Author: IMC

This post is the fourth example in an ongoing list of various sequences of functions which converge to different things in different ways.

Also in this series:

Example 1: converges almost everywhere but not in L1L1
Example 2: converges uniformly but not in L1L1
Example 3: converges in L1L1 but not uniformly
Example 5: converges pointwise but not in L1L1
Example 6: converges in L1L1 but does not converge anywhere

Example 4

A sequence of (Lebesgue) integrable functions fn:R→[0,∞)fn:R→[0,∞) so that {fn}{fn} converges to f:R→[0,∞)f:R→[0,∞) uniformly,  yet ff is not (Lebesgue) integrable.

‍Our first observation is that “ff is not (Lebesgue) integrable” can mean one of two things: either ff is not measurable or ∫f=∞∫f=∞. The latter tends to be easier to think about, so we’ll do just that. Now what function do you know of such that when you “sum it up” you get infinity? How about something that behaves like the divergent geometric series? Say, its continuous cousin f(x)=1xf(x)=1x? That should work since we know∫R1x=∫∞11x=∞.∫R1x=∫1∞1x=∞.Now we need to construct a sequence of integrable functions {fn}{fn} whose uniform limit is 1x1x. Let’s think simple: think of drawring the graph of f(x)f(x) one “integral piece” at a time. In other words, define:

This works because: It makes sense to define the fnfn as  f(x)=1xf(x)=1x “chunk by chunk” since this way the convergence is guaranteed to be uniform. Why? Because how far out we need to go in the sequence so that the difference f(x)−fn(x)f(x)−fn(x) is less than ϵϵ only depends on how small (or large) ϵϵ is. The location of xx doesn’t matter!

Also notice we have to define fn(x)=0fn(x)=0 for all x<1x<1 to avoid the trouble spot ln(0)ln⁡(0) in the integral ∫fn∫fn. This also ensures that the area under each fnfn is finite, guaranteeing integrability.

The details: Each fnfn is integrable since for a fixed nn,∫Rfn=∫n11x=ln(n).∫Rfn=∫1n1x=ln⁡(n).To see fn→ffn→f uniformly, let ϵ>0ϵ>0 and choose NN so that N>1/ϵN>1/ϵ. Let x∈Rx∈R. If x≤1x≤1, any nn will do, so suppose x>1x>1 and let n>Nn>N. If 1<x≤n1<x≤n, then we have |fn(x)−f(x)|=0<ϵ|fn(x)−f(x)|=0<ϵ. And if x>nx>n, then∣∣1xχ[1,∞)(x)−1xχ[1,n](x)∣∣=∣∣1x−0∣∣=1x<1n<1N<ϵ.|1xχ[1,∞)(x)−1xχ[1,n](x)|=|1x−0|=1x<1n<1N<ϵ.

‍For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Tai-Danae Bradley*

 

What do iPhones, Twitter, Netflix, cleaner cities, safer cars, state-of-the-art environmental management and modern medical diagnostics have in common? They are all made possible by Moore’s Law.

Moore’s Law stems from a seminal 1965 article by Intel founder Gordon Moore. He wrote:

“The complexity for minimum component costs has increased at a rate of roughly a factor of two per year … Certainly over the short term this rate can be expected to continue, if not to increase. Over the longer term, the rate of increase is a bit more uncertain, although there is no reason to believe it will not remain nearly constant for at least ten years. That means, by 1975, the number of components per integrated circuit for minimum cost will be 65,000.”

Moore noted that in 1965 engineering advances were enabling a doubling in semiconductor density every 12 months, but this rate was later modified to roughly 18 months. Informally, we may think of this as doubling computer performance.

In any event, Moore’s Law has now continued unabated for 45 years, defying several confident predictions it would soon come to a halt, and represents a sustained exponential rate of progress that is without peer in the history of human technology. Here is a graph of Moore’s Law, shown with the transistor count of various computer processors:

Where we’re at with Moore’s Law

At the present time, researchers are struggling to keep Moore’s Law on track. Processor clock rates have stalled, as chip designers have struggled to control energy costs and heat dissipation, but the industry’s response has been straightforward — simply increase the number of processor “cores” on a single chip, together with associated cache memory, so that aggregate performance continues to track or exceed Moore’s Law projections.

The capacity of leading-edge DRAM main memory chips continues to advance apace with Moore’s Law. The current state of the art in computer memory devices is a 3D design, which will be jointly produced by IBM and Micron Technology, according to a December 2011 announcement by IBM representatives.

As things stand, the best bet for the future of Moore’s Law are nanotubes — submicroscopic tubes of carbon atoms that have remarkable properties.

According to a recent New York Times article, Stanford researchers have created prototype electronic devices by first growing billions of carbon nanotubes on a quartz surface, then coating them with an extremely fine layer of gold atoms. They then used a piece of tape (literally!) to pick the gold atoms up and transfer them to a silicon wafer. The researchers believe that commercial devices could be made with these components as early as 2017.

