The study of algebraic structures and properties applying to large classes of such structures. For example, ideas from group theory and ring theory are extended and considered for structures with other signatures (systems of basic or fundamental operations).

learn more… | top users | synonyms

1
vote
2answers
127 views

What is a free $\Omega$-structure generated by $X$

I am reading the book "Notes on logic and set-theory". It gives the following definition: Given a set $X$ of variables and operational type $\Omega$ ($X \cap > \Omega = \emptyset $), the set $ ...
0
votes
1answer
76 views

Isomorphic two structures but different type.

If I have structure $(S, \cdot)$, where $\cdot$ has type $(2)$, i.e., $\cdot : S \times S \rightarrow S$ and $(S', \circ)$, where $\circ$ has type $(3)$, i.e., $\cdot : S' \times S' \times S' ...
1
vote
1answer
51 views

Existence of arbitrarily large ordinal subgroups in a group structure on a regular cardinal [duplicate]

Suppose $\kappa$ is an uncountable regular cardinal, and $(\kappa, \cdot, ^{-1}, e$) is a group. Prove that that $C = \{\alpha < \kappa: \alpha\, \textrm{is a subgroup of}\, \kappa)$ is unbounded ...
6
votes
1answer
121 views

What is a simple axiomatisation of groups using division?

I recall from an old exercise I did as an undergrad that groups can be axiomatised using division rather than multiplication: A group is a non-empty set equipped with a binary division operator / ...
4
votes
4answers
278 views

Preserving structures

Category theory abstracts the notion of the preservation of structure by means of morphisms. Is there a description of what it means to preserve structure of different types of mathematical structures ...
6
votes
4answers
228 views

Any commutative associative operation can be extended to a function on nonempty finite sets

This is a fact we use very frequently in general mathematics when we write such notations as $1+2+3+4$: since we know that $+$ is commutative and associative, we can just "drop the parentheses" and ...
6
votes
2answers
193 views

In a slice category C/A of a category C over a given object A, What is the role of the identity morphism of A in C with respect to C/A

In a slice category $C/A$ of a category $C$ over a given object $A$, what is the role of the $C$ identity morphism, $A\to A$ ($1_A$), in $C/A$, particularly with respect to composition? I ...
1
vote
1answer
218 views

Property: closure under an binary operation

We have the following definition: Definition: let $*_A:A^2 \rightarrow A$ and $B \subseteq A $, with $B \neq \emptyset$, $B$ is closed under $*_A$ if $a *_A|_Bb \in B$ $\forall a,b \in B$. Property: ...
1
vote
1answer
113 views

Algebraic substructure and restriction of a function

I am reading the follow pdf: http://www.math.uwaterloo.ca/~snburris/htdocs/UALG/univ-algebra2012.pdf in particular at pg 28 of pdf, and I think that, let $(A;f)$ an algebric structure and $B ...
0
votes
1answer
44 views

Do cancellative semigroups form a variety of algebras?

Sorry if this is a silly question. Define that a right-cancellative semigroup is a set $G$ together with an associative operation $*$ such that for all $a,b,x \in G$ it holds that $ax=bx \Rightarrow ...
3
votes
1answer
819 views

Modus Ponens: implication versus entailment

Would it be inconsistent to write Modus Ponens using only implication, not entailment? $(p \wedge (p \to q)) \to q$ The way I understand is that implication ($ \to$) is an operator that yields a new ...
0
votes
1answer
119 views

Represent the three element chain as a subdirect product of subdirectly irreducible lattices.

Represent the three element chain as a subdirect product of subdirectly irreducible lattices. I know the two element chain as either a Boolean algebra or a semilattice,is subdirectly irreducible.In ...
2
votes
1answer
135 views

Every group and ring is congruence-permutable , but not necessarily congruence-distributive

The problem is: Show that every group and ring is congruence-permutable , but not necessarily congruence-distributive. I know that in group every normal sub group has permutable property and in ...
0
votes
0answers
21 views

verify the claim that consequences of balanced identities are again balanced.

verify the claim that consequences of balanced identities are again balanced. An identity is p≈q balanced if each variable occurs the same number of times in p as in q.if ∑ is balanced set of ...
1
vote
1answer
132 views

Initial structures in the category of algebraic systems of the same type

From Handbook of Analysis and Its Foundations by Eric Schechter 9.21. Basic properties of subalgebras. We consider the category consisting of the algebraic systems of some type $(τ, \mathcal{J})$, ...
3
votes
4answers
656 views

why is a nullary operation a special element, usually 0 or 1?

