Showing posts with label ring theory. Show all posts
Showing posts with label ring theory. Show all posts

Motivating ring theory, domains with integer-polynomial analogies

When you were first introduced to polynomial long division, you were struck by how a process that worked for integers worked for abstract polynomials. Integer division seemed rather "specific" -- focused on details like the resulting quotient having to be an integer -- and it seems bizarre that even the notion of integer division could be generalized beyond the integers.

It's not like integer division is just polynomial division with $x=10$ or something -- the results of the division are different, because integer division does not assume a base of 10.

But polynomial division also focuses on an analogous detail: the resulting quotient having to be a polynomial. And the essential lesson of mathematics, and the idea of abstract mathematics, is that serious analogies are the sign of abstraction.

So what makes polynomials and integers similar -- in what sense are they similar? -- that we can perform a "long division" algorithm on them?

And you know: this is not the only analogy between integers and polynomials either. Here's a list, with a general "abstract" phrasing that works for both integers and polynomials:
  • Division with remainder: For any $a,b$, there is some "unique" representation $a=qb+r$, where "uniqueness" is with respect to the property that $r<b$ (this $<$ ordering on the integers refers to the the ordering of the absolute value, and on the polynomials refers to the degree).
  • Bezout's identity: For any $a, b$, the set $\{\lambda a + \mu b\}$ is precisely the multiples of $\mathrm{gcd}(a, b)$.
  • Unique factorization: For any $a$, there exists a unique representation $a=p_1\dots p_n$ among $p_i$ that are "prime" ("irreducible") in the sense of not having any further factors. Well, is that really true? Not exactly: prime numbers can be factored with $-1$s and $1$s, and irreducible polynomials can be factored with constant polynomials. Well, this caveat has to do with stuff having itself as a factor, e.g. $x-2=(1/2)\cdot(2)\cdot(x-2)$ or $37=(-1)\cdot(-1)\cdot 1\cdot 37$. This means the other elements of this "factorization of the prime number" multiply to 1/are each "invertible" ("units"). So we should say we have unique factorization is "up to units". 
  • Greatest common divisors: For any $a, b$, there is a $\delta$ that divides both and is divided by all $d$ that divide both $a$ and $b$ (note that the term "divides" can exist in more generality than the assumptions of "division with remainder").
Well, the first thing we observe is that we should assume the existence of some notions of addition, subtraction and multiplication -- these seem to be the "fundamental" structures present among integers and polynomials (as opposed to e.g. rational numbers, rational functions which also require division or natural numbers which don't have subtraction). 

We will omit discussing the properties of these operations for now, as we don't yet have enough to motivate them on. 

Next, each of these discussed properties can be considered as special axioms for special cases of rings, domains where specific important theorems hold -- we call them, respectively: 
  • Euclidean domain: The ring is equipped with a natural number-valued magnitude function $\|\cdot\|:R\to\mathbb{Z}^{\ge 0}$, called the Euclidean function. For all $a,b$ in $R$ with $b$ non-zero (intuit out this condition), there exist $q, r$ with $\|r\|<\|b\|$ such that $a=qb+r$. 
  • Principal Ideal Domain: A ring where all ideals (additive subgroups invariant under multiplication by a ring element) are principal (a set of multiples of a generating element). Another abstraction is a "Bezout domain", which only requires that linear combinations of principal ideals are principal, but a PID should be seen as a more "natural" generalization.
  • Unique factorization domain: A ring where every element has a unique factorization into irreducibles, modulo multiplication by a unit. 
  • GCD domain: A ring in which any two elements has a GCD. 
(Quick comments on why the Euclidean function must map to the naturals: in fact, they could map to any "well-ordered set" (a totally ordered set in which every subset has a least element). The reason why this property is needed -- why we can't, e.g. map to the nonnegative reals -- is to ensure Euclid's algorithm terminates.)

The abstractions of the basic theorems about integers and polynomials occur as relationships between these domains. As it turns out, we will see that:
\[{\rm{ED}} \Rightarrow {\rm{PID}} \Rightarrow {\rm{UFD}} \Rightarrow {\rm{GCD}}\]
Before that, though, we can already play with some basic results we'd like, to get a feel of what axioms about ring addition and multiplication we should assume.

E.g.
  • What should $\|0\|$ equal? Prove that 0 must have the least magnitude of any ring element, making up the axioms you need on the fly. You should require: additive identity, additive inverse, additive associativity.
  • Try to prove some obvious results regarding Bezout's identity, like with $a$ and $b$ equal. You should require: left-distributivity, right-distributivity. 
  • Consider generalizations of two-element properties, like the Bezout identity, to multiple elements. You should require additive associativity, multiplicative associativity, additive commutativity, multiplicative commutativity.
Well, to be honest these all seem like fairly elementary properties that would be useful outside the cases of these special domains. Out of the following properties:
  1. Left-distributivity
  2. Right-distributivity
  3. Additive associativity
  4. Additive identity
  5. Additive inverse
  6. Additive commutativity
  7. Multiplicative associativity
  8. Multiplicative identity
  9. Multiplicative commutativity
  10. Multiplicative inverse
10 is not a ring axiom (because integers and polynomials don't have it). 1-7 essentially always are. 8 sometimes is, but not if you want e.g. the even numbers to be a ring. 9 typically isn't, although this is once again mostly just a matter of convention -- you can't really "see" that non-commutative rings appear often enough to justify their classification of rings, etc.