Moore’s Law in science and maths

So what does this mean for researchers in science and mathematics?

Plenty, as it turns out. A scientific laboratory typically uses hundreds of high-precision devices that rely crucially on electronic designs, and with each step of Moore’s Law, these devices become ever cheaper and more powerful. One prominent case is DNA sequencers. When scientists first completed sequencing a human genome in 2001, at a cost of several hundred million US dollars, observers were jubilant at the advances in equipment that had made this possible.

Now, only ten years later, researchers expect to reduce this cost to only US$1,000 within two years and genome sequencing may well become a standard part of medical practice. This astounding improvement is even faster than Moore’s Law!

Applied mathematicians have benefited from Moore’s Law in the form of scientific supercomputers, which typically employ hundreds of thousands of state-of-the-art components. These systems are used for tasks such as climate modelling, product design and biological structure calculations.

Today, the world’s most powerful system is a Japanese supercomputer that recently ran the industry-standard Linpack benchmark test at more than ten “petaflops,” or, in other words, 10 quadrillion floating-point operations per second.

Below is a graph of the Linpack performance of the world’s leading-edge systems over the time period 1993-2011, courtesy of the website Top 500. Note that over this 18-year period, the performance of the world’s number one system has advanced more than five orders of magnitude. The current number one system is more powerful than the sum of the world’s top 500 supercomputers just four years ago.

Linpack performance over time.

Pure mathematicians have been a relative latecomer to the world of high-performance computing. The present authors well remember the era, just a decade or two ago, when the prevailing opinion in the community was that “real mathematicians don’t compute.”

But thanks to a new generation of mathematical software tools, not to mention the ingenuity of thousands of young, computer-savvy mathematicians worldwide, remarkable progress has been achieved in this arena as well (see our 2011 AMS Notices article on exploratory experimentation in mathematics).

In 1963 Daniel Shanks, who had calculated pi to 100,000 digits, declared that computing one billion digits would be “forever impossible.” Yet this level was reached in 1989. In 1989, famous British physicist Roger Penrose, in the first edition of his best-selling book The Emperor’s New Mind, declared that humankind would likely never know whether a string of ten consecutive sevens occurs in the decimal expansion of pi. Yet this was found just eight years later, in 1997.

Computers are certainly being used for more than just computing and analysing digits of pi. In 2003, the American mathematician Thomas Hales completed a computer-based proof of Kepler’s conjecture, namely the long-hypothesised fact that the simple way the grocer stacks oranges is in fact the optimal packing for equal-diameter spheres. Many other examples could be cited.

Future prospects

So what does the future hold? Assuming that Moore’s Law continues unabated at approximately the same rate as the present, and that obstacles in areas such as power management and system software can be overcome, we will see, by the year 2021, large-scale supercomputers that are 1,000 times more powerful and capacious than today’s state-of-the-art systems — “exaflops” computers (see NAS Report). Applied mathematicians eagerly await these systems for calculations, such as advanced climate models, that cannot be done on today’s systems.

Pure mathematicians will use these systems as well to intuit patterns, compute integrals, search the space of mathematical identities, and solve intricate symbolic equations. If, as one of us discussed in a recent Conversation article, such facilities can be combined with machine intelligence, such as a variation of the hardware and software that enabled an IBM system to defeat the top human contestants in the North American TV game show Jeopardy! we may see a qualitative advance in mathematical discovery and even theory formation.

It is not a big leap to imagine that within the next ten years tailored and massively more powerful versions of Siri (Apple’s new iPhone assistant) will be an integral part of mathematics, not to mention medicine, law and just about every other part of human life.

Some observers, such as those in the Singularity movement, are even more expansive, predicting a time just a few decades hence when technology will advance so fast that at the present time we cannot possibly conceive or predict the outcome.

Your present authors do not subscribe to such optimistic projections, but even if more conservative predictions are realised, it is clear that the digital future looks very bright indeed. We will likely look back at the present day with the same technological disdain with which we currently view the 1960s.

For more such insights, log into our website https://international-maths-challenge.com

Credit of the article given to Jonathan Borwein (Jon), University of Newcastle and David H. Bailey, University of California, Davis

 

When you hear the word closure, what do you think of? I think of wholeness – you know, tying loose ends, wrapping things up, filling in the missing parts. This same idea is behind the mathematician’s notion of closure, as in the phrase “taking the closure” of a set. Intuitively this just means adding in any missing pieces so that the result is complete, whole. For instance, the circle on the left is open because it’s missing its boundary. But when we take its closure and include the boundary, we say the circle is closed.

As another example, consider the set of all real numbers strictly between 0 and 1, i.e. the open interval (0,1). Notice that we can get arbitrarily close to 0, but we can’t quite reach it. In some sense, we feel that 0 might as well be included in set, right? I mean, come on, 0.0000000000000000000000000000000000000001 is basically 0, right? So by not considering 0 as an element in our set, we feel like something’s missing. The same goes for 1.

