuting graph of $R$. It remains to show that they are not only subrings but also division subrings.
Lemma 1. let $G$ be a group and $x$ an element of $G$ then the centralizer $C(x)$ is a subgroup of $G$.
Proof. the centralizer $C(x)$ is a submonoid, so let $y \in C(x)$ then $yx = xy$. Let $y^{-1}\in G$ then we want to show that $y^{-1}x = xy^{-1}$. By the fact that $G$ is a group, we can multiply both sides by $yx^{-1}$ to get $y^{-1}xyx^{-1} = 1$. Then by the fact that $x$ and $y$ commute this is equivalent to $y^{-1}yxx^{-1} = 1$ which is true, so that $y^{-1} \in C(x)$. $\square$
Lemma 2. let $S$ be a group with zero (respectively a Clifford 0-simple semigroup). Then centralizers of elements in $S$ are inverse closed.
Proof. let $x \in S$ and suppose that $x = 0$. Then the centralizer of $0$ is the entire semigroup, which is inverse closed. Suppose $x \not= 0$ then $x \in S-\{0\}$ which is a group, so the centralizer of $x$ in $S-\{0\}$ is a subgroup by lemma 1. It is therefore inverse closed. $\square$
Theorem. let $R$ be a division ring, then centralizers $C(x)$ are division subalgebras.
Proof. by commutativity necessary subrings, $C(x)$ is a subring and by lemma 2 it is inverse closed, so it is a division subalgebra. $\square$
The maximal cliques of the commuting graph of a division ring are precisely the maximal fields of the division ring. All the commutativity necessary subfields of a division ring are determined by maximal cliques and their intersections, which form a semilattice.
Centers of division rings:
The center of a division ring is precisely the intersection of all the maximal fields of the division ring. As it is the intersection of fields, it is a field. The division ring can then be seen to be a vector space over its central field. The measure of the commutativity of a division ring is the dimension of its extension over a central field: finite dimensional division rings (like the quaternions) are the most commutative.
Maximal subfields of a division ring:
An immediate corollary of this is that the maximal subfields of a division ring can be determined by the same general mechanism by which we determine maximal commutative subalgebras of semigroups, groups, rings, and semirings: they are maximal cliques of the commuting graph.
Proposition. let $D$ be a division ring then $F$ is a maximal subfield of $D$ iff $D$ is equal to its own centralizer: $C(D) = D$
It is not hard too see by the same reasoning that applies to any other algebraic structure like a group or ring, that every element of a division ring belongs to some maximal field.
Proposition. every element of a division ring belongs to some maximal field.
Maximal subfields of a division ring are a vital tool in the study of division rings. Their further properties can be studied using tensor products of division rings described as algebras over fields [1].
References:
[1] A first course in non-commutative rings
T.Y Lam
Subalgebras of division rings:
Let $R$ be a division ring, then $R$ as a ring is associated to a lattice of subrings. A meet subsemilattice of this lattice of subrings is the lattice of division subrings $Sub(R)$ of $R$, which consists of subrings that are also inverse closed with respect to non-zero elements.
The special case of centralizers:
By the theory of commutativity necessary subrings, we know that the centralizers of $R$ are subrings. The centralizers of $R$ together form a lattice: the centralizer lattice of the commuting graph of $R$. It remains to show that they are not only subrings but also division subrings.
Lemma 1. let $G$ be a group and $x$ an element of $G$ then the centralizer $C(x)$ is a subgroup of $G$.
Proof. the centralizer $C(x)$ is a submonoid, so let $y \in C(x)$ then $yx = xy$. Let $y^{-1}\in G$ then we want to show that $y^{-1}x = xy^{-1}$. By the fact that $G$ is a group, we can multiply both sides by $yx^{-1}$ to get $y^{-1}xyx^{-1} = 1$. Then by the fact that $x$ and $y$ commute this is equivalent to $y^{-1}yxx^{-1} = 1$ which is true, so that $y^{-1} \in C(x)$. $\square$
Lemma 2. let $S$ be a group with zero (respectively a Clifford 0-simple semigroup). Then centralizers of elements in $S$ are inverse closed.
