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Why does the carrier of an algebraic structure has to be closed under the operations of the algebraic structure?

One could also consider $(\mathbb{N}^*, \div)$. But why isn't that an algebraic structure?

asdfusername
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2 Answers2

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It is perfectly valid to call (informally) algebraic structure a set endowed with a family of partial operations. In the example you give, division is a partial operation on $\mathbb {N}$ (defined for pairs $s, t $ such that $t $ divides $s $). In this sense, for instance, a category might be called an algebraic structure with a partial operation $\circ $ and several units.

Nevertheless, when you have operations defined on the whole base set, you can work more smoothly. As in Stefan's answer, you may define straightforwardly the notions of subalgebra, homomorphism, isomorphism, etc. Otherwise, you should be careful to check what happens if one expression is defined on one side of an equation and the other doesn't, and the like.

Finally, you might apply your idea of "the result is in another set" and go to multi-sorted structures. In this case, your operations may jump from one set to another, but you have to gather in advance all sets that will ever appear in this process (the sorts).

Several constructions in universal algebra lead to sets with partially defined operations (v.g, the clone of term operations). This can also be considered as a multi-sorted structure (one sort for each number of arguments).

Pedro Sánchez Terraf
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  • Thanks. I've read that you are a research mathematician in universal algebra. Does one also consider 'structures' with two (possibly disjoint) carriers (and operations going from one carrier to the other) in algebra? – asdfusername Jun 18 '15 at 12:21
  • Where is the point in the definition of a substructure where we had to check what happens if one expression is defined on one side of an equation and the other doesn't? – asdfusername Jun 18 '15 at 12:28
  • @asdfusername One of the examples I gave -the clone of term operations- has countably many disjoint carriers. For example, one carrier contains all unary $f(x) $ and another all binary $g(x, y)$; one operation of composition takes $f $ and $g $ and returns $f(g(x, y)) $ in the second carrier. There are actually many compositions, one for each type of the arguments. – Pedro Sánchez Terraf Jun 18 '15 at 15:19
  • @asdfusername Concerning your second question, I'm not so sure that the problem lies in considering substructures but the meaning of identities $\forall\bar x. t_1(\bar x)=t_2(\bar x)$ is no longer a straightforward business. One is interested in studying classes with axioms of that shape, and if your structure doesn't have total operations, it'll be not so clear how to interpret the former. Does $(\mathbb{N},+, \div)$ satisfy the distributive law? How do you write an axiom for the class of structures with this version of this axiom? – Pedro Sánchez Terraf Jun 19 '15 at 12:51
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As your initial question may be answered by saying that this is part of the definition, let me instead give you a reason for this. Say, we start with the structure $(\mathbb R^*; \cdot, \ ^{-1}, 1)$. Then $(\mathbb Q^* , \cdot, \ ^{-1}, 1)$ is a substructure, but what exactly do we mean by this?

We mean that

  1. $\mathbb Q^*$ is a subset of $\mathbb R^*$
  2. Given two rational numbers $a,b \in \mathbb Q^*$ it doesn't matter whether we build their product in the one or the other structure - the answer will be the same.
  3. The same holds true if we build the inverse $a^{-1}$ of a rational number $a \in \mathbb Q^*$.
  4. The neutral element $1$ is the same element in both structures.

Now, consider $\mathbb N^*$. The first point still holds true and so does the 2nd and 4th. But there is a problem with the 3rd point of our list. If we want to know what $2^{-1}$ is, we cannot find such a value in $\mathbb N^*$, but we can in the other two structures. So there is no way to define a function $$ ^{-1} \colon \mathbb N^* \rightarrow \mathbb N^* $$ that agrees with the one on $\mathbb Q^*$ or $\mathbb R^*$ on their common domain.

Stefan Mesken
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  • My pinky must have slipped. I corrected the typo. – Stefan Mesken Jun 14 '15 at 19:51
  • Excuse me, I'm a bit supid. I don't see, why this is an answer of my question. Can you explain why? – asdfusername Jun 15 '15 at 14:24
  • Nah, it really isn't your fault. I suspect that this requirement puzzles you, because you didn't deal with structures in a more general setting. While dealing with natural numbers, it appears very natural to pick missing elements (like $\frac 2 3$) in a canonical superset, like the rational numbers. This is a reasonable approach (see Pedros) answer, but we then have to specify in advance how exactly we deal with such problems (in more abstract example there may be no canonical choice for such a superset). The definition of a substructure is one way to avoid such trouble. – Stefan Mesken Jun 15 '15 at 14:40