We say an element is a limit point of a given set if that element is “close” to the set,* and we say the set’s closure is the set together with its limit points. (So 0 and 1 are both limit points of (0,1) and its closure is [0,1].) It turns out the word closure is also used in algebra, specifically the algebraic closure of a field, but there it has a completely different definition which has to do with roots of polynomials, called algebraic elements. Now why would mathematicians use the same word to describe two seemingly different things? The purpose of today’s post is to make the observation that they’re not so different after all! This may be somewhat obvious, but it wasn’t until after a recent conversation with a friend that I saw the connection:

 

‍algebraic elements of a field

are like

limit points of a sequence!

‍(Note: I’m not claiming any theorems here, this is just a student’s simple observation.)

 

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Tai-Danae Bradley*

 

This post is the third example in an ongoing list of various sequences of functions which converge to different things in different ways.

‍Example 3

A sequence of continuous functions {fn:R→[0,∞)}{fn:R→[0,∞)} which converges to 0 in the L1L1 norm, but does not converge to 0 uniformly.

There are four criteria we want our functions to satisfy:

1. First off is the uniform convergence. Observe that “{fn}{fn} does not converge to 0 uniformly” can mean one of three things:

• converges to 0 pointwise only
• converges to something other than 0 (pointwise or uniformly)
• does not converge at all

So it’s up to you to decide which one feels more comfortable to work with. Here we’ll choose the second option.

2. Next, “{fn}{fn} converges to 0 in the L1L1 norm” means that we want to choose our sequence so that the area under the curve of the fnfn gets smaller and smaller as n→∞n→∞.
3. Further, we also want the fnfn to be positive (the image of each fnfn must be [0,∞)[0,∞)) (notice this allows us to remove the abosolute value sign in the L1L1 norm: ∫|fn|⇒∫fn∫|fn|⇒∫fn)
4. Lastly, the functions must be continuous.

A slick* but very simple solution is a sequence of triangles of decreasing area with height 1!

This works because: At x=0x=0, fn(x)=1fn(x)=1 for all nn, so there’s no way it can converge to zero (much less uniformly). In fact we have fn→ffn→f pointwise wheref(x)={1,if x=00otherwise.f(x)={1,if x=00otherwise.The area of each triangle is 1n1n which clearly goes to zero for nn large. Also, it’s clear to see visually that the area is getting smaller. This guarantees fn→0fn→0 in the L1L1 norm. Further, each fnfn is positive since we’ve defined it to equal zero as soon as the edges of the triangle reach the xx-axis. And lastly we have piecewise continuity.

The details: Let ϵ>0ϵ>0 and x∈Rx∈R. If x=0x=0, then fn(x)=1fn(x)=1 for all n and so fn→1fn→1. Otherwise x>0x>0 or x<0x<0 If x>0x>0 and x>1x>1, then fn(x)=0fn(x)=0 for all nn. Otherwise if x∈(0,1]x∈(0,1] choose N>1xN>1x. Then whenever n>Nn>N we have fn(x)=1−nx<1−1xx=0<ϵ.fn(x)=1−nx<1−1xx=0<ϵ. The case when x<0x<0 follows a similar argument.

Lastly fn→0fn→0 in the L1L1 norm since, as we mentioned, the areas are decreasing to 0. Explicitly:  ∫R|fn|=∫0−1n1+nx+∫1n01−nx=2n→0.∫R|fn|=∫−1n01+nx+∫01n1−nx=2n→0.

‍*I can brag because this particular example came from a friend. My own attempt at a solution was not nearly as intuitive.

Constructing the Tensor Product of Modules

The Basic Idea

Today we talk tensor products. Specifically this post covers the construction of the tensor product between two modules over a ring. But before jumping in, I think now’s a good time to ask, “What are tensor products good for?” Here’s a simple example where such a question might arise:

Suppose you have a vector space VV over a field FF. For concreteness, let’s consider the case when VV is the set of all 2×22×2 matrices with entries in RR and let F=RF=R. In this case we know what “FF-scalar multiplication” means: if M∈VM∈V is a matrix and c∈Rc∈R, then the new matrix cMcM makes perfect sense. But what if we want to multiply MM by complex scalars too? How can we make sense of something like (3+4i)M(3+4i)M? That’s precisely what the tensor product is for! We need to create a set of elements of the form(complex number) “times” (matrix)(complex number) “times” (matrix)so that the mathematics still makes sense. With a little massaging, this set will turn out to be C⊗RVC⊗RV.

So in general, if FF is  an arbitrary field and VV an FF-vector space, the tensor product answers the question “How can I define scalar multiplication by some larger field which contains FF?” And of course this holds if we replace the word “field” by “ring” and consider the same scenario with modules.

Now this isn’t the only thing tensor products are good for (far from it!), but I think it’s the most intuitive one since it is readily seen from the definition (which is given below).

So with this motivation in mind, let’s go!

