# The Universe of Discourse

Thu, 06 Aug 2020

Lately my favorite read has been Matt Levine’s Money Stuff articles from Bloomberg News. Bloomberg's web site requires a subscription but you can also get the Money Stuff articles as an occasional email. It arrives at most once per day.

Almost every issue teaches me something interesting I didn't know, and almost every issue makes me laugh.

Example of something interesting: a while back it was all over the news that oil prices were negative. Levine was there to explain what was really going on and why. Some people manage index funds. They are not trying to beat the market, they are trying to match the index. So they buy derivatives that give them the right to buy oil futures contracts at whatever the day's closing price is. But say they already own a bunch of oil contracts. If they can get the close-of-day price to dip, then their buy-at-the-end-of-the-day contracts will all be worth more because the counterparties have contracted to buy at the dip price. How can you get the price to dip by the end of the day? Easy, unload 20% of your contracts at a bizarre low price, to make the value of the other 80% spike… it makes my head swim.

But there are weird second- and third-order effects too. Normally if you invest fifty million dollars in oil futures speculation, there is a worst-case: the price of oil goes to zero and you lose your fifty million dollars. But for these derivative futures, the price could in theory become negative, and for short time in April, it did:

If the ETF’s oil futures go to -$37.63 a barrel, as some futures did recently, the ETF investors lose$20—their entire investment—and, uh, oops? The ETF runs out of money when the futures hit zero; someone else has to come up with the other $37.63 per barrel. One article I particularly remember discussed the kerfuffle a while back concerning whether Kelly Loeffler improperly traded stocks on classified coronavirus-related intelligence that she received in her capacity as a U.S. senator. I found Levine's argument persuasive: “I didn’t dump stocks, I am a well-advised rich person, someone else manages my stocks, and they dumped stocks without any input from me” … is a good defense! It’s not insider trading if you don’t trade; if your investment manager sold your stocks without input from you then you’re fine. Of course they could be lying, but in context the defense seems pretty plausible. (Kelly Loeffler, for instance, controversially dumped about 0.6% of her portfolio at around the same time, which sure seems like the sort of thing an investment adviser would do without any input from her? You could call your adviser and say “a disaster is coming, sell everything!,” but calling them to say “a disaster is coming, sell a tiny bit!” seems pointless.) He contrasted this case with that of Richard Burr, who, unlike Loeffler, remains under investigation. The discussion was factual and informative, unlike what you would get from, say, Twitter, or even Metafilter, where the response was mostly limited to variations on “string them up” and “eat the rich”. Money Stuff is also very funny. Today’s letter discusses a disclosure filed recently by Nikola Corporation: More impressive is that Nikola’s revenue for the second quarter was very small, just$36,000. Most impressive, though, is how they earned that revenue:

During the three months ended June 30, 2020 and 2019 the Company recorded solar revenues of $0.03 million and$0.04 million, respectively, for the provision of solar installation services to the Executive Chairman, which are billed on time and materials basis. …

“Solar installation projects are not related to our primary operations and are expected to be discontinued,” says Nikola, but I guess they are doing one last job, specifically installing solar panels at founder and executive chairman Trevor Milton’s house? It is a \$13 billion company whose only business so far is doing odd jobs around its founder’s house.

A couple of recent articles that cracked me up discussed clueless day-traders pushing up the price of Hertz stock after Hertz had declared bankruptcy, and how Hertz diffidently attempted to get the SEC to approve a new stock issue to cater to these idiots. (The SEC said no.)

One recurring theme in the newsletter is “Everything is Securities Fraud”. This week, Levine asks:

Is it securities fraud for a public company to pay bribes to public officials in exchange for lucrative public benefits?

Of course you'd expect that the executives would be criminally charged, as they have been. But is there a cause for the company’s shareholders to sue? If you follow the newsletter, you know what the answer will be:

Oh absolutely…

because Everything is Securities Fraud.

Still it is a little weird. Paying bribes to get public benefits is, you might think, the sort of activity that benefits shareholders. Sure they were deceived, and sure the stock price was too high because investors thought the company’s good performance was more legitimate and sustainable than it was, etc., but the shareholders are strange victims. In effect, executives broke the law in order to steal money for the shareholders, and when the shareholders found out they sued? It seems a little ungrateful?

I recommend it.

Wed, 05 Aug 2020

I'm not sure if this is interesting, trivial, or both. You decide.

Let's divide the numbers from 1 to 30 into the following six groups:

 A 1 2 4 8 16 B 3 6 12 17 24 C 5 9 10 18 20 D 7 14 19 25 28 E 11 13 21 22 26 F 15 23 27 29 30

Choose any two rows. Chose a number from each row, and multiply them mod 31. (That is, multiply them, and if the product is 31 or larger, divide it by 31 and keep the remainder.)