Proof. let $x \in S$ and suppose that $x = 0$. Then the centralizer of $0$ is the entire semigroup, which is inverse closed. Suppose $x \not= 0$ then $x \in S-\{0\}$ which is a group, so the centralizer of $x$ in $S-\{0\}$ is a subgroup by lemma 1. It is therefore inverse closed. $\square$
Theorem. let $R$ be a division ring, then centralizers $C(x)$ are division subalgebras.
Proof. by commutativity necessary subrings, $C(x)$ is a subring and by lemma 2 it is inverse closed, so it is a division subalgebra. $\square$
The maximal cliques of the commuting graph of a division ring are precisely the maximal fields of the division ring. All the commutativity necessary subfields of a division ring are determined by maximal cliques and their intersections, which form a semilattice.
Centers of division rings:
The center of a division ring is precisely the intersection of all the maximal fields of the division ring. As it is the intersection of fields, it is a field. The division ring can then be seen to be a vector space over its central field. The measure of the commutativity of a division ring is the dimension of its extension over a central field: finite dimensional division rings (like the quaternions) are the most commutative.
Maximal subfields of a division ring:
An immediate corollary of this is that the maximal subfields of a division ring can be determined by the same general mechanism by which we determine maximal commutative subalgebras of semigroups, groups, rings, and semirings: they are maximal cliques of the commuting graph.
Proposition. let $D$ be a division ring then $F$ is a maximal subfield of $D$ iff $D$ is equal to its own centralizer: $C(D) = D$
It is not hard too see by the same reasoning that applies to any other algebraic structure like a group or ring, that every element of a division ring belongs to some maximal field.
Proposition. every element of a division ring belongs to some maximal field.
Maximal subfields of a division ring are a vital tool in the study of division rings. Their further properties can be studied using tensor products of division rings described as algebras over fields [1].
References:
[1] A first course in non-commutative rings
T.Y Lam
Wednesday, July 14, 2021
Commutativity necessary subrings
The commuting graphs of semigroups induce a number of different commutativity necessary subsemigroups: centralizers, maximal commuting cliques, commutative principal filters, the center, etc. As it it happens all of these concepts can be transferred to non-commutative rings, so that we can understand non-commutative algebra using graph-theoretic techniques.
Definition. let $R$ be a ring then the commuting graph $Com(R)$ of $R$ is the commuting graph of its multiplicative semigroup. \[ Com(R) = \{(x,y) : xy = yx \} \] The first step towards proving the existence of commutativity necessary subrings for a ring $R$ is to demonstrate that centralizers are subrings. The existence of the other types of commutativity necessary subrings immediately follows.
Theorem. let $R$ be a ring, $x$ an element of $R$, then the centralizer $S$ of $x$ is a subring. \[ S = \{ c : cx = xc \}\] Proof. (1) by basic semigroup theory, then centralizer is a multiplicative subsemigroup.
(2) by the nullary distributive law we have \[ \forall x*0 = 0 = 0*x \] It follows that $0$ is a central element which implies $0 \in S$.
(3) by the binary distributive law we have \[ x(a+b) = xa + xb = ax + bx = (a+b)x \] So the centralizer $S$ is additively closed.
(4) let $c \in R$ then $(-c)x$ is equal to $-(cx)$ by the distributive law and the definition of additive inverses. \[ cx + (-c)x = (c-c)x = 0*x = 0 \] In the other direction, $x(-c)$ is equal to $-(xc)$ for the same reasons. \[ xc + x(-c) = x(c-c) = x*0 = 0 \] Thus if $c$ is in the centralizer $S$ then so is $-c$ \[ (-c)*x = -(cx) = -(xc) = x*(-c)\] Thus $S$ is also negation closed, so that $S$ is a subring. $\square$
The first three conditions are equally applicable to semirings. Semirings axiomatically have both the binary and nullary distributive laws, which ensures that centralizers are additive submonoids. In the ring case, we see that centralizers also form additive subgroups.