‍From English to Math

Let RR be a ring with 1 and let MM be a right RR-module and NN a left RR-module and suppose AA is any abelian group. Our goal is to create an abelian group M⊗RNM⊗RN, called the tensor product of MM and NN, such that if there is an RR-balanced map i:M×N→M⊗RNi:M×N→M⊗RN and any RR-balanced map φ:M×N→Aφ:M×N→A, then there is a unique abelian group homomorphism Φ:M⊗RN→AΦ:M⊗RN→A such that φ=Φ∘iφ=Φ∘i, i.e. so the diagram below commutes.

Notice that the statement above has the same flavor as the universal mapping property of free groups!

Definition: Let XX be a set. A group FF is said to be a free group on XX if there is a function i:X→Fi:X→F such that for any group GG and any set map φ:X→Gφ:X→G, there exists a unique group homomorphism Φ:F→GΦ:F→G such that the following diagram commutes: (i.e. φ=Φ∘iφ=Φ∘i)

set map, so in particular we just want our’s to be RR-balanced:

: Let RR be a ring with 1. Let MM be a right RR-module, NN a left RR-module, and AA an abelian group. A map φ:M×N→Rφ:M×N→R is called RR-balanced if for all m,m1,m2∈Mm,m1,m2∈M, all n,n1,n2∈Nn,n1,n2∈N and all r∈Rr∈R,
φ(m1+m2,n)=φ(m1,n)+φ(m2,n)φ(m1+m2,n)=φ(m1,n)+φ(m2,n)φ(m,n1+n2)=φ(m,n1)+φ(m,n2)φ(m,n1+n2)=φ(m,n1)+φ(m,n2)φ(mr,n)=φ(m,rn)φ(mr,n)=φ(m,rn)

By “replacing” F by a certain quotient group F/HF/H! (We’ll define HH precisely below.)
These observations give us a road map to construct the tensor product. And so we begin:

‍Step 1

Let FF be a free abelian group generated by M×NM×N and let AA be an abelian group. Then by definition (of free groups), if φ:M×N→Aφ:M×N→A is any set map, and M×N↪FM×N↪F by inclusion, then there is a unique abelian group homomorphism Φ:F→AΦ:F→A so that the following diagram commutes.

Step 2

that the inclusion map M×N↪FM×N↪F is not RR-balanced! To fix this, we must “modify” the target space FF by replacing it with the quotient F/HF/H where H≤FH≤F is the subgroup of FF generated by elements of the form

(m1+m2,n)−(m1,n)−(m2,n)(m1+m2,n)−(m1,n)−(m2,n)

• (m,n1+n2)−(m,n1)−(m,n2)(m,n1+n2)−(m,n1)−(m,n2)
• (mr,n)−(m,rn)(mr,n)−(m,rn)

where m1,m2,m∈Mm1,m2,m∈M, n1,n2,n∈Nn1,n2,n∈N and r∈Rr∈R. Why elements of this form? Because if we define the map i:M×N→F/Hi:M×N→F/H byi(m,n)=(m,n)+H,i(m,n)=(m,n)+H,we’ll see that ii is indeed RR-balanced! Let’s check:

So, are we done now? Can we really just replace FF with F/HF/H and replace the inclusion map with the map ii, and still retain the existence of a unique homomorphism Φ:F/H→AΦ:F/H→A? No! Of course not. F/HF/H is not a free group generated by M×NM×N, so the diagram below is bogus, right?

Not totally. We haven’t actually disturbed any structure!

How can we relate the pink and blue lines? We’d really like them to be the same. But we’re in luck because they basically are!

‍Step 3

H⊆ker(f)H⊆ker⁡(f), that is as long as f(h)=0f(h)=0 for all h∈Hh∈H. And notice that this condition, f(H)=0f(H)=0, forces ff to be RR-balanced!

Let’s check:

Sooooo… homomorphisms f:F→Af:F→A such that H⊆ker(f)H⊆ker⁡(f) are the same as RR-balanced maps from M×NM×N to AA! (Technically, I should say homomorphisms ff restricted to M×NM×N.) In other words, we have

In conclusion, to say “abelian group homomorphisms from F/HF/H to AA are the same as (isomorphic to) RR-balanced maps from M×NM×N to AA” is the simply the hand-wavy way of saying

Whenever i:M×N→Fi:M×N→F is an RR-balanced map and φ:M×N→Aφ:M×N→A is an RR-balanced map where AA is an abelian group, there exists a unique abelian group homomorphism Φ:F/H→AΦ:F/H→A such that the following diagram commutes:

And this is just want we want! The last step is merely the final touch:

‍Step 4

the abelian quotient group F/HF/H to be the tensor product of MM and NN,

whose elements are cosets,

where m⊗nm⊗n for m∈Mm∈M and n∈Nn∈N is referred to as a simple tensor. And there you have it! The tensor product, constructed.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Tai-Danae Bradley*

This post is the second example in an ongoing list of various sequences of functions which converge to different things in different ways.

‍Example 2

A sequence of functions {fn:R→R}{fn:R→R} which converges to 0 uniformly but does not converge to 0 in L1L1.