Regardless of which two numbers you chose, the result will always be in the same row. For example, any two numbers chosen from rows B and D will multiply to yield a number in row E. If both numbers are chosen from row F, their product will always appear in row A.

Mon, 03 Aug 2020

Gulliver's Travels (1726), Part III, chapter 2:

I observed, here and there, many in the habit of servants, with a blown bladder, fastened like a flail to the end of a stick, which they carried in their hands. In each bladder was a small quantity of dried peas, or little pebbles, as I was afterwards informed. With these bladders, they now and then flapped the mouths and ears of those who stood near them, of which practice I could not then conceive the meaning. It seems the minds of these people are so taken up with intense speculations, that they neither can speak, nor attend to the discourses of others, without being roused by some external action upon the organs of speech and hearing… . This flapper is likewise employed diligently to attend his master in his walks, and upon occasion to give him a soft flap on his eyes; because he is always so wrapped up in cogitation, that he is in manifest danger of falling down every precipice, and bouncing his head against every post; and in the streets, of justling others, or being justled himself into the kennel.

When I first told Katara about this, several years ago, instead of “the minds of these people are so taken up with intense speculations” I said they were obsessed with their phones.

Now the phones themselves have become the flappers:

Y. Tung and K. G. Shin, "Use of Phone Sensors to Enhance Distracted Pedestrians’ Safety," in IEEE Transactions on Mobile Computing, vol. 17, no. 6, pp. 1469–1482, 1 June 2018, doi: 10.1109/TMC.2017.2764909.

Our minds are not even taken up with intense speculations, but with Instagram. Dean Swift would no doubt be disgusted.

Sat, 01 Aug 2020

I know this question has been asked many times and there is good information out there which has clarified a lot for me but I still do not understand how the addition and multiplication tables for !!GF(4)!! is constructed?

I've seen [links] but none explicity explain the construction and I'm too new to be told "its an extension of !!GF(2)!!"

The only “reasonable” answer here is “get an undergraduate abstract algebra text and read the chapter on finite fields”. Because come on, you can't expect some random stranger to appear and write up a detailed but short explanation at your exact level of knowledge.

But sometimes Internet Magic Lightning strikes  and that's what you do get! And OP set themselves up to be struck by magic lightning, because you can't get a detailed but short explanation at your exact level of knowledge if you don't provide a detailed but short explanation of your exact level of knowledge — and this person did just that. They understand finite fields of prime order, but not how to construct the extension fields. No problem, I can explain that!

I had special fun writing this answer because I just love constructing extensions of finite fields. (Previously: [1] [2])

For any given !!n!!, there is at most one field with !!n!! elements: only one, if !!n!! is a power of a prime number (!!2, 3, 2^2, 5, 7, 2^3, 3^2, 11, 13, \ldots!!) and none otherwise (!!6, 10, 12, 14\ldots!!). This field with !!n!! elements is written as !!\Bbb F_n!! or as !!GF(n)!!.

Suppose we want to construct !!\Bbb F_n!! where !!n=p^k!!. When !!k=1!!, this is easy-peasy: take the !!n!! elements to be the integers !!0, 1, 2\ldots p-1!!, and the addition and multiplication are done modulo !!n!!.

When !!k>1!! it is more interesting. One possible construction goes like this:

1. The elements of !!\Bbb F_{p^k}!! are the polynomials $$a_{k-1}x^{k-1} + a_{k-2}x^{k-2} + \ldots + a_1x+a_0$$ where the coefficients !!a_i!! are elements of !!\Bbb F_p!!. That is, the coefficients are just integers in !!{0, 1, \ldots p-1}!!, but with the understanding that the addition and multiplication will be done modulo !!p!!. Note that there are !!p^k!! of these polynomials in total.

2. Addition of polynomials is done exactly as usual: combine like terms, but remember that the the coefficients are added modulo !!p!! because they are elements of !!\Bbb F_p!!.

3. Multiplication is more interesting:

a. Pick an irreducible polynomial !!P!! of degree !!k!!. “Irreducible” means that it does not factor into a product of smaller polynomials. How to actually locate an irreducible polynomial is an interesting question; here we will mostly ignore it.

b. To multiply two elements, multiply them normally, remembering that the coefficients are in !!\Bbb F_p!!. Divide the product by !!P!! and keep the remainder. Since !!P!! has degree !!k!!, the remainder must have degree at most !!k-1!!, and this is your answer.

Now we will see an example: we will construct !!\Bbb F_{2^2}!!. Here !!k=2!! and !!p=2!!. The elements will be polynomials of degree at most 1, with coefficients in !!\Bbb F_2!!. There are four elements: !!0x+0, 0x+1, 1x+0, !! and !!1x+1!!. As usual we will write these as !!0, 1, x, x+1!!. This will not be misleading.