Does a nullary operation mean an operation not taking any argument? Then why is a nullary operation a special element, usually 0 or 1, in an algebraic structure? Thanks!
0
votes
0answers
43 views

Question about a property of lattice-morphism

I would like to know if there is a name for the class of commutative (i.e., $\phi(x,y)=\phi(y,x)$) lattice-morphisms $\phi : L_1\times L_{1} \rightarrow L_2$ with the following property: $\phi(x ...
1
vote
0answers
70 views

Which “conditions” generate subalgebras?

While looking at this question I suddenly wondered about a more general question. Think of your favorite class of algebra (groups or rings, say), and forgive me for vocabulary mistakes while I try to ...
18
votes
2answers
482 views

A structural proof that $ax=xa$ forms a monoid

During the discussion on this problem I found the following simple observation: If $M$ is a monoid and $a \in M$ then $\{x: ax = xa\}$ is a submonoid. This is trivial to prove by checking ...
3
votes
3answers
168 views

Free object is a coproduct: $F_{A\cup B}\cong F_A \coprod F_B$

Let $A,B$ be sets, and $A\sqcup B$ the disjoint union. Suppose that in a (concrete) category, the free objects $F_A,F_B,F_{A\sqcup B}$ exist, and that the coproduct $F_A \coprod F_B$ exists. How can I ...
-2
votes
1answer
177 views

presentation of the direct sum of commutative rings / algebras

If $I,J$ are index sets, $R$ a commutative unital ring, $\mathfrak{a},\mathfrak{b}$ ideals of polynomial rings $R[x_i; i\!\in\!I]$, $R[y_j; j\!\in\!J]$, and $\langle\langle\ldots\rangle\rangle$ is the ...
-1
votes
1answer
92 views

Describing all subdirectly irreducible mono-unary algebras.

(Wenzel). Describe all subdirectly irreducible mono-unary algebras. [In particular show that they are countable.] Thanks!
0
votes
1answer
93 views

Why $Z(A)$ is an equivalence relation on $A$?

For every algebra $A$, the center $Z(A)$ is a congruence on $A$. Why is $Z(A)$ an equivalence relation on $A$?
1
vote
1answer
137 views

homomorphisms and congruence relations

Do compositions of homomorphisms in universal algebra correspond to joins of congruence relations? That is- is the congruence relation $g \circ f(a ) = g \circ f( b) \Leftrightarrow a \sim b $ the ...
1
vote
2answers
71 views

Quotients of quotients in universal algebra

In universal algebra, when is the quotient of a quotient of an algebra $\mathcal{A} $, a quotient of $\mathcal{A} $?
0
votes
1answer
61 views

Proof for Tarski theorem in universal Algebra page 108

Given a variety V and a set of variables X, IrB(Idv(X)) is a convex set. I need a complete proof for this theorem. If anyone can help me it would be wonderful.
0
votes
1answer
49 views

Equational theories & their relation to fully invarient congruences on T(X)?

The equational theories of type F over X form an algebraic lattice which is isomorphic to the lattice of fully invarient congruences on T(X). I need the proof of above theorem which is in page 103 ...
0
votes
2answers
113 views

Why fully invariant congruence is an algebraic closure operator?

If we have an algebra $A$ of type $F$ then congruence of fully invariant is an algebraic closure structure operator on $A\cdot A$. Actually it's in Universal Algebra Sankappanavar page $100$ (Lemma ...
1
vote
1answer
42 views

My question is why the set of fully invariant congruence of a set A is closed under arbitrary intersection?

Why ($\operatorname{Con}_{FI}(A))$ is closed under arbitrary intersection?
1
vote
0answers
80 views

Another way of saying that algebraic objects are isomorphic

From a universal algebraic perspective, let's say we have two isomorphic groups. Then can I speak of their isomorphic nature by saying the binary operations of multiplication of the two groups are ...
8
votes
1answer
226 views

An exercise in infinite combinatorics from Burris and Sankappanavar

Exercise 6.7 in chapter IV of Burris and Sankappanavar's A Course in Universal Algebra starts as follows: Show that for $I$ countably infinite there is a subset $S$ of the set of functions from ...
2
votes
1answer
159 views

Unification of expressions involving sets

Let's let $\def\OP#1#2{\left\langle#1,#2\right\rangle}\OP xy$ represent the set $\{\{x\},\{x,y\}\}$ as is usual, per Kuratowski. Then: $$ \begin{eqnarray} \OP{\OP ab}c & = & \{\{\{\{a\}, ...
4
votes
1answer
69 views

Associativity, Jacobi, and self-action representations

About a year and a half ago, I was at a talk by Martin Hyland where he suggested that the Jacobi identity is to the associative law as the anticommutative law is to the commutative law. I think this ...
3
votes
1answer
171 views

Do filters on a Boolean algebra also make a Boolean algebra?

