Residue class of $a$ modulo $I$ notation clarification?

Question

In the Wikipedia page it has been stated that,

Given a ring $R$ and a two-sided ideal $I$ in R, we may define an equivalence relation $\sim$ on $R$ as follows:

$$ a \sim b \iff a − b \in I. $$

Using the ideal properties, it is not difficult to check that $\sim$ is a congruence relation. In case $a \sim b$, we say that $a$ and $b$ are congruent modulo $I$. The equivalence class of the element $a$ in $R$ is given by $$ [a] = a + I := \{ a + r : r \in I \}. $$ This equivalence class is also sometimes written as a mod I and called the "residue class of a modulo $I$"

Question Does the above statement mean we use $\frac {a}{I}$ instead of $\frac{a}{\sim}$? If Yes, why? Since $I$ is a set and $\sim$ is a relation which means it is a set of ordered pairs! Can someone clarify this notation and conclude it from equivalence relation? What does $\frac{R}{I}$ mean exactly being deduced from equivalence relation? and why is the equivalence relation defined that way?

Answer 1

The ideal-based construction of quotient rings (& groups) is a special case of the congruence-based construction of a quotient algebra of equational algebra $A.\,$ Here a congruence is an equivalence relation that is compatible with all operations of $A,\,$ i.e. $\,a'\equiv a,\ b'\equiv b\,\Rightarrow\, a'\circ b'\equiv b\circ b\,$ for all binary operations $\circ,\,$ and analogously for all $n$-ary operations of $A.\,$ Equivalently, one can view a congruence via subalgebras of powers of $A,\,$ e.g. see here where that is described in detail for rings.

In algebras (like rings & groups) enjoying "subtraction" we can $0-$normalize equations $\,a \equiv b \iff a-b\equiv 0,\,$ which means that congruences are completely determined by the equivalence class of $0\,$ (called ideal-determined). In this case we can simplify the congruence-based construction of the quotient object to an ideal-based construction - just as for rings.

The two constructions yield isomorphic quotient objects hence it is possible to abuse notation by mixing and matching both notations (or completely omitting the quotient annotation), e.g. in $\,\Bbb Z/n\Bbb Z\cong \Bbb Z\bmod n\,$ with $\,2 = 2\bmod n = [2]_n = 2+n\Bbb Z\,$ among other abuses). Usually it is clear from context precisely what is intended (but when first learning these topics one should use rigorous complete notation till one has a strong enough grasp that no confusion can arise due to such abuse).

There is further discussion in the links above, e.g. conditions implying isomorphism of the congruence lattice and ideal lattice for general algebras. Any textbook on "universal algebra" will discuss congruences in general (an excellent [and free] choice is George Bergman's An Invitation to General Algebra and Universal Constructions).

Answer 2

I can't say much about the notation, other than since it is used so frequently, it is probably a good idea to keep track on the ideal you are dividing by. Hence $R/I$ gives you more information than $R/\sim$ at first sight. But this is just my take on this, it would be nice to know the historical reasons.

As for the definition, I can offer you some motivation. One disclaimer, though: I am far from knowledgeable in this topic. This is just how quotient rings were presented to me, and I find it to be interesting, if anything. Surely people here can add some more enlightening reasons as to why we study this construction.

After defining what a ring and a ring morphism are, and looking at some examples, one may ask the following question:

Let $I \subset R$ be a subset of a ring. When is $I = \ker f$ for some morphism $f : R \to S$?

Let's look at necessary conditions first: let $I$ be such that $I = \ker f$ for some $f: R \to S$. Now if $a,b \in I$ and $r \in R$,

$$ f(a+b) = f(a) + f(b) = 0 + 0 = 0 $$

and $f(ra) = f(r)f(a) = 0 = f(a)f(r) = f(ar)$. So necessarily we get $a+b, ra, ar \in \ker f = I$ and so any such $I$ needs to be closed under summation and absorb products.

Are these to properties enough to conclude that $I$ is the kernel of some morphism?

If $I$ were the kernel of some function $f$, then in particular for any $a \in R$ we should have $$ f(a+I) = f(a) + f(I) = \{f(a)\}. $$ Rephrasing this, if $a,b \in R$ differ only by an element of $I$ then their image via $f$ should be the same, in the following sense: if $a = b + i$ with $i \in I$, this is equivalent to $a-b \in I$. And in that case,

$$ f(a) = f(a-b + b) = f(a-b) + f(b) = f(b) $$

because $a-b \in I = \ker f$. So a natural idea would be to group elements of $R$ depending on whether they 'differ by an element of $I$ '. Concretely, we define the relation $a \sim b \iff (a-b) \in I$. This is seen to be an equivalence relation from the properties of $I$, and we always have a canonical mapping from $R$ to the classes of $R/I := R/\sim$ (which I will interchangeably note $[a]$ or $a+I$) via

$$ \begin{align*} \pi : \ & R \to R/I \\ & r \mapsto r + I \end{align*} $$

So far $R/I$ has no ring structure, since $\pi$ always exists as a function between sets (for any equivalence relation, actually). But $R$ is a ring, so maybe we can make sense of $R/I$ as a ring itself. In effect, with a bit of work, one can show that this is the case, by defining

$$ [a] + [b] := [a+b] \text{ and } [a] \cdot [b] := [ab]. $$

with $[0]$ and $[1]$ as identities. These could be ill defined, meaning that different representatives of a class could give different results, but it is not the case: I will prove so for summation, and I'll leave you to complete the other: if $a \sim a'$ and $b \sim b'$, we have

$$ [a']+[b'] = [a'+b'] = [a+b + (a'-a) + (b'-b)] = [a+b] = [a]+[b], $$

since $(a'-a),(b'-b) \in I$. In particular, with this structure $\pi$ is a ring morphism.

Okay, we have seen that $I$ needs to have some special properties, and that these motivate $R/I$. This is all assuming that $I$ is the kernel of some morphism, but so far there is no indication that this should be the case: in particular, we need to have a suitable ring $S$ as the codomain, which in principle should be related to our subset $I$. Well, we do have one newly constructed morphism at hand, and this one seems to do the trick: if $a \in R$, then $a \in \ker \pi$ if and only if $\pi(a) = [0]$. That is $a \sim 0$, or equivalently, $a = a -0 \in I$. This concludes the answer to our question:

Proposition. A subset $I$ of a ring $R$ is the kernel of some morphism $f: R \to S$ if and only if $I$ is an ideal.

I hope this at least motivates why one would make such a construction: ideals are closely related to kernels, and kernels appear everywhere naturally, because they are intrinsic to morphisms between rings.