All other commutativity necessary subrings can be formed by intersections of centralizers. The center is the intersection of all centralizers. Maximal commuting cliques are equal to the intersection of the centralizers of all their elements. Commutative principal filters are formed by the centralizer of the centralizer.
Centralizers:
Let $Mat_n(F)$ be the non-commutative ring of matrices over a field $F$. Then the centralizers of $Mat_n(F)$ can be computed by solving systems of linear polynomial equations. In general, the previous theorem shows that centralizers are always subrings.
Centers:
The center of any non-commutative ring is a commutative ring, over which the non-commutative ring is a ring extension. For example, the center of the quaternions $\mathbb{H}$ are the reals $\mathbb{R}$, so that $\mathbb{}$ is a four dimensional vector space over $\mathbb{R}$. In a non-commutative ring of matrices like $Mat_n(F)$ the center consists of the commutative ring of scalar matrices.
Commutative principal filters:
The commuting preorder of a ring is the adjacency preorder of its commuting graph. Then the multiplicative iteration preorder is a suborder of the commuting preorder. The principal filters of this preorder are all commutativity necessary subrings. The center is the minimal commutative principal filter.
Maximal commuting cliques:
Let $G$ be a finite graph, then every element of $G$ is contained in some maximal clique. In the infinite case we need Zorn's lemma, and the fact that the union of a chain of cliques is a clique. Then every element of an infinite graph is in a maximal clique as well. This implies that every element in a ring is contained in some commutative subring.
Theorem. monogenic rings are commutative
Proof. By Zorn's lemma every element is embedded in a maximal clique, which is a commutative subring. Then every element of a ring is embedded in a commutative subring. A monogenic subring is a minimal subring containing a given element, so it must be embedded in some greater then or equal commutative subring. The property of commutativity is hereditary, so monogenic rings are commutative. $\square$
This should work unless there is some issue with Zorn's lemma in the infinite case. This generalizes the well known fact that all monogenic semigroups are commutative, and it shows that rings can be built up from commutative building blocks.
Implications of Lagrange's theorem:
Let $R$ be a finite ring, then Lagrange's theorem implies that all the different commutativity necessary subrings have orders that divide the order of the ring $R$. In particular, the degrees of each element of the commuting graph are divisors of its order. This means that commuting graphs of rings are a lot like those for groups, but they aren't completely the same.
For example, the commuting graphs of rings don't necessarily have to have the same degree of commuting equality which is implied for in groups by Cauchy's theorem and the number of iteration equal elements $\phi(n)$ of prime cyclic groups of order $n$. Even so, Lagrange's theorem alone places a greater constraint on the set of possible multiplicative semigroups of rings.
In the case of finite division rings, we get that subalgebras must satisfy Lagrange's theorem in two different ways, corresponding to the two types of subgroups in a division ring: \[ (d)|(n) \] \[ (d-1)|(n-1) \] This places even greater constraints on the possible commuting graphs of division rings, but we can get around worrying about that altogether by Wedderburn's little theorem [1] which shows that all finite division rings are commutative. The double Lagrange's theorem still applies to finite fields.
See also:
Commutativity necessary subsemigroups
Subalgebra lattices of finite fields
References:
Weddernburn's theorem
Definition. let $R$ be a ring then the commuting graph $Com(R)$ of $R$ is the commuting graph of its multiplicative semigroup. \[ Com(R) = \{(x,y) : xy = yx \} \] The first step towards proving the existence of commutativity necessary subrings for a ring $R$ is to demonstrate that centralizers are subrings. The existence of the other types of commutativity necessary subrings immediately follows.
Theorem. let $R$ be a ring, $x$ an element of $R$, then the centralizer $S$ of $x$ is a subring. \[ S = \{ c : cx = xc \}\] Proof. (1) by basic semigroup theory, then centralizer is a multiplicative subsemigroup.
