Analogy for Cosets, Ideals, and Quotient Rings

Edit: FYI my textbook starts with rings instead of groups so I don't really understand Normal Subgroups either.

Hi all, I have read the definition of Ideals and Quotient rings many times but don't have much intuition here. Would someone be able to help me visualize a 2D geometric example so these concepts feel more natural?

How strange/likely is it for a subring to be an Ideal?

I see that an Ideal maps to the 0 of a Quotient ring, so you are basically reducing the ring to a simpler ring with less elements. (like how $\Bbb Z \rightarrow \Bbb Z_8 \rightarrow \Bbb Z_4$ and each step simplifies the group while also getting rid a lot of structure as well)

In the book problems, I find identifying the quotient group using an isomorphism is less straightforward than the ring itself (and rather tedious and nontrivial for large sets when figuring addition/multiplication by hand). for example, $\Bbb F[x]/ g(x)$ where $g(x) \in \Bbb F[x]$

and why do we even want to know the remainder classes with polynomials anyway?

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You're asking us to write a book. This site works better with one question per question, with a specific answer expected. – Gerry Myerson Mar 18 '13 at 12:21
What makes you think 2D geometry is related to ring theory? We could explain how algebraic varieties in $\Bbb R^2$ correspond to certain ideals in $\Bbb R[X,Y]$, but I don't really think that is what you are asking for... – Marc van Leeuwen May 24 '13 at 8:34

Quotient rings allow you a rougher look at the ring, which might be preferable in some situations.

Consider the example of counting hours. This is basically like arithmetic in $\mathbb{Z}$. Say you know it is $5000$ hours from new year. Now if you ask the question "what time of the day is it", the number $5000$ carries in some sense to much information. For the time of the day, $24$ is the same as $0$. So for this question, the best suited algebraic model is not $\mathbb{Z}$, but the quotient ring of $\mathbb{Z}$ modulo the ideal generated by the element $24$. In fact, clocks are counters modulo 12 or modulo 24, depending on the model.

A more mathematical argument is that quotient rings allow you the construction of new rings. Like in the above example, the ring $\mathbb{Z}$ gives you the rings $\mathbb{Z}/n$. For a less trivial example, the finite fields are usually represented as quotient rings of $\mathbb{Z}/p\mathbb{Z}[X]$.

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Just to pick an easy part from your question: "How strange/likely is it for a subring to be an Ideal?", the answer is (assuming your textbook is about unitary rings, requiring the existence of a neutral element $1$ for multiplication) that this is extremely unlikely: it only happens when the ideal$~I$ is the whole ring (so it is not a proper ideal and the quotient $R/I$ is the trivial ring). Because a subring is required to contain the element $1$, and any ideal that contains $1$ must contain all its multiples, in other words be the whole ring.

This is a marked difference with group theory, where the notion of Normal Subgroup (the analogue of Ideal) is a refinement of the notion of subgroup (the analogue of subring). In ring theory this does not hold: subrings and ideals are fairly unrelated notions, and are used for rather different purposes.

Just for illustration: on one hand the ring $\Bbb Z$ has no proper subrings but infinitely many ideals (the multiples of any fixed $n$ is one), but on the other hand the ring $\Bbb Q$ has (uncountably) infinitely many subrings (among which $\Bbb Z$, but also for instance the rationals with odd denominator) but (since $\Bbb Q$ is a field) only two ideals, $\{0\}$ and $\Bbb Q$.

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Quotients always have to do with equivalences. If you have an algebraic structure $A$ of some kind, and $\approx$ is an equivalence relation on $A$ that is appropriate for that algebraic structure (a "congruence" relation), then you can talk about the quotient $A/{\approx}$. The elements of the quotient are the equivalence classes under $\approx$, i.e., for any element $a \in A$, there is an equivalence class $[a]$ of all the elements equivalent to $a$. The construction guarantees that all the algebraic operations of the original structure $A$ are inherited by the quotient $A/{\approx}$. For example, if we have multiplication $a \cdot b$ in $A$, then the corresponding multiplication of equivalence classes is $[a] \cdot [b] = [a \cdot b]$.

(What I have said here is called "universal algebra", which most mathematicians find too abstract to worry about. But I think it provides a much easier way to understand algebra, without getting bogged down by specifics. It also explains the reasons why algebra is the way it is.)

A congruence relation means an equivalence relation that respects all the algebraic operations. So, if $a$ and $a'$ are equivalent and $b$ and $b'$ are equivalent, then $a+b$ should be equivalent to $a'+b'$ and $ab$ should be equivalent to $a'b'$. Since addition has inverses in a ring, $-a$ should be equivalent to $-a'$ too. In a ring, it turns out that the entire equivalence relation can be represented by the subset of elements that are all equivalent to $0$. That is precisely what an ideal is.

Your text book would probably define an ideal as an additive subgroup that is closed under multiplication by the ring elements. Why does that make sense? If $a \approx 0$ and $b \approx 0$, then clearly $a + b \approx 0$, $0 \approx 0$ and $-a \approx 0$. That takes care of the "additive subgroup" business. If $a \approx 0$ and $k$ is an element of the ring, then $ka \approx k 0 = 0$ and $ak \approx 0 k = 0$. That tells you that the ideal should be closed under multiplication by all the ring elements. Thus it is that the algebraists like to represent the congruence relations of rings as ideals.

To find the quotient of the ring by an ideal $I$, you have to work backwards and reconstruct the equivalence relation that the ideal represents. It is not too hard. Since $I$ contains precisely the elements that are equivalent to $0$, $k + a$ should be equivalent to $k$ for any element $a \in I$. So, the equivalence class $[k]$ can therefore be written as $k + I$. (Algebraists tend to use the fancy term "coset" to talk about such equivalence classes. I think that term is not important here. Rather, "equivalence class" is the important concept.)

In azimut's excellent example in an earlier answer, the equivalence relation being talked about is equality modulo 24, i.e., $k = k' \pmod {24}$. That is what it means for $k$ and $k'$ to represent the same hour of the day. The ideal then says that $24$ is equivalent to $0$ and so is any integer multiple of $24$. The equivalence classes are $[0]$, ..., $[23]$, the hours of the day.

[PS. Normal subgroups of groups follow the same pattern as above. You think of group congruences and then notice that you can represent them by the subset of elements congruent to $1$ (the unit of the group). It turns out that this subset is closed under the group operations, ergo a subgroup, and it is also closed under conjugation, ergo a normal subgroup.]

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