this is an odd question, but I am quite curious about this. I have only slightly more than a quarter's worth of understanding of abstract algebra, but I was wondering, when we define an object such as a ring or group, is it enough to check that the group/ring axioms hold over an equivalence relation, or is equality necessary? For example, on a ring, is it necessary that $a\cdot (b+c)=(a\cdot b)+(a\cdot c),$ or is it possible that if we have some equivalence relation $\sim$ that constitutes an alternate notion of equality, would it be enough for $a\cdot (b+c)\sim(a\cdot b)+(a\cdot c)$ for the structure to still be a ring? Otherwise, what would it be? Is there a way to generalize definitions over equivalences instead of equalities? Thank you!
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1No, you need actual equality. – Malady Jun 12 '25 at 03:41
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3The question is a little bit vaguely phrased, but if you have a set $S$ with an equivalence relation $\sim$ on which (for example) $a \cdot (b + c) \sim (a \cdot b) + (a \cdot c)$, you might be able to define a ring structure on the set of equivalence classes. There are lots of things you need to check to make sure the operations are well-defined but in many cases this works out. – mweiss Jun 12 '25 at 03:53
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2To add to @mweiss's comment: If you want give the set $S$ an algebraic structure, you need actual equality. If you want an algebraic structure on $S / \sim$ (the set of equivalence classes), you need the axioms on $S$ only up to $\sim$. The reason is that equality in $S / \sim$ is just the relation $\sim$ on $S$. You have to make sure, that your operations are well-defined, though! – Tzimmo Jun 12 '25 at 09:12
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Moreover, there is always an equivalence relation $\sim$ on $S$ such that $a\cdot(b+c)\sim a\cdot b + a\cdot c$: just take $\sim$ to be $S^2$, that is, the equivalence by which every element of $S$ is equivalent to each other (so there is only one equivalence class). That corresponds, following the preceding comments, to $S/{\sim}$ to be a trivial ring (one single element), and likewise you could define it to be a group, a Boolean algebra, whatever... – amrsa Jun 13 '25 at 08:48
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To define a group structure on a set $G$, you need to specify a binary operation on $G$, i.e. you need to exhibit a function $\star:G\times G\to G$. You can use equivalence classes or whatever else to come up with your definition of $\star$, but ultimately, whether $(G,\star)$ is a group depends solely on whether $\star$ is associative, unital, and whether inverses exist. – Joe Jun 14 '25 at 21:48
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2It seems you may be attempting to rediscover the notion of a congruence used in defining quotient algebras - a topic not normally treated until a course in Universal Algebra. See here and here for more. $\ \ $ – Bill Dubuque Jun 29 '25 at 21:48
1 Answers
Short Answer: No, you cannot replace every "$=$" with an arbitrary equivalence relation. Instead, if $\sim$ is a congruence relation, $X/\sim$ has a well-defined structure, and $[x]=[y]$ is the same as $x\sim y$ for $[x],[y] \in X/\sim$. However note we are not working in $X$ anymore, but in $X/\sim$.
The axioms usually presented for groups or rings use the strict equality ($=$) symbol defined in first-order logic. Therefore we need that the elements we compare are equal in that sense, not in a different one. However, if we have an arbitrary equivalence relation, we can take the quotient of the set modulo the equivalence relation, $X/\sim$, which is the set of all equivalence classes of $X$ with respect to $\sim$. One can show that two equivalence classes are equal $[x] = [y]$ if and only if $x \sim y$; that is, equality in $X/\sim$ is the same as equivalence in the sense of $\sim$. (Note $[x]$ as a set is not the same as $x$ itself)
If we insist on staying inside $X$, we cannot just replace strict equality "$=$" for an arbitrary equivalence relation in the axioms. To see why this does not make sense, take our set $X$ and define a relation $\sim$ such that $x \sim y$ for all $x,y \in X$. You can check that this relation satisfies all the axioms of an equivalence relation (see Appendix for the axioms), therefore it is an equivalence relation on $X$. If you changed every equality relation with this equivalence relation from each group/ring axiom, they would become vacuous and the definition ultimately meaningless.
That said, we can without issue define axioms that use an equivalence relation instead of strict equality. For instance, let $X$ be a (non-empty) set and $\cdot: X \times X \to X$ a binary operation. Then, define a relation $\sim$ such that for all $x,y \in X$, $x \sim y$ if and only if $x \cdot a = y \cdot a$ for every $a \in X$. You can intuitively think that this relation "sees" two elements being equal whenever they give the same element when right-multiplying (in the sense of our binary operation) by any other element in our set. You can check that this relation is an equivalence relation. With these definitions, we will state the axioms: for every $x,y,z \in X$ $$ 1. \ x \cdot (y \cdot z) \sim (x \cdot y) \cdot z $$ $$ 2. \ \text{There exists } e \in X \text{ such that } e \cdot x \sim x \cdot e \sim x $$ First axiom can be seen as that $(x \cdot (y \cdot z)) \cdot a = ((x \cdot y) \cdot z) \cdot a$ and the second axiom can be seen as there is an element $e \in X$ so that $(e \cdot x) \cdot a = (x \cdot e) \cdot a = x \cdot a$. So this is similar to a monoid but taking equality of elements as equality when right-multiplied by every element of the set.
Notice that now we can start proving results from this. For instance that if there is some $1 \in X$ so that $x \cdot 1 = x$ for all $x \in X$, $\sim$ is equivalent to strict equality.
Appendix
Equivalence relation axioms
Given a relation on a set $X$, it is an equivalence relation if and only if for all $x,y,z \in X$ it satisfies the following $$ 1. \ x \sim x \ \text{(Reflexivity)} $$ $$ 2. \ x \sim y \iff y \sim x \ \text{(Symmetry)} $$ $$ 3. \ (x \sim y \wedge y \sim z) \implies x \sim z \ \text{(Transitivity)} $$
Compatible relation
A relation on a set $X$ is compatible with a structure if and only if for each $n$-ary operation $\mu$ the structure has, if $a_k \sim b_k \ (k=1,2,3,\dots,n)$ then $\mu(a_1,a_2,a_3,\dots,a_n) \sim \mu(b_1,b_2,b_3,\dots,b_n)$. When the relation is both an equivalence relation and compatible with the structure, it is called a congruence relation (ordinary/strict equality is always a congruence relation for any operation).
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