(2) by the nullary distributive law we have \[ \forall x*0 = 0 = 0*x \] It follows that $0$ is a central element which implies $0 \in S$.
(3) by the binary distributive law we have \[ x(a+b) = xa + xb = ax + bx = (a+b)x \] So the centralizer $S$ is additively closed.
(4) let $c \in R$ then $(-c)x$ is equal to $-(cx)$ by the distributive law and the definition of additive inverses. \[ cx + (-c)x = (c-c)x = 0*x = 0 \] In the other direction, $x(-c)$ is equal to $-(xc)$ for the same reasons. \[ xc + x(-c) = x(c-c) = x*0 = 0 \] Thus if $c$ is in the centralizer $S$ then so is $-c$ \[ (-c)*x = -(cx) = -(xc) = x*(-c)\] Thus $S$ is also negation closed, so that $S$ is a subring. $\square$
The first three conditions are equally applicable to semirings. Semirings axiomatically have both the binary and nullary distributive laws, which ensures that centralizers are additive submonoids. In the ring case, we see that centralizers also form additive subgroups.
All other commutativity necessary subrings can be formed by intersections of centralizers. The center is the intersection of all centralizers. Maximal commuting cliques are equal to the intersection of the centralizers of all their elements. Commutative principal filters are formed by the centralizer of the centralizer.
Centralizers:
Let $Mat_n(F)$ be the non-commutative ring of matrices over a field $F$. Then the centralizers of $Mat_n(F)$ can be computed by solving systems of linear polynomial equations. In general, the previous theorem shows that centralizers are always subrings.
Centers:
The center of any non-commutative ring is a commutative ring, over which the non-commutative ring is a ring extension. For example, the center of the quaternions $\mathbb{H}$ are the reals $\mathbb{R}$, so that $\mathbb{}$ is a four dimensional vector space over $\mathbb{R}$. In a non-commutative ring of matrices like $Mat_n(F)$ the center consists of the commutative ring of scalar matrices.
Commutative principal filters:
The commuting preorder of a ring is the adjacency preorder of its commuting graph. Then the multiplicative iteration preorder is a suborder of the commuting preorder. The principal filters of this preorder are all commutativity necessary subrings. The center is the minimal commutative principal filter.
Maximal commuting cliques:
Let $G$ be a finite graph, then every element of $G$ is contained in some maximal clique. In the infinite case we need Zorn's lemma, and the fact that the union of a chain of cliques is a clique. Then every element of an infinite graph is in a maximal clique as well. This implies that every element in a ring is contained in some commutative subring.
Theorem. monogenic rings are commutative
Proof. By Zorn's lemma every element is embedded in a maximal clique, which is a commutative subring. Then every element of a ring is embedded in a commutative subring. A monogenic subring is a minimal subring containing a given element, so it must be embedded in some greater then or equal commutative subring. The property of commutativity is hereditary, so monogenic rings are commutative. $\square$
This should work unless there is some issue with Zorn's lemma in the infinite case. This generalizes the well known fact that all monogenic semigroups are commutative, and it shows that rings can be built up from commutative building blocks.
Implications of Lagrange's theorem:
Let $R$ be a finite ring, then Lagrange's theorem implies that all the different commutativity necessary subrings have orders that divide the order of the ring $R$. In particular, the degrees of each element of the commuting graph are divisors of its order. This means that commuting graphs of rings are a lot like those for groups, but they aren't completely the same.
For example, the commuting graphs of rings don't necessarily have to have the same degree of commuting equality which is implied for in groups by Cauchy's theorem and the number of iteration equal elements $\phi(n)$ of prime cyclic groups of order $n$. Even so, Lagrange's theorem alone places a greater constraint on the set of possible multiplicative semigroups of rings.
In the case of finite division rings, we get that subalgebras must satisfy Lagrange's theorem in two different ways, corresponding to the two types of subgroups in a division ring: \[ (d)|(n) \] \[ (d-1)|(n-1) \] This places even greater constraints on the possible commuting graphs of division rings, but we can get around worrying about that altogether by Wedderburn's little theorem [1] which shows that all finite division rings are commutative. The double Lagrange's theorem still applies to finite fields.
