GaryP

Congrats! Can I still buy the Journal of Algebra at Wal-Mart?

Goliath

Thanks, everyone!

Helen,

What?! Then what do you read? You don't keep your nose stuck in those stuffy old academic journals like McCall's or Better Homes and Gardens, do you?

ROFL! I doubt it would sell at Wal-Mart (especially since the price for an individual subscription is $2,300 per year).. However, most university libraries subscribe to the Journal of Algebra, even if it's an electronic subscription.

Sincerely,

Goliath

nogods4me

Goliath.
congrats on being published


I almost got published in the Journal of Geometry, but they said the angle i took in my article was too obtuse.

Jobar

Congratulations- that sounds like a very prestigious addition to your list of publications!

I'm not a complete ignoramus about math- but I confess that "Boundary Valuation Domains" tells me zip. Can you give a short and simple sketch of what you are writing about?

Goliath

Jobar,

Well, I study Factorization Theory (specifically, Factorization in Integral Domains). First, some preliminary stuff:

A ring is basically a set where you can add and multiply things (yes, that's a very informal definition--to be more precise, a ring R is a nonempty set with two binary operations, + and x (addition and multiplication) such that R is an abelian group under addition, multiplication is associative, and multiplication distributes over addition). The most natural example of a ring is the integers, which I will denote by Z (ie Z={...,-2,-1,0,1,2,...}).

The integers form a very nice ring. Specifically, Z has the property that for every two nonzero elements x and y in Z, xy is nonzero (such rings are called integral domains, or domains for short).

Here's an example of a ring that is not an integral domain: Consider the integers mod 4, which I will denote as Z_4. Z_4={0*,1*,2*,3*}. In Z_4, we add two elements--say a* and b* by taking the remainder when a+b is divided by 4. So, for example, 3*+2*=5*=1*, 3*+1*=4*=0*.

Multiplication works in much the same way: to get (a*)(b*), just take the remainder when ab is divided by 4. In particular, we have (2*)(2*)=4*=0*. So, we have two nonzero elements whose product is zero, whence Z_4 is not a domain.

In domains, there are different "building blocks:"

1. zero (yes, every ring has a zero element...it's merely the additive identity).

2. Units--ie nonzero elements that are invertible in R. For example, if R=Z, then the only units are 1 and -1 (1/2, for example, is not an integer, so 2 is not a unit in Z).

3. Irreducibles--ie nonzero nonunits that don't "break down" any more under multiplication. To be slightly more technical, an element x in a domain R is said to be irreducible if whenever x=ab for some a,b in R, then one of a or b is a unit of R. If R=Z, then the irreducibles are all of the primes (ie 2, 3, 5, etc). (I should say that techincally, "prime" and "irreducible" are two different concepts, although they happen to mean the same thing if our domain has unique factorization).

Now, in the integers--as well as a lot of other "nice" domains, every nonzero nonunit element can be factored into irreducibles. The world would be a nice place if this were the case for all domains, but unfortunately it isn't. Domains in which all nonzero nonunits can be factored into irreducibles are called atomic domains (irreducibles are sometimes called atoms). For the sake of brevity, I won't mention any examples of non-atomic domains, but I'll be glad to post some examples if requested (most--if not all--domains that are encountered in an undergraduate Abstract Algebra course are atomic), but there are a lot of them out there. In fact, there exist domains for which no nonzero nonunits can be factored into irreducibles (these are called antimatter domains).

When you memorized your multiplication tables as a kid, you learned that every integer (except 0, 1, or -1) can be uniquely factored uniquely into irreducibles (primes). This, in fact, is the statement of the Fundamental Theorem of Arithmetic, which was originally proven by Euclid about 2500 years ago. Now, what do I mean by "uniquely"? Well, if you take 12, for example, you can write 12=2*2*3, or 12=(-3)*(-2)*2. But what did you do to get the second factorization? You simply reordered the primes (no big deal, since we're dealing strictly with rings that have commutative multiplication), and you multiplied by a couple of units (ie the -1's). No big deal! These factorizations are essentially the same. So, this idea leads to the definition of a Unique Factorization Domain.

Definition: A domain R is a Unique Factorization Domain (UFD) if R is atomic, and whenever we have

a_1 a_2....a_m=b_1 b_2 ...b_n

for a_1, a_2,...,a_m,b_1,b_2,...,b_n irreducibles of R, then:

1. n=m (in other words, each irredicuble factorization has the same number of irreducibles)

2. There exist units of R--call them u_1, u_2, ..., u_n-- such that (possibly after reordering), a_i=u_i b_i for each i between 1 and n (in other words, we can "pair up" irreducibles in a way that one varies from the other by a unit in R).

Now, UFD's are very, very nice. A lot is known about them. Not nearly as much is known about the following kinds of domains, which are generalizations of UFD's.

Definition: A domain R is a Half Factorial Domain (HFD) if R is atomic and whenever we have

a_1 a_2....a_m=b_1 b_2 ...b_n

for a_1, a_2,...,a_m,b_1,b_2,...,b_n irreducibles of R, then n=m

(note that we have half the axioms, hence the term "half factorial domain). HFD's were first studied in a 1960 paper by Carlitz, although the term "Half Factorial Domain" wasn't coined until 1976 in a paper by Zaks.

Anyways, I primarily study HFD's. My advisor, in a 1999 paper, introduced a nifty little tool called the boundary map. My entire thesis is about using this tool in a more general context than he used it in, as well as generalizing the boundary map.

Boundary valuation domains are a special class of HFD's that I defined using the boundary map.

I could go on, but I've rambled enough. If anyone reading this has any questions, feel free to reply or PM me.

Sincerely,

Goliath (who should probably go back to grading his Calc III finals....blech...)

mongrel

Huh?

But seriously, congrats on getting published, even though I have absolutely no clue what you're on about.

Goliath

mongrel,

Thanks for the congratulations.

As for the algebra....well, take a few years of undergraduate Mathematics, then we'll talk.

Sincerely,

Goliath

(who is gonna scream if he sees that someone else has written that the integral of (ln(x))^8 dx is ((ln(x))^9)/9 + C).

Vespertine

I'd tell my uncle (he's got a doctorate in math and teaches at EWU) but he's a fundie and probably wouldn't like that I got it from this website.

KitKit

That's wonderful!!! Congratulations!

Pensee

I swear I thought the title read:
"Wohoo! I'm gonna be punished!"

I was like, "I thought Goliath was older than that."

Anyway, now I'm feeling like a dunce. Congrats!