This works because:  The sequence tends to 0 as n→∞n→∞ since the height of each function tends to 0 and the the region where fnfn is taking on this decreasing height is tending towards all of R+R+ ((0,n)(0,n) as n→∞n→∞) (and it’s already 0 on R−∪{0}R−∪{0}). The convergence is uniform because the number of times we have to keep “squishing” the rectangles until their height is less than ϵϵ does not depend on xx.

The details: Let ϵ>0ϵ>0 and choose N∈NN∈N so that N>1ϵN>1ϵ and let n>Nn>N. Fix x∈Rx∈R.

Case 1 (x≤0x≤0 or x≥nx≥n) Then fn(x)=0fn(x)=0 and so |fn(x)−0|=0<ϵ|fn(x)−0|=0<ϵ.

• Case 2 (0<x<n0<x<n ) Then fn(x)=1nfn(x)=1n and so |fn(x)−0|=1n<1N<ϵ|fn(x)−0|=1n<1N<ϵ

Finally, fn↛0fn↛0 in L1L1 since∫R|fn|=∫(0,n)1n=1nλ((0,n))=1.∫R|fn|=∫(0,n)1n=1nλ((0,n))=1.

‍Remark: Here’s a question you could ask: wouldn’t fn=nχ(0,1n)fn=nχ(0,1n) work here too? Both are tending to 0 everywhere and both involve rectangles of area 1. The answer is “kinda.” The problem is that the convergence of nχ(0,1n)nχ(0,1n) is pointwise. BUT Egoroff’s Theorem gives us a way to actually “make” it uniform!.

‍On the notation above:   For a measurable set X⊂RX⊂R, denote the set of all Lebesgue integrable functions f:X→Rf:X→R by L1(X)L1(X). Then a sequence of functions {fn}{fn} is said to converge in L1L1  to a function ff if limn→∞∫|fn−f|=0limn→∞∫|fn−f|=0.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Tai-Danae Bradley*

 

Given a sequence of real-valued functions {fn}{fn}, the phrase, “fnfn converges to a function ff” can mean a few things:

• fnfn converges uniformly
• fnfn converges pointwise
• fnfn converges almost everywhere (a.e.)
• fnfn converges in L1L1 (set of Lebesgue integrable functions)
• and so on…

Other factors come into play if the fnfn are required to be continuous, defined on a compact set, integrable, etc.. So since I do not have the memory of an elephant (whatever that phrase means…), I’ve decided to keep a list of different sequences that converge (or don’t converge) to different functions in different ways. With each example I’ll also include a little (and hopefully) intuitive explanation for why. Having these sequences close at hand is  especially useful when analysing the behavior of certain functions or constructing counterexamples.

The first sequence we’ll look at is one which converges almost everywhere, but does not converge in L1L1 (the set of Lebesgue integrable functions).

‍Example 1

A sequence of functions {fn:R→R}{fn:R→R} which converges to 0 almost everywhere but does not converge to 0 in L1L1.       

This works because: Recall that to say fn→0fn→0 almost everywhere means fn→0fn→0 pointwise on RR except for a set of measure 0. Here, the set of measure zero is the singleton set {0}{0} (at x=0x=0, fn(x)=nfn(x)=n and we can’t make this less than ϵϵ for any ϵ>0ϵ>0). So fnfn converges to 0 pointwise on (0,1](0,1]. This holds because if x<0x<0 or x>1x>1 then fn(x)=0fn(x)=0 for all nn. Otherwise, if x∈(0,1]x∈(0,1], we can choose nn appropriately:

The details:  Let ϵ>0ϵ>0 and x∈(0,1]x∈(0,1] and choose N∈NN∈N so that N>1xN>1x. Then whenever n>Nn>N, we have n>1xn>1x which implies x>1nx>1n and so fn(x)=0fn(x)=0. Hence |fnx−0|=0<ϵ|fnx−0|=0<ϵ.

Further*, fn↛0fn↛0 in L1L1 since∫R|fn|=∫[0,1n]n=nλ([0,1n])=1.∫R|fn|=∫[0,1n]n=nλ([0,1n])=1.

Remark: Notice that Egoroff’s theorem applies here! We just proved that fn→0fn→0 pointwise a.e. on RR, but Egoroff says that we can actually get uniform convergence a.e. on a bounded subset of RR, say (0,1](0,1].

In particular for each ϵ>0ϵ>0 we are guaranteed the existence of a subset E⊂(0,1]E⊂(0,1] such that fn→0fn→0 uniformly and λ((0,1]∖E)<ϵλ((0,1]∖E)<ϵ. In fact, it should be clear that that subset must be something like (ϵ2,1](ϵ2,1] (the “zero region” in the graph above). Then no matter where xx is in (0,1](0,1], we can always find nn large enough – namely all nn which satisfy 1n<ϵ21n<ϵ2 – so that fn(x)=0fn(x)=0, i.e. fn→ffn→f uniformly. And indeed, λ((0,1]∖(ϵ2,1]=ϵ/2<ϵλ((0,1]∖(ϵ2,1]=ϵ/2<ϵ as claimed.