Let $\mathfrak{B}=(B,\bot,\top,\lnot,\wedge,\vee)$ be a boolean algebra. $B_F$ be the set of all filters on $\mathfrak B$. And for all filter $F$, $G$, $F \wedge_{B_F} G \colon= \mathbf C(F \cup G)$ ...
5
votes
2answers
195 views

Stone duality for ideals and filters (exercise)

In A Course in Universal Algebra (Burris, Sankapannavar), the exercise 4.4.7-8, p.158, says: Let $A$ be a Boolean algebra. Denote $A^\ast:=\{\text{ultrafilters of }A\}$, and give $A^\ast$ the ...
10
votes
4answers
432 views

Are isomorphic structures really indistinguishable?

I always believed that in two isomorphic structures what you could tell for the one you would tell for the other... is this true? I mean, I've heard about structures that are isomorphic but different ...
2
votes
1answer
149 views

correspondence for universal subalgebras of $U/\vartheta$

Let $U$ be a universal algebra of type $T$, and denote $\mathrm{Con}(U)\!=\!\{\text{congruence relations on }U\}$ and $\mathrm{Sub}(U)\!=\!\{\text{subalgebras of }U\}$. Let "$\leq$" mean "subalgebra". ...
3
votes
2answers
255 views

What is the name of the structure Z4 under subtraction?

If we consider $\mathbb{Z_4}$ under addition, then it forms a cyclic group of order 4. However if we change the binary operation to subtraction on $\mathbb{Z_4}$, we get a different structure $J$ with ...
3
votes
1answer
165 views

$M_3$ is a simple lattice

I'd like to prove (exercise 9.5 in Roman's Lattices and Ordered Sets, p.203) that the lattice $M_3$ is simple, meaning that the only congruences on $M_3$ are the trivial ones (the 'equality' ...
2
votes
1answer
140 views

ideals of a ring form a modular lattice

We know that if $M$ is a left $R$-module, then $(\{\text{submodules of }M\},\subseteq)$ is a modular lattice. Taking $M\!=\!R$, we deduce that $(\{\text{ideals of }R\},\subseteq)$ is a modular ...
4
votes
2answers
295 views

Variety generated by finite fields

Let $K_1,\dotsc,K_n$ be finite fields and let $V$ be the variety of rings, generated by the $K_i$ (rings aren't necessarily unital). I want to figure out what $V$ looks like. By a theorem of Tarski, ...
6
votes
1answer
269 views

Lattices are congruence-distributive

$\newcommand{\r}[1]{\mathrel{#1}}$ First, a few definitions. Given a lattice $L$, a congruence on $L$ is an equivalence relation $\theta$, compatible with the lattice operations, i.e. if ...
2
votes
3answers
219 views

Are groups algebras over an operad?

I'm trying to understand a little bit about operads. I think I understand that monoids are algebras over the associative operad in sets, but can groups be realised as algebras over some operad? In ...
4
votes
2answers
253 views

What is the smallest variety of algebras containing all fields?

A field is a ring whose nonzero elements form a commutative group under multiplication. A field is also a commutative inverse semigroup with respect to multiplication. The unique multiplicative ...
1
vote
0answers
85 views

direct product of different algebras?

Is it possible to define a "direct product" of two algebras with different signatures? For example, boolean lattice $\otimes$ monoid? Perhaps we need to take some quotient to make sense of it (e.g Q ...
1
vote
0answers
150 views

amalgam of structures

Trying to refine my question here. This is a response to the questions here: Homomorphisms between structures My objective is to take a set of $S-$structures and form an amalgam object out of that ...
11
votes
1answer
281 views

Why do free monoids have a “trivial” automorphism group and free groups don't?

Let $X$ be a set and $M$ the free monoid over $X$. Then an automorphism $f$ of $M$ satisfies $f(X)=X$ and so $\text{Aut}(M)$ is canonically isomorphic to $\mathfrak{S}_X$. My Proof: For every word ...
2
votes
1answer
344 views

Injective Homomorphism and direct products

This question relates to products of structures all with the same symbol set $S$. After I give a little background the question follows. Direct Products This definition of the direct product is ...
1
vote
2answers
225 views

Existence of universal enveloping inverse semigroup (similar to “Grothendieck group”)

Context In its simplest form, the Grothendieck group construction associates an abelian group to a commutative semigroup in a "universal way". Now I'm interested in the following nilpotent ...
3
votes
1answer
125 views

Is the rank of a relatively free group… ill-defined in general?

A relatively free algebra $F$ has a free generating set (basis) $X$ such that any map $f : X \to F$ can be extended to an endomorphism of $F$. It is known that, in general the notion of rank of $F$ ...