Presumably 6 (additive commutativity) is hardest to see the importance of, but it's relevant to note the relationships between these axioms. In fact, it is fairly simple to show that 1, 2 and 8 imply 6 (consider $(1+1)(x+y)$). As a result, addition -- the operation that multiplication distributes over -- is just generically seen as commutative.

Another important property often seen in algebraic problems is the ability to factor to find roots, i.e. to say that if $ab=0$, either $a$ or $b$ should be 0. This is known as an integral domain. The full sequence of inclusions, as we will see, is actually given by:
\[{\rm{ED}} \Rightarrow {\rm{PID}} \Rightarrow {\rm{UFD}} \Rightarrow {\rm{GCD}} \Rightarrow {\rm{ID}} \Rightarrow {\rm{Ring}}\]
The broader point of all this is that you should start often thinking of a lot of basic mathematical facts in the language of abstract algebra -- what kind of ring/domain is this result valid in? etc. -- because this is the most general setting in which a result is valid in, and you know exactly what it is "saying", i.e. what implies what. 

Understanding polynomial-ish differential equations

This is a rather simple idea, perhaps not one you really had too many problems understanding to begin with. Given you know that $e^{\lambda x}$ solves first-order polynomial differential equations, it's not too much of a stretch to imagine it solves higher-order polynomial differential equations too. But let's talk about this anyway.

So suppose you have differential equation like:

$$y''-3y'+2=0$$
A more interesting way of writing this would be:

$$(D-1)(D-2)y=0$$
The fact that you can do such a factoring is a consequence of the fact that polynomials in $D$ form a commutative ring. The idea behind rings and fields and other such objects is to look for a bunch of properties that a familiar set -- like the integers or the real numbers -- satisfies, then drilling those properties down to the basic axioms that imply them, to generalise them to objects other than the integers or real numbers. Differentiation operators are a great example of such a ring.

Now, your first instinct may to look at the factorisation and claim that $(D-1)y=0$ or $(D-2)y=0$. But this isn't right -- you assumed, here, incorrectly, that $(D-1)^{-1}$ and $(D-2)^{-1}$ existed (and that when applied on 0, they give you 0). This is not right, though -- we know there are in fact multiple functions that give 0 when you take $(D-1)$ of them. Which functions, specifically? The functions that are in the null space of $D-1$, i.e. the functions which satisfy:

$$(D-1)f=0$$
And 0 isn't the only such function. Ok, I've been giving you silly tautologies for about three lines now, but the point I'm making is that when you take the inverse operator of $(D-1)$ of both sides, what you really get is:

$$(D-2)y=(D-1)^{-1}0=ce^{x}$$
For arbitrary $c$.

The way to think about this kind of a $c$ is that you don't really have an equal to relation, i.e. an equation, you have an equivalence relation -- the "=" sign there is really abuse of notation. And you're saying that $(D-2)y$ belongs in an equivalence class where all elements are of the form $ce^{x}$ (and your quotient group's "representative element" can be $e^x$. The same applies, for example, for _____ in calculus -- fill in the blank. Well, fill it in.

Anyway, what you now have is a first-order differential equation (or really differential equivalence) in $y$.

$$(D-2)y=c_1e^x$$
But it isn't homogenous. I don't really know how to motivate a solution for a non-homogenous differential equation, really -- all I can say is that because the right-hand-side is an exponential, we just know that we can get some hints as to what $(D-2)^{-1}(ce^x)$ is by applying $(D-2)(ce^x)$ -- and if the right-hand-side isn't an exponential, then you can make it a sum or integral of exponentials, which is what Laplace and Fourier transforms are all about.

In any case, performing $(D-2)$ on $c_1e^x$ gives us $(c_1-2)e^x$, which immediately gives us an example solution, or a particular solution, $(c_1+2)e^x$ -- and all other solutions can be formed by adding linear combinations of the elements of the null space, i.e. solutions to the homogenous equation $(D-2)y=0$. These elements we know to take the form $c_2e^{2x}$.

$$y=(D-2)^{-1}c_1e^x=(c_1+2)e^x+c_2e^{2x}$$
Or transforming arbitrary constants,

$$y=c_1e^x+c_2e^{2x}$$

Use this method to find a general form for the solution to $(D-\alpha_1)(D-\alpha_2)...(D-\alpha_n)y=0$. Formalise our method with induction, and prove this general form with induction.