‍On the notation above:   For a measurable set X⊂RX⊂R, denote the set of all Lebesgue integrable functions f:X→Rf:X→R by L1(X)L1(X). Then a sequence of functions {fn}{fn} is said to converge in L1L1  to a function ff if limn→∞∫|fn−f|=0limn→∞∫|fn−f|=0.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Tai-Danae Bradley*

In any area of math, it’s always good idea to keep a few counterexamples in your back pocket. Examples/non-examples from some of the different subsets of integral domains.

Z[i√5]Z[i5] is an integral domain which is not a UFD

That Z[i√5]Z[i5] an integral domain is easy to check (just computation).

• It’s not a UFD since we can write 6=2⋅3=(1+i√5)(1−i√5)6=2⋅3=(1+i5)(1−i5) as two distinct facorizations into irreducibles*

‍Z[x]Z[x] is a UFD which is not a PID

We know Z[x]Z[x] is a UFD because ZZ is a UFD (recall, a commutative ring RR is a UFD iff R[x]R[x] is a UFD).

• The ideal (2,x)={2f(x)+xg(x):f(x),g(x)∈Z[x]}(2,x)={2f(x)+xg(x):f(x),g(x)∈Z[x]} (polynomials with even constant term) is not principal**

‍Z[12+i√192]Z[12+i192] is a PID which is not a Euclidean domain

• This is a PID since it has a Dedekind-Hasse norm (see Dummit and Foote, 3rd ed., §8.2§8.2).
• It is not a Euclidean domain since it has no universal side divisors (ibid.).

ZZ is a Euclidean domain which is not a field

ZZ is a Euclidean domain via the absolute value norm (which gives the familiar division algorithm).

• It is not a field since the only elements which are units are 11 and −1−1.

‍  (*) Check 2,3,1+i√52,3,1+i5, and 1−i√51−i5 are indeed irreducible in Z[i√5]Z[i5]:

Write 2=αβ2=αβ for α,β∈Z[i√5]α,β∈Z[i5]. Then α=a+ib√5α=a+ib5 and N(α)=a2+5b2N(α)=a2+5b2 for some integers a,ba,b. Since 4=N(2)=N(α)N(β)4=N(2)=N(α)N(β), we must have a2+5b2=1,2a2+5b2=1,2 or 44. Notice b=0b=0 must be true (since a2+5b2∉{1,2,4}a2+5b2∉{1,2,4} for b≥1b≥1 and for any aa). Hence either α=a=1α=a=1 or 22. If α=1α=1 then αα is a unit. If α=2α=2, then we must have β=1β=1 and so ββ is a unit.

• Showing 3 is irreducible follows a similar argument.

‍Write 1+i√5=αβ1+i5=αβ with α=a+ib√5α=a+ib5 so that N(α)=a2+5b2∈{1,2,3,6}N(α)=a2+5b2∈{1,2,3,6} since 6=N(α)N(β)6=N(α)N(β). Consider two cases:  (case 1) If b=0b=0, then a2∈{1,2,3,6}a2∈{1,2,3,6} which is only true if a2=1a2=1 and so α=a=±1α=a=±1 is a unit. (case 2) If b>0b>0, we can only have b2=1b2=1 (since b2>1b2>1 gives a contradiction), and so a2+5∈{1,2,3,6}a2+5∈{1,2,3,6}, which implies a2=1a2=1. Hence α=±1±i√5α=±1±i5 and so N(α)=6N(α)=6. This implies N(β)=1N(β)=1 and so β=±1β=±1, which is a unit.

‍Showing 1−i√51−i5 is irreducible follows a similar argument.

principal in Z[x]Z[x]:

• Suppose to the contrary (2,x)=(f(x))(2,x)=(f(x)) for some polynomial f(x)∈Z[x]f(x)∈Z[x]. Since 2∈(f(x))2∈(f(x)), we must have 2=f(x)p(x)2=f(x)p(x) for some p(x)∈Z[x]p(x)∈Z[x]. Hence 0=degf(x)+degp(x)0=deg⁡f(x)+deg⁡p(x) which implies both f(x)f(x) and p(x)p(x) are constants. In particular, since 2=±1⋅±22=±1⋅±2, we need f(x),p(x)∈{±1,±2}f(x),p(x)∈{±1,±2}. If f(x)=±1f(x)=±1, then (f(x))=Z[x](f(x))=Z[x] which is a contradiction since (f(x))=(2,x)(f(x))=(2,x) mustbe a proper ideal (not every polynomial over Z[x]Z[x] has even constant term). It follows that f(x)=±2f(x)=±2. But since x∈(f(x))x∈(f(x)) as well, x=2r(x)x=2r(x) for some r(x)∈Z[x]r(x)∈Z[x]. But of course this is impossible for any polynomial with integer coefficients, r(x)r(x). Thus (2,x)(2,x) is not principal.