Addition is straightforward: combine like terms, remembering that !!1+1=0!! because the coefficients are in !!\Bbb F_2!!:

$$\begin{array}{c|cccc} + & 0 & 1 & x & x+1 \\ \hline 0 & 0 & 1 & x & x+1 \\ 1 & 1 & 0 & x+1 & x \\ x & x & x+1 & 0 & 1 \\ x+1 & x+1 & x & 1 & 0 \end{array}$$

The multiplication as always is more interesting. We need to find an irreducible polynomial !!P!!. It so happens that !!P=x^2+x+1!! is the only one that works. (If you didn't know this, you could find out easily: a reducible polynomial of degree 2 factors into two linear factors. So the reducible polynomials are !!x^2, x·(x+1) = x^2+x!!, and !!(x+1)^2 = x^2+2x+1 = x^2+1!!. That leaves only !!x^2+x+1!!.)

To multiply two polynomials, we multiply them normally, then divide by !!x^2+x+1!! and keep the remainder. For example, what is !!(x+1)(x+1)!!? It's !!x^2+2x+1 = x^2 + 1!!. There is a theorem from elementary algebra (the “division theorem”) that we can find a unique quotient !!Q!! and remainder !!R!!, with the degree of !!R!! less than 2, such that !!PQ+R = x^2+1!!. In this case, !!Q=1, R=x!! works. (You should check this.) Since !!R=x!! this is our answer: !!(x+1)(x+1) = x!!.

Let's try !!x·x = x^2!!. We want !!PQ+R = x^2!!, and it happens that !!Q=1, R=x+1!! works. So !!x·x = x+1!!.

I strongly recommend that you calculate the multiplication table yourself. But here it is if you want to check:

$$\begin{array}{c|cccc} · & 0 & 1 & x & x+1 \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & x & x+1 \\ x & 0 & x & x+1 & 1 \\ x+1 & 0 & x+1 & 1 & x \end{array}$$

To calculate the unique field !!\Bbb F_{2^3}!! of order 8, you let the elements be the 8 second-degree polynomials !!0, 1, x, \ldots, x^2+x, x^2+x+1!! and instead of reducing by !!x^2+x+1!!, you reduce by !!x^3+x+1!!. (Not by !!x^3+x^2+x+1!!, because that factors as !!(x^2+1)(x+1)!!.) To calculate the unique field !!\Bbb F_{3^2}!! of order 27, you start with the 27 third-degree polynomials with coefficients in !!{0,1,2}!!, and you reduce by !!x^3+2x+1!! (I think).

The special notation !!\Bbb F_p[x]!! means the ring of all polynomials with coefficients from !!\Bbb F_p!!. !!\langle P \rangle!! means the ring of all multiples of polynomial !!P!!. (A ring is a set with an addition, subtraction, and multiplication defined.)

When we write !!\Bbb F_p[x] / \langle P\rangle!! we are constructing a thing called a “quotient” structure. This is a generalization of the process that turns the ordinary integers !!\Bbb Z!! into the modular-arithmetic integers we have been calling !!\Bbb F_p!!. To construct !!\Bbb F_p!!, we start with !!\Bbb Z!! and then agree that two elements of !!\Bbb Z!! will be considered equivalent if they differ by a multiple of !!p!!.

To get !!\Bbb F_p[x] / \langle P \rangle!! we start with !!\Bbb F_p[x]!!, and then agree that elements of !!\Bbb F_p[x]!! will be considered equivalent if they differ by a multiple of !!P!!. The division theorem guarantees that of all the equivalent polynomials in a class, exactly one of them will have degree less than that of !!P!!, and that is the one we choose as a representative of its class and write into the multiplication table. This is what we are doing when we “divide by !!P!! and keep the remainder”.

A particularly important example of this construction is !!\Bbb R[x] / \langle x^2 + 1\rangle!!. That is, we take the set of polynomials with real coefficients, but we consider two polynomials equivalent if they differ by a multiple of !!x^2 + 1!!. By the division theorem, each polynomial is then equivalent to some first-degree polynomial !!ax+b!!.

Let's multiply $$(ax+b)(cx+d).$$ As usual we obtain $$acx^2 + (ad+bc)x + bd.$$ From this we can subtract !!ac(x^2 + 1)!! to obtain the equivalent first-degree polynomial $$(ad+bc) x + (bd-ac).$$

Now recall that in the complex numbers, !!(b+ai)(d + ci) = (bd-ac) + (ad+bc)i!!. We have just constructed the complex numbers,with the polynomial !!x!! playing the role of !!i!!.

[ Note to self: maybe write a separate article about what makes this a good answer, and how it is structured. ]