Localization

All intelligent thoughts have already been thought; what is necessary is only to try to think them again – Goethe

And it comes to be that only at 2:20 AM on a Wednesday morning of  a surprisingly tolerable  Bostonian November that I finally realize that a ring R needn’t always sit inside its localization.  It’s discomfiting to know how one can have a “wrong picture” even when you can glibly recite the definition in the middle of your sleep.  Having given the climax away, let me begin at the beginning.

Age 5 :  Exasperated K.G teacher tries to teach counting. “One apple, two apples, many apples”

Age 10 : Tearful student grapples with the concept of negative numbers. “What is negative money ?!”

Age 15 : Having accepted that God made integers, students are drilled with the work of man – The rationals. “Irony ?”

Age 20: Rationals are just localizations with respect to the prime ideal (0) of ring .  “Whatever”

To begin at the beginning again (This is how a book on mathematics is read. You flip to the end and read the main theorem. It will generally resemble something like this : “ Image of a constructible set under a morphism of algebraic varieties is constructible “. Having understood not one word of this, you begin at the beginning (where generally most authors define set operations like union and intersection) again and again. Finally after having gotten thoroughly sick of the whole thing, you give up and watch a movie), there are the natural numbers. You can add two natural numbers and still stay within natural numbers, but you can’t subtract. 1-3 is not a natural number. So people threw in negative numbers and the zero and observed that now you could subtract and add without any qualms. Mathematicians promptly abstracted this out and fondly called it a group.

It turns out that even when you multiply any number with any other number, you still stay with the numbers. This makes the integers a ring. On a side note, one wonders how the word ring ever came about. A group is understandable, but a ring ?

And where there is multiplication, you’d like division to be present. Alas, you can’t divide two numbers and still stay within the integers. So you throw in elements of the form , b non zero (of – course) to get the rationals, where you can add, subtract, multiply and divide and so you’re happy. And if you still remember your high school math, you would know that . In fact, you say any two elements , if ad-bc=0.

The integers are nice rings.  That is, ab=0 would mean a=0 or b=0. And as is expected, things are not as hunky-dory as usual, but this process of “inverting elements” can be generalized as follows :

Let R be a ring. P is a prime ideal if  . Then you define localization at P to be the ring with an equivalence relation, namely if there exists an such that s(ad-bc)=0.

This is just saying that you take elements of R, throw in inverses of elements not in P and make it into a ring. And this is exactly where one should have realized that R does not sit in , the equivalence relation makes weird things equal.

Look at the ring Z/6Z, (ie) {0,1,2,3,4,5} with the usual operations + and * , where you go modulo 6 all the time. {0,2,4} is a prime ideal of this ring. Let’s localize our ring with respect to this prime ideal . So we get elements of the form where b is from the set {1,3,5}. Now, ladies and gentlemen, in our localized ring 2=0 —- [3*(2-0)=0]

So what ?

Well …Z/6Z is not a field (2*3=0) but the localizations at its prime ideals are fields. Yes, it’s trivial, but I just realized that.

..at a time when many eminent scholars, endowed with a great geometric talent, make a point of never disclosing the simple and direct ideas that guided them, subordinating their elegant results to abstract general theories which often have no application outside the particular case in question. Geometry was becoming a study of algebraic, differential or partial differential equations, thus losing all the charm that comes from its being an art.

- Henri Lebesgue

Thedi vanda lakshmi

Our very own charade , just a decade later but in black and white.

vs

Grumpy ?

Great

It (even has raravenu bits incorporated)

Nostalgia

..
valarndhida thaneerai vaarthavan
verai , paripadhazhago ?

This is why everyone should know how to skate.

The problem with ..

Ashwin’s one line solution to this:

To prove that if ( ) () = 0 then either all or all are zero, consider the subgroup H generated by and . H is finitely generated and torsion free and hence free.

The problem with generalizations

My office mate stumbled upon this innocuous statement in a paper,

Note that because P is an abelian multiplicative group with no torsion elements , the group ring ZP has no zero divisors

Breaking down the fancy terminology, apparently is nothing but the ring of all formal integral linear combinations of elements of P with the condition that any set of elements of P are linearly independent in this ring. More simply,

any element of looks like where and and this element can be 0 if and only of all the ’s are 0.

After trying in vain to come up with counter examples, we decided to prove it. Note that a nice example is , the ring of laurent polynomials in one variable, which sort of convinces one why this must be true. And what else to appeal to first, but induction ? So here goes.

Let us induct on the number of  basis elements (k) needed to express a divisor.

k=1

That is, the zero divisor looks like rp where r is a non-zero integer and p is an element of P. Let

,

where are distinct elements of P and s, non zero integers. So you have

,

and as s are distinct, so are s, and you get , which gives you r=0. Cheap, yes. And we don’t use the fact about torsionlessness (er) at all, but then, you hit case number 2,

k=2

Let where are integers and are elements of P. Expanding, you get

Note that for any and similarly for s too. But to get 0, they must cancel out! So given any i, there exists a j such that . Also note that this j is unique, as would imply, . And also, no two i, i’ can have the same j (By that I mean is not possible if )

So all we are finally saying is that, there exists a permutation of the set {1,2,..n} such that

———————————————————–(*).

Now, for the trick. Take product of  (*) over all . We get,

, which simplifies to or . Since P is torsion free (aha!), this would mean p=q, a contradiction to our initial assumption.

This looks more like it (or so we thought). However hopes for an inductive proof were lost at this point of time. But then, whatever works here must generalize, right ? Nope. I mean, I don’t know if it does or doesn’t :P. After two days of conscientiously avoiding googling the problem, I gave up and found this on wikipedia.

A long-standing conjecture of Kaplansky (~1940) says that if G is a torsion-free group, and K is a field, then the group ring K[G] has no non-trivial zero divisors. In fact, the condition that K is a field can be relaxed to any ring that can be embedded into an integral domain.The conjecture remains open in full generality, however some special cases of torsion-free groups have been shown to satisfy the zero divisor conjecture.

So much for that. It just seemed so doable.

However, as a consolation, for any finitely generated abelian group P, torsion free would mean that by structure theorem, and so the ring ZP would just be the ring of laurent polynomials in finitely many variables, (ie)

,

which is a domain. The proof  I expect follows on the similar lines of proving that a polynomial ring of  several variables is a domain, which incidentally I always have trouble proving. Can someone give me a proper proof which is also not very messy :P ?

Ask rajini

Are full or empty beer bottles sturdier and does their fracture-threshold suffice to break the human skull?

Watch this . One wonders how many ignoble awards the tamil film makers missed.