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Tai-Danae Bradley*

 

Here is a list of some of the subsets of integral domains, along with the reasoning (a.k.a proofs) of why the bullseye below looks the way it does. Part 2 of this post will include back-pocket examples/non-examples of each.

Integral Domain: a commutative ring with 1 where the product of any two nonzero elements is always nonzero

Unique Factorization Domain (UFD): an integral domain where every nonzero element (which is not a unit) has a unique factorization into irreducibles

Principal Ideal Domain (PID): an integral domain where every ideal is generated by exactly one element

Euclidean Domain: an integral domain RR with a norm NN and a division algorithm (i.e. there is a norm NN so that for every a,b∈Ra,b∈R with b≠0b≠0, there are q,r∈Rq,r∈R so that a=bq+ra=bq+r with r=0r=0 or N(r)<N(b)N(r)<N(b))

Field: a commutative ring where every nonzero element has an inverse

Because… We can just choose the zero norm: N(r)=0N(r)=0 for all r∈Fr∈F.

Proof: Let FF be a field and define a norm NN so that N(r)=0N(r)=0 for all r∈Fr∈F. Then for any a,b∈Fa,b∈F with b≠0b≠0, we can writea=b(b−1a)+0.a=b(b−1a)+0.

Because… If I◃RI◃R is an arbitrary nonzero ideal in the Euclidean domain RR, then I=(d)I=(d), where d∈Id∈I such that dd has the smallest norm among all elements in II. Prove this using the division algorithm on dd and some a∈Ia∈I.

Proof: Let RR be a Euclidean domain with respect to the norm NN and let I◃RI◃R be an ideal. If I=(0)I=(0), then II is principle. Otherwise let d∈Id∈I be a nonzero element such that dd has the smallest norm among all elements in II. We claim I=(d)I=(d). That (d)⊂I(d)⊂I is clear so let a∈Ia∈I. Then by the division algorithm, there exist q,r∈Rq,r∈R so that a=dq+ra=dq+r with r=0r=0 or N(r)<N(d)N(r)<N(d). Then r=a−dq∈Ir=a−dq∈I since a,d∈Ia,d∈I. But my minimality of dd, this implies r=0r=0. Hence a=dq∈(d)a=dq∈(d) and so I⊂(d)I⊂(d).

Because…Every PID has the ascending chain condition (acc) on its ideals!* So to prove PID ⇒⇒ UFD, just recall that an integral domain RR is a UFD if and only if 1) it has the acc on principal ideals** and 2) every irreducible element is also prime.

Proof: Let RR be a PID. Then 1) RR has the ascending chain condition on principal ideals and 2) every irreducible element is also a prime element. Hence RR is a UFD.

Because… By definition.

Proof: By definition.

‍*Def: In general, an integral domain RR has the acc on its principal ideals if these two equivalent conditions are satisfied:

1. Every sequence I1⊂I2⊂⋯⊂⋯I1⊂I2⊂⋯⊂⋯ of principal ideals is stationary (i.e. there is an integer n0≥1n0≥1 such that In=In0In=In0 for all n≥n0n≥n0).
2. For every nonempty subset X⊂RX⊂R, there is an element m∈Xm∈X such that whenever a∈Xa∈X and (m)⊂(a)(m)⊂(a), then (m)=(a)(m)=(a).

**To see this, use part 1 of the definition above. If I1⊂I2⊂⋯I1⊂I2⊂⋯ is an acsending chain, consider their union I=⋃∞n=1InI=⋃n=1∞In. That guy must be a principal ideal (check!), say I=(m)I=(m). This implies that mm must live in some In0In0  for some n0≥1n0≥1 and so I=(m)⊂In0I=(m)⊂In0. But since II is the union, we have for all n≥n0n≥n0(m)=I⊃In⊃In0=(m).(m)=I⊃In⊃In0=(m).Voila!

Every field FF is a PID

because the only ideals in a field are (0)(0) and F=(1)F=(1)! And every field is vacuously a UFD since all elements are units. (Recall, RR is a UFD if every non-zero, non-invertible element (an element which is not a unit) has a unique factorzation into irreducibles).

‍In an integral domain, every maximal ideal is also a prime ideal. 

(Proof: Let RR be an integral domain and M◃RM◃R a maximal ideal. Then R/MR/M is a field and hence an integral domain, which implies M◃RM◃R is a prime ideal.)

Butut the converse is not true (see counterexample below). However, the converse is true in a PID because of the added structure!

(Proof: Let RR be a PID and (p)◃R(p)◃R a prime ideal for some p∈Rp∈R. Then pp is a prime – and hence an irreducible – element (prime ⇔⇔ irreducible in PIDs). Since in an integral domain a principal ideal is maximal whenever it is generated by an irreducible element, we conclude (p)(p) is maximal.)