See also:
Commutativity necessary subsemigroups
Subalgebra lattices of finite fields
References:
Weddernburn's theorem
Sunday, April 25, 2021
Pre-additive categories
Previously, we talked about groups with additonal structure. A more general concept is that of a category with additional structure, the most important example of which is a pre-additive category. Pre-additive categories are categories whose hom classes are commutative groups, such that the group distributes over composition. A (not necessarily commutative) ring is a pre-additive category with a single object.
Non-commutative rings of actions:
Let $C$ be a concrete category and $X \in Ob(C)$ then $End(X)$ is a monoid action on the underlying set of $X$. A wide variety of different monoid actions emerge from concrete categories like posets, graphs, etc. We can also explore the endomorphism monoid of a commutative group. In this case, pre-additive categories add additional perspective to the theory of commutative groups.
By taking the category of commutative groups to be an abelian category, we get that $End(G)$ for any commutative group is an endomorphism ring. Therefore the actions which previously only took on the structure of a monoid, now have the structure of a ring. This is the basic context in which non-commutative rings emerge and become indispensible. We naturally consider rings to be commutative starting with the ring of integers $\mathbb{Z}$, but here we see how some non-commutative rings emerge even in problems of commutative algebra.
Let $R$ be a ring, then the category of $R$-modules is an abelian category. In this case, for any $R$-module $M$ we have that $End(M)$ is the matrix ring of the module $M$. A more familiar case might be the matrix ring of a vector space over a field. These matrix rings are amongst the most important non-commutative rings, and they emerge as rings of actions on a set. Therefore, non-commutative rings are an indispensible part of the modern algebraic theory of actions.
Concrete rings:
A concrete pre-additive category is a pre-additive category $C$ with a faithful set-valued functor $F : C \to Sets$. This makes it so that all the elements of the ring are functions acting on a set. In this case, we have a multiplicative monoid action on the underlying set of the concrete ring. We see that the matrix ring $End(V)$ is a concrete ring whose elements are functions acting on the underlying set of vectors. Concrete rings proide a natural categorical description of non-commutative rings of actions.
Links:
Preadditive and additive categories
Non-commutative rings of actions:
Let $C$ be a concrete category and $X \in Ob(C)$ then $End(X)$ is a monoid action on the underlying set of $X$. A wide variety of different monoid actions emerge from concrete categories like posets, graphs, etc. We can also explore the endomorphism monoid of a commutative group. In this case, pre-additive categories add additional perspective to the theory of commutative groups.
By taking the category of commutative groups to be an abelian category, we get that $End(G)$ for any commutative group is an endomorphism ring. Therefore the actions which previously only took on the structure of a monoid, now have the structure of a ring. This is the basic context in which non-commutative rings emerge and become indispensible. We naturally consider rings to be commutative starting with the ring of integers $\mathbb{Z}$, but here we see how some non-commutative rings emerge even in problems of commutative algebra.
Let $R$ be a ring, then the category of $R$-modules is an abelian category. In this case, for any $R$-module $M$ we have that $End(M)$ is the matrix ring of the module $M$. A more familiar case might be the matrix ring of a vector space over a field. These matrix rings are amongst the most important non-commutative rings, and they emerge as rings of actions on a set. Therefore, non-commutative rings are an indispensible part of the modern algebraic theory of actions.
Concrete rings:
A concrete pre-additive category is a pre-additive category $C$ with a faithful set-valued functor $F : C \to Sets$. This makes it so that all the elements of the ring are functions acting on a set. In this case, we have a multiplicative monoid action on the underlying set of the concrete ring. We see that the matrix ring $End(V)$ is a concrete ring whose elements are functions acting on the underlying set of vectors. Concrete rings proide a natural categorical description of non-commutative rings of actions.
Links:
Preadditive and additive categories
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