This suggests that if you want to find a counterexample – an integral domain with a prime ideal which is not maximal – try to think of a ring which is not a PID:   In Z[x]Z[x], consider the ideal (p)(p) for a prime integer pp. Then (p)(p) is a prime ideal, yet it is not maximal since(p)⊂(p,x)⊂Z[x].(p)⊂(p,x)⊂Z[x].

‍If FF is a field, then F[x]F[x] – the ring of polynomials in xx with coefficients in FF – is a Euclidean domain with the norm N(p(x))=degp(x)N(p(x))=deg⁡p(x) where p(x)∈F[x]p(x)∈F[x].

By the integral domain hierarchy above, this implies every ideal in F[x]F[x] is of the form (p(x))(p(x)) (i.e. F[x]F[x] is a PID) and every polynomial can be factored uniquely into a product of prime polynomials (just like the integers)! The next bullet gives an “almost converse” statement.

‍If R[x]R[x] is a PID, the RR must be a field.

To see this, simply observe that R⊂R[x]R⊂R[x] and so RR must be an integral domain (since a subset of a integral domain inherets commutativity and the “no zero divisors” property). Since R[x]/(x)≅RR[x]/(x)≅R, it follows that R[x]/(x)R[x]/(x) is also an integral domain. This proves that (x)(x) is a prime ideal. But prime implies maximal in a PID! So R[x]/(x)R[x]/(x) – and therefore RR – is actually a field.

• This is how we know, for example, that Z[x]Z[x] is not a PID (in the counterexample a few bullets up) – ZZ is not a field!

‍For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Tai-Danae Bradley*

After some exposure to group theory, you quickly learn that when trying to prove a group GG is abelian, checking if xy=yxxy=yx for arbitrary x,yx,y in GG is not always the most efficient – or helpful! – tactic. Here is a (not comprehensive) running tab of other ways you may be able to prove your group is abelian:

Show the commutator [x,y]=xyx−1y−1[x,y]=xyx−1y−1of two arbitary elements x,y∈Gx,y∈G must be the identity

• Show the group is isomorphic to a direct product of two abelian (sub)groups
• Check if the group has order p2p2 for any prime pp OR if the order is pqpq for primes p≤qp≤q with p∤q−1p∤q−1.
• Show the group is cyclic.
• Show |Z(G)|=|G|.|Z(G)|=|G|.
• Prove G/Z(G)G/Z(G) is cyclic. (e.g. does G/Z(G)G/Z(G) have prime order?)
• Show that GG has a trivial commutator subgroup, i.e. is [G,G]={e}[G,G]={e}.

Here’s a thought map which is (probably) more fun than practical. Note, pp and qq denote primes below:

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Tai-Danae Bradley*

Yes! I’m writing about math. No! Don’t close your browser window. Hear me out first…

I know very well that math has a bad rap. It’s often taught or thought of as a dry, intimidating, unapproachable, completely boring, who-in-their-right-mind-would-want-to-think-about-this-on-purpose kind of subject. I get it. Math was the last thing on earth I thought I’d study. Seriously.

But my understanding of math has since changed. I used to think it was a mess of equations and formulas, only enjoyed by a small number of masochists. But oh how I was wrong! Mathematics is not just numbers. It is not just strange symbols. And it is certainly not something reserved only for the few elite geniuses of the world.

Mathematics is a language –

a language of ideas, concepts, and notions.

‍

It’s true! Math is a language just like English, French, or Mandarin. And just like some ideas are best communicated in a particular language, other ideas are best communicated “in math.” This is why I’ve started a SaiBlog – as an aid in my own pursuit of becoming more proficient at thinking/speaking/reading mathematics.

One of the main challenges I face in this pursuit is the ability to strip away the intimidation factor

– the cryptic symbols, the elaborate vocabulary, the fancy formalities –

and unveil the true meaning of the text at hand. For me, this unveiling comes by reading and rereading, by working through problem after problem, and by writing. Quite often while learning new (and recalling old) mathematics, I have to stop and ask, “What is the text really saying behind all that jargon?” And if I can proceed to write down the idea in English (i.e. in lingo that’s easy on the brain) then that bit of information becomes engrained in my mind. Or at least it gets stored away in my brain somewhere. And if (or when) I forget it, I find that looking at my own handwritten notes conjures up the memory and the blood, sweat, and tears that went into learning that bit of info, and it all comes right back.

So Math3ma is my online repository as I make my way through this journey. Here’s the plan for now: some of the SaiBlog posts will be divided into two sections, in keeping with the aforementioned thought process:

And some posts will fall into “The Back Pocket” where I’ll keep little tidbits of math for a rainy day (or, perhaps, an exam). As for the actual content, I’m focusing on material found in the initial years of a graduate math program because, well, passing the qualifying exams is next on my agenda. But I think I’ll include some include undergrad material too. And as for future content, who knows? I’m excited to see what Math3ma can turn into.

Thanks for taking the time to peak into my journey as I work to see mathematics for what it really is–a very powerful, very beautiful language inherent in the world all around us!

For more such insights, log into www.international-maths-challenge.com.

*Credit for article given to Tai-Danae Bradley*